Differential equations

Class 12 Mathematics, unit 4, book pages 120–139

A differential equation links an unknown function with its derivatives, so it describes how a quantity changes. This unit shows how to classify a DE by order, degree and linearity, how to form one by eliminating arbitrary constants, and what general and particular solutions mean. It then solves first order, first degree equations by separating the variables or, for homogeneous equations, by the substitution . The same few methods model population growth, radioactive decay, Newton's law of cooling, falling bodies, electric circuits and continuous compound interest.

Learning outcomes (4)
  • Identify and construct first order differential equations from practical situations.
  • Give the concept of solution of a differential equation.
  • Solve differential equations of first order and first degree of the form: separable variables equations and homogeneous equations.
  • Solve real life problems related to differential equations, such as population growth and decay, the cooling/warming law, flow of electricity, series circuits, economics and finance, and radioactive decay.

Introduction and mathematical models

A differential equation (DE) is an equation that contains the derivative of an unknown function. It links a quantity with its rate of change. In real life we often know how fast something changes (a population grows by a fixed fraction each year, a hot cup of tea cools faster when it is much hotter than the room). A DE turns that knowledge into a formula for the quantity itself.

DEs are used wherever quantities change continuously in time or space:

  • Physics: motion of bodies under forces (Newton's laws) and wave propagation.
  • Biology: population growth, the spread of diseases, interaction between species.
  • Engineering: electric circuits, control systems and mechanical vibrations.
  • Chemistry: reaction rates and diffusion.
  • Economics: growth of investments, market behaviour and optimisation.

Mathematical models

A mathematical model is a mathematical description, such as a differential equation, that imitates a real situation so that we can predict how it will behave. Most applications of DEs are mathematical models.

Example: a falling object. Newton's second law says that the net force on a body equals its mass times its acceleration. Acceleration is the rate of change of velocity with time , that is . So

This law is a differential equation because it contains a derivative. If we ignore air resistance, the only force on an object falling close to Earth's surface is its weight , directed downward, where . The model of the free-falling object is therefore

The modelling cycle. The assumption here is that air resistance can be ignored; the prediction is compared with reality and the model improved if needed.

Translating words into a differential equation

The first SLO of this unit is to construct first order DEs from practical situations. Three phrases do most of the work: **rate of change of ** means ; is proportional to means "equals a constant times"; decreases or decays means the rate is negative.

Each statement becomes a first order DE. The constant is found later from the data.
StatementDifferential equation
A population grows at a rate proportional to its size.
A radioactive mass decays at a rate proportional to the mass remaining.
A body cools at a rate proportional to the difference between its temperature and the room temperature .
Money grows at a continuous interest rate .
The velocity of a falling body increases at the constant rate .
The slope of a curve at any point is .

4.1Differential equation

A differential equation is an equation containing one or more derivatives of an unknown function. The book's definition: a mathematical equation containing the derivatives of one dependent variable with respect to one independent variable is called a differential equation (DE).

If the unknown function depends on only one variable, the DE is an ordinary differential equation (ODE). In this unit every DE is ordinary, so we simply call it a DE. Two things must be present: a derivative and an equation (an equals sign).

Dependent variable
The unknown function whose derivatives appear, usually (or , , in applications).
Independent variable
The variable with respect to which we differentiate, usually (or time ).

Where a differential equation comes from

Take the function , where is an arbitrary constant (it may have any fixed value). Differentiate with respect to . The constant disappears:

The result is a differential equation. Every curve , for every value of , satisfies it. So one DE describes a whole family of curves.

The simplest differential equations have the form

where is a known function of . Some more examples, in each of which is the dependent variable and the independent variable:

4.2Order and degree of a differential equationOften examined

Order. The order of a differential equation is the order of the highest order derivative present in the equation.

Types of DE with respect to order
First order DE
A DE whose highest derivative is the first derivative. It can be written . Example: .
Second order DE
A DE whose highest derivative is the second derivative . Example: .
Degree

Degree. The degree of a differential equation is the power (exponent) of the highest order derivative in the equation, after the equation has been made free of radicals and fractional powers of the derivatives.

Order is decided by the highest derivative; degree is the power of that same derivative.
Reading order and degree
EquationHighest derivativeOrderDegree
, power 111
, power 222
, power 323
Example 1
Determine the order and degree of the following differential equations. (i) (ii) (iii)
  1. (i) The only derivative is , of first order, and it appears to the power 1.
  2. (ii) The highest derivative is , and it appears to the power 1.
  3. (iii) The highest derivative is , and it is squared. The lower derivatives do not affect the degree.

Some more types of differential equations are: ordinary DEs (ODE), partial DEs (PDE), linear DEs, non-linear DEs, homogeneous DEs and non-homogeneous DEs. They are described in the next section.

4.2.3Ordinary, linear and non-linear equations

Ordinary differential equation (ODE)

If only one independent variable is involved, the DE is called an ordinary differential equation. It contains one or more derivatives of the unknown function with respect to that single independent variable. The general form of an th order ODE, solved for its highest derivative, is

where is a known function. This is called an explicit ODE of order . Examples of ODEs:

Linear and non-linear DEs

An th order differential equation is linear if it can be written in the form

that is, it satisfies the following two conditions:

  • (a) the dependent variable and all its derivatives are of power one (and are not multiplied together);
  • (b) all the coefficients and the function are constants or depend only on the independent variable .

If any one of these two conditions is not satisfied, the DE is non-linear. Signs of non-linearity: a power of or of a derivative (, ), a product such as , or a function of such as or .

Example 2
Identify the linear and non-linear differential equations: (i) (ii) (iii) (iv) (v) (vi)
EquationTypeReason
(i)linear has power 1; the right side is a constant.
(ii)linear, and all have power 1 with constant coefficients.
(iii)non-linearthe coefficient of is (a product ), which breaks condition (b).
(iv)non-linearwritten as , it contains the product .
(v)linearthe coefficient of depends only on .
(vi)non-linear is squared, which breaks condition (a).

4.3Concept of solution of a differential equation

Differential equations are studied mainly through their solutions: the functions that satisfy them.

A solution of a differential equation is any function y=f(x), defined on some interval I, that reduces the equation to an identity (a statement true for every x in I) when y and its derivatives are substituted.

To solve a DE such as \dfrac{dy}{dx}-x=0 means to find the unknown function, either explicitly as y=f(x) or implicitly as a relation between x and y.

The simplest case: integrate

Consider the first order DE \dfrac{dy}{dx}=f(x). If f(x) is continuous, integrating both sides gives

y=\int f(x)\,dx=F(x)+c

where F(x) is an anti-derivative of f(x) and c is the constant of integration.

Example: solve \dfrac{dy}{dx}=1+e^{2x}
Find the solution of the differential equation \dfrac{dy}{dx}=1+e^{2x}.
  1. The right side contains only x, so integrate both sides with respect to x.
    y=\int\left(1+e^{2x}\right)dx
  2. Use \int e^{ax}\,dx=\dfrac{1}{a}e^{ax} with a=2, and add one constant.
    y=x+\dfrac{1}{2}e^{2x}+c
y=x+\tfrac{1}{2}e^{2x}+c
Example 3
Show that x^{2}+y^{2}=c is a solution of the differential equation y\dfrac{dy}{dx}+x=0. Also plot the graph of the solution.
  1. Differentiate the given relation implicitly with respect to x. The constant c gives 0.
    2x+2y\dfrac{dy}{dx}=0
  2. Divide by 2 and rearrange. This is exactly the given DE, so the relation is a solution.
    y\dfrac{dy}{dx}+x=0
  3. Graph. The solution depends on the arbitrary constant c. Choosing c=1,4,9,16,\dots gives different solutions.
    x^{2}+y^{2}=1^{2},\quad x^{2}+y^{2}=2^{2},\quad x^{2}+y^{2}=3^{2},\quad x^{2}+y^{2}=4^{2},\ \dots
  4. These are circles with centre (0,0) and radii 1,2,3,4,\dots. So the solution x^{2}+y^{2}=c represents a family of infinitely many circles.
Solutions x^{2}+y^{2}=c
The general solution is a family of concentric circles with centre (0,0) and radius \sqrt{c}.
x^{2}+y^{2}=c\ \text{ is a family of concentric circles}
Slope field of \dfrac{dy}{dx}=-\dfrac{x}{y}. The accent circles x^{2}+y^{2}=c fit the slope marks everywhere.

This example shows that a DE has two types of solutions: the general solution and a particular solution.

4.3.1General and particular solutionsOften examined

General solution
The solution that contains as many arbitrary constants as the order of the differential equation. It is a relation between the independent variable x and the dependent variable y, obtained by removing the derivatives (by integration). In Example 3, x^{2}+y^{2}=c is the general solution.
Particular solution
A solution free from arbitrary constants. It is obtained by giving particular values to the arbitrary constants. In Example 3, x^{2}+y^{2}=4 and x^{2}+y^{2}=9 are particular solutions.
General solutionParticular solution
Arbitrary constantsas many as the ordernone
Grapha family of curvesone curve of the family
Example for \dfrac{dy}{dx}=2xy=x^{2}+cy=x^{2}+1
Example 4
Verify that y=\dfrac{x^{4}}{16} is a solution of the differential equation \dfrac{dy}{dx}-xy^{1/2}=0.
  1. Differentiate the given solution y=\dfrac{x^{4}}{16} with respect to x.
    \dfrac{dy}{dx}=\dfrac{4x^{3}}{16}=\dfrac{x^{3}}{4}
  2. Substitute y and \dfrac{dy}{dx} in the left side of the DE. Note \left(\dfrac{x^{4}}{16}\right)^{1/2}=\dfrac{x^{2}}{4}.
    \dfrac{dy}{dx}-xy^{1/2}=\dfrac{x^{3}}{4}-x\left(\dfrac{x^{4}}{16}\right)^{1/2}=\dfrac{x^{3}}{4}-x\left(\dfrac{x^{2}}{4}\right)=\dfrac{x^{3}}{4}-\dfrac{x^{3}}{4}=0
  3. The left side is 0 for every real x, which is the right side. So the function is a solution on \mathbb{R}.
    \sqrt{x^{4}/16}=x^{2}/4 is true for every real x because x^{2}/4\ge0.
y=\dfrac{x^{4}}{16}\ \text{ is a solution for all } x\in\mathbb{R}
Check point
Find the DE corresponding to the equation y=3x^{2}+c. Which type of family of curves does the solution represent?
  1. There is one arbitrary constant, so differentiate once. The constant c disappears.
    \dfrac{dy}{dx}=6x
  2. For each value of c, y=3x^{2}+c is the parabola y=3x^{2} moved c units up (or down if c<0). All of them have the y-axis as axis and open upward.
y=3x^{2}+c: the same parabola shifted up or down by c.
\dfrac{dy}{dx}=6x;\ \text{ a family of parabolas } y=3x^{2}+c
Example 5
Is the function y=xe^{x} a solution of the differential equation y''-2y'+y=0 on the interval (-\infty,+\infty)?
  1. Differentiate y=xe^{x} with the product rule.
    y'=xe^{x}+e^{x}
  2. Differentiate again.
    y''=xe^{x}+e^{x}+e^{x}=xe^{x}+2e^{x}
  3. Substitute in the left side of the DE and simplify.
    y''-2y'+y=xe^{x}+2e^{x}-2\left(xe^{x}+e^{x}\right)+xe^{x}=xe^{x}+2e^{x}-2xe^{x}-2e^{x}+xe^{x}=0
  4. This is true for every x\in\mathbb{R}, so y=xe^{x} is a solution on (-\infty,+\infty).
\text{Yes: } y''-2y'+y=0\ \text{ for all } x
Trivial solution
A solution of a DE that is identically zero on an interval, such as y=0 in Examples 4 and 5.

4.4Formation of differential equationsOften examined

We can form a differential equation by eliminating the arbitrary constants from an algebraic equation (which is then a solution of the DE). Each differentiation gives one more equation, so to remove n constants we differentiate n times.

Method
  1. Count the arbitrary constants in the given equation; call the number n.
  2. Differentiate the equation n times with respect to x.
  3. Use the n+1 equations (the given one and its n derivatives) to eliminate all n constants.
  4. The result is a DE of order n, the lowest possible order.
Example: the DE of y=e^{x}
Find the differential equation corresponding to y=e^{x}.
  1. Differentiate.
    y=e^{x}\quad\Longrightarrow\quad y'=e^{x}
  2. Both right sides are e^{x}, so y'=y.
    y'-y=0
y'-y=0
Example 6
Find the DE corresponding to the equation y=a\cos x+b\sin x.
  1. There are two constants, a and b, so differentiate twice.
    y'=-a\sin x+b\cos x
  2. Differentiate again.
    y''=-a\cos x-b\sin x=-\left(a\cos x+b\sin x\right)
  3. The bracket is y itself, so both constants disappear.
    y''=-y\qquad\text{or}\qquad y''+y=0

So y=a\cos x+b\sin x is a solution of the differential equation y''+y=0.

y''+y=0
Example 6 (second example with this number)
Eliminate the arbitrary constants from y=A\sin(2x-B) and form a differential equation of the lowest order.
  1. Two constants A and B, so differentiate twice. Use the chain rule.
    \dfrac{dy}{dx}=2A\cos(2x-B)
  2. Differentiate again.
    \dfrac{d^{2}y}{dx^{2}}=-4A\sin(2x-B)=-4\left[A\sin(2x-B)\right]
  3. The bracket is y.
    \dfrac{d^{2}y}{dx^{2}}=-4y\qquad\Longrightarrow\qquad \dfrac{d^{2}y}{dx^{2}}+4y=0

This is a second order ODE. Its physical meaning: if y is a displacement and x is time, the acceleration \dfrac{d^{2}y}{dx^{2}} is proportional to the displacement and opposite to it. This is the differential equation of simple harmonic motion.

\dfrac{d^{2}y}{dx^{2}}+4y=0
Check point
Check whether the equation of the parabola y^{2}=4a(x-b), where a and b are arbitrary constants, is a solution of the differential equation y\dfrac{d^{2}y}{dx^{2}}+\left(\dfrac{dy}{dx}\right)^{2}=0.
  1. Differentiate y^{2}=4a(x-b) implicitly. The constant b disappears.
    2y\dfrac{dy}{dx}=4a\quad\Longrightarrow\quad y\dfrac{dy}{dx}=2a
  2. Differentiate again with the product rule. The constant a disappears.
    y\dfrac{d^{2}y}{dx^{2}}+\dfrac{dy}{dx}\cdot\dfrac{dy}{dx}=0\quad\Longrightarrow\quad y\dfrac{d^{2}y}{dx^{2}}+\left(\dfrac{dy}{dx}\right)^{2}=0
  3. This is the given DE, so every parabola y^{2}=4a(x-b) is a solution of it.
\text{Yes, } y^{2}=4a(x-b)\text{ is a solution}

4.4.1Explicit and implicit solutions

Explicit solution
A solution that can be written in the form y=f(x), with y alone on one side.
Implicit solution
A solution in the form of a relation f(x,y)=0, not solved for y.
Example 7
Prove that, for -2<x<2, the relation x^{2}+y^{2}-4=0 is an implicit solution of the differential equation \dfrac{dy}{dx}=-\dfrac{x}{y}.
  1. Differentiate the relation implicitly with respect to x.
    2x+2y\dfrac{dy}{dx}=0
  2. Solve for \dfrac{dy}{dx} (possible because y\neq0 when -2<x<2). This is the given DE.
    \dfrac{dy}{dx}=-\dfrac{x}{y}

Note. On the interval (-2,2) the relation defines two explicit differentiable functions, y=\sqrt{4-x^{2}} and y=-\sqrt{4-x^{2}}. Each is an explicit solution of the DE.

The implicit solution x^{2}+y^{2}-4=0 splits into two explicit solutions on (-2,2). The end points are excluded because \dfrac{dy}{dx}=-\dfrac{x}{y} is not defined when y=0.
x^{2}+y^{2}-4=0\ \text{ is an implicit solution on } (-2,2)

4.4.2Number of solutions

A given differential equation usually has an infinite number of solutions, one for each value of the arbitrary constant(s).

Illustration (i)
Show that, for any value of c, the function y=\dfrac{c}{x}+1 is a solution of x\dfrac{dy}{dx}+y=1 on the interval (0,\infty).
  1. Differentiate.
    \dfrac{dy}{dx}=-\dfrac{c}{x^{2}}
  2. Substitute in the left side.
    x\left(-\dfrac{c}{x^{2}}\right)+\dfrac{c}{x}+1=-\dfrac{c}{x}+\dfrac{c}{x}+1=1
  3. This holds for every c, so there are infinitely many solutions. For c=0 we get the constant solution y=1.
Some members of the family y=\dfrac{c}{x}+1, all solutions of x\dfrac{dy}{dx}+y=1 on (0,\infty).
Illustration (ii)
The functions y=c_1\cos4x and y=c_2\sin4x, where c_1,c_2 are arbitrary constants, are solutions of y''+16y=0. Show that their sum y=c_1\cos4x+c_2\sin4x is also a solution.
  1. Differentiate the sum twice.
    y'=-4c_1\sin4x+4c_2\cos4x,\qquad y''=-16c_1\cos4x-16c_2\sin4x
  2. So y''=-16y, that is y''+16y=0. The sum is a solution for all c_1,c_2.
    y''+16y=-16y+16y=0

4.5Solution of first order, first degree equations

The solution of an algebraic equation such as x^{2}-5x+6=0 is a number: the values x=2 and x=3 satisfy it. The solution of a differential equation such as y\,dy=x\,dx is a function. To find it we remove the derivative by some technique, and every technique involves integration. After removing the derivative, the dependent variable y is given in terms of the independent variable x, either explicitly (y=f(x)) or implicitly (g(x,y)=0).

A DE of first order and first degree can be written as \dfrac{dy}{dx}=f(x,y) or as M(x,y)\,dx+N(x,y)\,dy=0. Some standard forms have ready methods:

  • variable separable differential equations,
  • equations reducible to variable separable form,
  • homogeneous differential equations,
  • equations reducible to homogeneous form.

4.5.1Variable separable equationsOften examined

A first order differential equation of the form

\dfrac{dy}{dx}=f(x)\,g(y)

where f is a function of x only and g is a function of y only, is called separable, or is said to have separable variables.

For example, \dfrac{dy}{dx}=y^{3}xe^{2x+y} is separable, because it can be written as \dfrac{dy}{dx}=\left(xe^{2x}\right)\left(y^{3}e^{y}\right)=f(x)\,g(y). But \dfrac{dy}{dx}=y+\sin x is not separable: y+\sin x cannot be written as a product f(x)\,g(y).

Derivation Solving a separable equation

Aim. Find the general solution of \dfrac{dy}{dx}=f(x)\,g(y).

Assumptions.
  • g(y)\neq0, so that we may divide by it.
  1. Divide by g(y) and multiply by dx. Now every y is on the left with dy, and every x is on the right with dx.
    \dfrac{dy}{g(y)}=f(x)\,dx
  2. Integrate both sides, the left with respect to y and the right with respect to x.
    \int\dfrac{dy}{g(y)}=\int f(x)\,dx
  3. Each integral produces a constant; combine them into a single constant c written on one side.
    G(y)=F(x)+c
    G and F are anti-derivatives of \dfrac{1}{g(y)} and f(x).
  4. If possible, solve for y to get an explicit solution. Otherwise leave the implicit relation.
Result\int\dfrac{dy}{g(y)}=\int f(x)\,dx+c

One integration gives one arbitrary constant, as expected for a first order DE. Any value y_0 with g(y_0)=0 gives a constant solution y=y_0, which may be lost by the division.

Example 8
Solve \dfrac{dy}{dx}=\dfrac{x^{2}}{y}.
  1. Separate the variables: multiply both sides by y\,dx.
    y\,dy=x^{2}\,dx
  2. Integrate both sides.
    \int y\,dy=\int x^{2}\,dx\quad\Longrightarrow\quad \dfrac{y^{2}}{2}=\dfrac{x^{3}}{3}+c_1
  3. Multiply by 6 to clear the fractions.
    3y^{2}=2x^{3}+6c_1
  4. Write c=6c_1, which is still an arbitrary constant.
    3y^{2}=2x^{3}+c
3y^{2}=2x^{3}+c
Example 9
Solve (1+x)\,dy-y\,dx=0.
  1. Move the dx term to the right.
    (1+x)\,dy=y\,dx
  2. Divide by y(1+x) to separate the variables.
    \dfrac{dy}{y}=\dfrac{dx}{1+x}
  3. Integrate both sides; write the constant as \ln c.
    \int\dfrac{dy}{y}=\int\dfrac{dx}{1+x}\quad\Longrightarrow\quad \ln y=\ln(x+1)+\ln c
  4. Combine the logarithms with \ln a+\ln b=\ln(ab).
    \ln y=\ln\left[c(x+1)\right]
  5. Take antilog of both sides.
    y=c(x+1)
The general solution y=c(x+1) is a family of straight lines through (-1,0).
y=c(x+1)
Example 10
Solve \dfrac{dy}{dx}=\dfrac{1}{x\tan y}.
  1. Separate the variables.
    \tan y\,dy=\dfrac{1}{x}\,dx
  2. Integrate. Use \int\tan y\,dy=-\ln(\cos y) and write the constant as \ln c.
    -\ln(\cos y)=\ln x+\ln c
  3. Bring everything to one side and combine.
    0=\ln(\cos y)+\ln x+\ln c=\ln(cx\cos y)
  4. Take antilog: e^{\ln(cx\cos y)}=e^{0}=1.
    cx\cos y=1
  5. Divide by c and rename \dfrac{1}{c}=C.
    x\cos y=C
x\cos y=C

4.5.1Initial condition and initial value problemOften examined

The general solution of a DE contains as many arbitrary constants as the order of the DE. Often we need the one solution that satisfies an extra (supplementary) condition, such as a known value of y at a given x.

Suppose we want the solution of \dfrac{dy}{dx}=f(x,y) with y=y_0 at x=x_0. Substituting x=x_0 and y=y_0 in the general solution gives a particular value of the constant, and so a particular solution. The condition y(x_0)=y_0 is called the initial condition, and the DE together with it is an initial value problem (IVP):

\dfrac{dy}{dx}=f(x,y),\qquad y(x_0)=y_0
Method for an IVP
  1. Solve the DE to get the general solution, containing c.
  2. Substitute x=x_0 and y=y_0 to find c.
  3. Put this value of c back into the general solution.
  4. If you then solve for y, choose the sign or branch that agrees with the initial condition.
Example 11
Solve the initial value problem \dfrac{dy}{dx}=-\dfrac{x}{y}, \ y(1)=3.
  1. Separate the variables.
    y\,dy=-x\,dx
  2. Integrate both sides.
    \int y\,dy=\int -x\,dx\quad\Longrightarrow\quad \dfrac{y^{2}}{2}=-\dfrac{x^{2}}{2}+c\quad\Longrightarrow\quad x^{2}+y^{2}=2c\qquad(1)
  3. Use the initial condition x=1, y=3 in (1).
    1^{2}+3^{2}=2c\quad\Longrightarrow\quad 2c=10\quad\Longrightarrow\quad c=5
  4. Substitute 2c=10 back into (1).
    x^{2}+y^{2}=10

This is the solution of the IVP. It is the circle with centre (0,0) and radius \sqrt{10}.

The IVP picks the circle through (1,3). As a function, the solution is the upper half (solid), because y(1)=3>0.
x^{2}+y^{2}=10
Example 12
Solve \dfrac{dy}{dx}=2x such that y(2)=4.
  1. Separate: dy=2x\,dx. Integrate both sides.
    \int dy=\int2x\,dx\quad\Longrightarrow\quad y=2\left(\dfrac{x^{2}}{2}\right)+c\quad\Longrightarrow\quad y-x^{2}=c\qquad(1)
  2. Use x=2, y=4 in (1).
    4-2^{2}=c\quad\Longrightarrow\quad c=0
  3. Substitute c=0 in (1).
    y-x^{2}=0\qquad\text{that is}\qquad y=x^{2}\qquad(2)

Note. The general solution (1) represents a family of parabolas for different values of c, whereas the particular solution (2) is the member of the family that passes through (2,4).

The general solution y=x^{2}+c is a family of parabolas; the particular solution y=x^{2} is the member through (2,4).
y=x^{2}
Check point
Show that the solution of the differential equation \dfrac{dy}{dx}=2 represents a family of parallel lines. Draw some of the parallel lines. Also solve the differential equation for the initial condition y(0)=1.
  1. Separate and integrate.
    dy=2\,dx\quad\Longrightarrow\quad y=2x+c
  2. For every c this is a straight line with slope 2 and y-intercept c. Equal slopes mean the lines are parallel.
  3. Use y(0)=1: 1=2(0)+c.
    c=1\quad\Longrightarrow\quad y=2x+1
Solutions of \dfrac{dy}{dx}=2 are parallel lines of slope 2. The initial condition y(0)=1 selects y=2x+1.
y=2x+c;\qquad y=2x+1

4.6.1Homogeneous functions

Before homogeneous differential equations we need homogeneous functions. Roughly, a polynomial in x and y is homogeneous when every term has the same total degree (power of x plus power of y).

Definition. If a function f has the property

f(tx,\,ty)=t^{n}f(x,y)

where t\in\mathbb{R}^{+} and n\in\mathbb{R}, then f is a **homogeneous function of degree n**. To test a function, replace x by tx and y by ty, simplify, and see whether a power t^{n} comes out as a common factor.

Example 13
Check whether the functions (i) f(x,y)=\sqrt{x^{3}+y^{3}} (ii) f(x,y)=x^{2}+y^{2}+2 (iii) f(x,y)=\dfrac{x}{2y}+4 are homogeneous or not. If homogeneous, find the degree.
  1. (i) Replace x with tx and y with ty; take t^{3} out of the root.
    f(tx,ty)=\sqrt{(tx)^{3}+(ty)^{3}}=\sqrt{t^{3}\left(x^{3}+y^{3}\right)}=t^{3/2}\sqrt{x^{3}+y^{3}}=t^{3/2}f(x,y)
  2. So f is homogeneous of degree \dfrac{3}{2} (the degree need not be a whole number).
  3. (ii) The constant 2 does not get a factor t^{2}.
    f(tx,ty)=t^{2}x^{2}+t^{2}y^{2}+2\neq t^{2}f(x,y)
  4. So f is not homogeneous.
  5. (iii) The t cancels in the fraction.
    f(tx,ty)=\dfrac{tx}{2ty}+4=\dfrac{x}{2y}+4=t^{0}f(x,y)
  6. So f is homogeneous of degree 0.
\text{(i) degree } \tfrac32;\quad\text{(ii) not homogeneous};\quad\text{(iii) degree } 0
Check point
Check whether the functions are homogeneous or not. If homogeneous, find the degree. (a) f(x,y)=x^{2}-3xy+y^{2} (b) f(x,y)=x-\sqrt{xy}+5y
  1. (a) Every term has total degree 2, so t^{2} factors out.
    f(tx,ty)=t^{2}x^{2}-3t^{2}xy+t^{2}y^{2}=t^{2}f(x,y)\quad(\text{degree }2)
  2. (b) Since t>0, \sqrt{t^{2}xy}=t\sqrt{xy}, so t factors out of every term.
    f(tx,ty)=tx-\sqrt{t^{2}xy}+5ty=t\left(x-\sqrt{xy}+5y\right)=t\,f(x,y)\quad(\text{degree }1)
\text{(a) degree }2;\qquad\text{(b) degree }1

4.6.2Homogeneous differential equationsOften examined

A differential equation of the form

M(x,y)\,dx+N(x,y)\,dy=0\qquad(1)

is homogeneous if M and N are homogeneous functions of the same degree n, that is M(tx,ty)=t^{n}M(x,y) and N(tx,ty)=t^{n}N(x,y).

Equivalently, the equation \dfrac{dy}{dx}=f(x,y) is homogeneous when f is homogeneous of degree 0; then f depends only on the ratio \dfrac{y}{x}. Such an equation is reduced to a separable one by substituting y=ux (or x=vy), where u (or v) is a new dependent variable.

Derivation Why the substitution y=ux works

Aim. Show that y=ux turns the homogeneous equation (1) into an equation with separable variables x and u.

Assumptions.
  • M and N are homogeneous of the same degree n.
  • x>0 (for x<0 the same steps work with |x|).
  1. Put y=ux, where u is a function of x. Differentiate with the product rule.
    \dfrac{dy}{dx}=u+x\dfrac{du}{dx}\qquad\text{or}\qquad dy=u\,dx+x\,du
  2. Substitute y=ux and dy=u\,dx+x\,du into (1).
    M(x,ux)\,dx+N(x,ux)\left[u\,dx+x\,du\right]=0
  3. Use homogeneity with t=x: M(x\cdot1,\,x\cdot u)=x^{n}M(1,u) and likewise for N.
    x^{n}M(1,u)\,dx+x^{n}N(1,u)\left[u\,dx+x\,du\right]=0
  4. Divide by x^{n} and collect the dx terms.
    \left[M(1,u)+u\,N(1,u)\right]dx+x\,N(1,u)\,du=0
  5. Divide by x\left[M(1,u)+uN(1,u)\right]. The first term contains only x and the second only u: the variables are separated.
    \dfrac{dx}{x}+\dfrac{N(1,u)\,du}{M(1,u)+u\,N(1,u)}=0\qquad(2)
Result\dfrac{dx}{x}+\dfrac{N(1,u)\,du}{M(1,u)+u\,N(1,u)}=0

After y=ux the variables separate. Integrate, then replace u by \dfrac{y}{x} to return to x and y.

Method for a homogeneous equation
  1. Check that the equation is homogeneous (M and N of the same degree, or \dfrac{dy}{dx} a function of \dfrac{y}{x}).
  2. Put y=ux and \dfrac{dy}{dx}=u+x\dfrac{du}{dx}.
    u+x\dfrac{du}{dx}=F(u)
  3. Cancel the powers of x and move u to the right.
    x\dfrac{du}{dx}=F(u)-u
  4. Separate the variables and integrate, with one constant.
    \int\dfrac{du}{F(u)-u}=\int\dfrac{dx}{x}
  5. Replace u by \dfrac{y}{x} and simplify. For an IVP, use the initial condition last.
Example 14
Solve \dfrac{dy}{dx}=\dfrac{x^{2}+y^{2}}{2xy}.
  1. Numerator and denominator are both of degree 2, so the equation is homogeneous. Put y=ux, \dfrac{dy}{dx}=u+x\dfrac{du}{dx}.
    u+x\dfrac{du}{dx}=\dfrac{x^{2}+u^{2}x^{2}}{2x\cdot ux}=\dfrac{1+u^{2}}{2u}
  2. Subtract u from both sides.
    x\dfrac{du}{dx}=\dfrac{1+u^{2}}{2u}-u=\dfrac{1+u^{2}-2u^{2}}{2u}=\dfrac{1-u^{2}}{2u}
  3. Separate the variables.
    \dfrac{2u}{1-u^{2}}\,du=\dfrac{dx}{x}
  4. Integrate. The numerator 2u is minus the derivative of 1-u^{2}, so the left side is -\ln(1-u^{2}).
    -\ln\left(1-u^{2}\right)=\ln x+\ln c\quad\Longrightarrow\quad \ln\left(1-u^{2}\right)+\ln x+\ln c=0
  5. Combine the logarithms and take antilog.
    \ln\left[cx\left(1-u^{2}\right)\right]=0\quad\Longrightarrow\quad cx\left(1-u^{2}\right)=1
  6. Replace u by \dfrac{y}{x}.
    cx\left(1-\dfrac{y^{2}}{x^{2}}\right)=1\quad\Longrightarrow\quad cx\left(\dfrac{x^{2}-y^{2}}{x^{2}}\right)=1\quad\Longrightarrow\quad x^{2}-y^{2}=\dfrac{x}{c}
  7. Rename \dfrac{1}{c}=C.
    x^{2}-y^{2}=Cx
x^{2}-y^{2}=Cx
Example 15
Solve the initial value problem x\dfrac{dy}{dx}=y+xe^{y/x}, \ y(1)=1.
  1. Divide by x. The right side depends only on \dfrac{y}{x}, so the equation is homogeneous of degree zero.
    \dfrac{dy}{dx}=\dfrac{y}{x}+e^{y/x},\qquad y(1)=1
  2. Put y=ux, \dfrac{dy}{dx}=u+x\dfrac{du}{dx}. The u cancels.
    u+x\dfrac{du}{dx}=\dfrac{ux}{x}+e^{ux/x}=u+e^{u}\quad\Longrightarrow\quad x\dfrac{du}{dx}=e^{u}
  3. Separate and integrate.
    e^{-u}\,du=\dfrac{dx}{x}\quad\Longrightarrow\quad \int e^{-u}\,du=\int\dfrac{dx}{x}\quad\Longrightarrow\quad -e^{-u}=\ln x+c
  4. Replace u by \dfrac{y}{x}.
    -e^{-y/x}=\ln x+c\qquad(\text{i})
  5. Use x=1, y=1 in (i). Since \ln1=0:
    -e^{-1}=\ln1+c\quad\Longrightarrow\quad c=-e^{-1}
  6. Substitute back into (i) and rearrange.
    -e^{-y/x}=\ln x-e^{-1}\quad\Longrightarrow\quad e^{-1}-e^{-y/x}=\ln x

Solving for y gives the explicit form y=-x\ln\left(e^{-1}-\ln x\right), valid for 0<x<e^{1/e}\approx1.44 (where e^{-1}-\ln x>0).

e^{-1}-e^{-y/x}=\ln x

4.7Applications of differential equationsOften examined

Differential equations are used to solve problems in engineering and the physical sciences, such as physics, chemistry, biology and economics. First order ODEs describe the flow of electricity, the motion of a pendulum and of a falling object, heat flow (thermodynamics) and population growth. In medicine they are used to follow the growth of diseases through graphs.

Steps for any application
  1. Choose symbols and units: the independent variable (usually time t) and the unknown quantity.
  2. Translate the statement into a DE (see the table in the Introduction).
  3. Solve the DE, usually by separating the variables.
  4. Use the given data (an initial value and one more reading) to find the constants.
  5. Answer the question with units, and check that the answer is sensible.
Example 16
A ball is thrown downward from a tower of height 20\,\mathrm{m}. Develop a differential equation representing the flow phenomenon and find the velocity of the ball after 1 second. Also find the velocity with which the ball hits the ground. Neglect the air resistance.
Downward is positive. s is the distance fallen at time t.
  1. The DE. For downward motion the acceleration a=\dfrac{dv}{dt} equals g.
    \dfrac{dv}{dt}=g\qquad\text{or}\qquad dv=g\,dt
  2. Integrate both sides.
    v=gt+c_1
  3. At t=0, v=0 (released from rest), so c_1=0.
    v=gt\qquad(\text{i})
  4. Velocity after 1 s. Substitute g=9.8\,\mathrm{m\,s^{-2}} and t=1\,\mathrm{s}.
    v=9.8\times1=9.8\,\mathrm{m\,s^{-1}}
  5. Distance. Velocity is the rate of change of the distance s: \dfrac{ds}{dt}=gt, so ds=gt\,dt. Integrate.
    s=g\dfrac{t^{2}}{2}+c_2\qquad(\text{ii})
  6. At t=0, s=0, so c_2=0 and s=\dfrac{gt^{2}}{2}. The ball reaches the ground when s=20\,\mathrm{m}.
    20=9.8\times\dfrac{t^{2}}{2}\quad\Longrightarrow\quad t^{2}=\dfrac{40}{9.8}=4.08\quad\Longrightarrow\quad t=2.02\,\mathrm{s}
  7. Velocity at the ground. From (i):
    v=9.8\times2.02=19.8\,\mathrm{m\,s^{-1}}
v(1)=9.8\,\mathrm{m\,s^{-1}},\qquad v_{\text{ground}}\approx19.8\,\mathrm{m\,s^{-1}}

4.7Population growth and decayOften examined

In 1798 Thomas Malthus proposed that the rate of increase of the population of a country is proportional to the population at that time. If P(t) is the population at time t,

\dfrac{dP}{dt}\propto P\qquad\Longrightarrow\qquad \dfrac{dP}{dt}=kP

where k is the constant of proportionality, called the growth constant. k>0 gives growth and k<0 gives decay. The same DE describes bacteria in a culture, money at continuous compound interest and radioactive decay.

Derivation Solution of the growth equation

Aim. Solve \dfrac{dP}{dt}=kP with P=P_0 at t=0.

Assumptions.
  • P>0, so \ln P is defined.
  • k is constant.
  1. Separate the variables.
    \dfrac{dP}{P}=k\,dt
  2. Integrate both sides; write the constant as \ln c.
    \ln P=kt+\ln c
  3. Move \ln c to the left and take antilog.
    \ln\dfrac{P}{c}=kt\quad\Longrightarrow\quad P=c\,e^{kt}
  4. At t=0, P=P_0 and e^{0}=1, so c=P_0.
    P=P_0e^{kt}
ResultP=P_0\,e^{kt}

In every time interval of the same length \tau, the quantity is multiplied by the same factor e^{k\tau}. This is exponential growth (k>0) or exponential decay (k<0).

P=P_0e^{kt}
Both curves start at P_0. Growth speeds up as P increases; decay slows down as P decreases.

**Finding k from two readings.** If P=P_1 at t=t_1, then P_1=P_0e^{kt_1}, so

k=\dfrac{1}{t_1}\ln\dfrac{P_1}{P_0}

Doubling time. The population doubles when e^{kt}=2, that is after t_d=\dfrac{\ln2}{k}, whatever the starting value.

Example (extra practice)
A culture of bacteria grows at a rate proportional to the number present. There are 400 bacteria at the start and 1200 after 3 hours. Find the number after 5 hours and the doubling time.
  1. Model: \dfrac{dN}{dt}=kN, so N=400e^{kt} with t in hours.
  2. Use N(3)=1200.
    1200=400e^{3k}\quad\Longrightarrow\quad e^{3k}=3\quad\Longrightarrow\quad k=\dfrac{\ln3}{3}=0.366\ \text{per hour}
  3. Number after 5 hours: e^{5k}=\left(e^{3k}\right)^{5/3}=3^{5/3}.
    N(5)=400\times3^{5/3}=400\times6.240=2496
  4. Doubling time.
    t_d=\dfrac{\ln2}{k}=\dfrac{3\ln2}{\ln3}=1.89\ \text{hours}
N(5)\approx2496,\qquad t_d\approx1.89\ \text{h}

4.7Radioactive decay and half-life

Radioactive substances disintegrate at a rate proportional to the amount present. If A(t) is the amount (mass or number of atoms) at time t,

\dfrac{dA}{dt}=-\lambda A,\qquad \lambda>0

The minus sign shows that A decreases, and \lambda is the decay constant. Solving exactly as in the growth equation (with k=-\lambda) gives A=A_0e^{-\lambda t}, where A_0 is the amount at t=0.

Derivation Half-life

Aim. Find the half-life T_{1/2}, the time in which half of the substance decays, and write A in terms of it.

  1. At t=T_{1/2} the amount is \dfrac{A_0}{2}.
    \dfrac{A_0}{2}=A_0e^{-\lambda T_{1/2}}\quad\Longrightarrow\quad e^{-\lambda T_{1/2}}=\dfrac12
  2. Take natural logarithms. \ln\tfrac12=-\ln2.
    -\lambda T_{1/2}=-\ln2\quad\Longrightarrow\quad T_{1/2}=\dfrac{\ln2}{\lambda}
  3. Write e^{-\lambda t}=\left(e^{-\lambda T_{1/2}}\right)^{t/T_{1/2}} and use step 1.
    A=A_0\left(\dfrac12\right)^{t/T_{1/2}}
ResultT_{1/2}=\dfrac{\ln2}{\lambda},\qquad A=A_0\left(\tfrac12\right)^{t/T_{1/2}}

The half-life does not depend on how much substance there is at the start.

Radioactive decay
After 1, 2, 3 half-lives, \tfrac12, \tfrac14, \tfrac18 of the substance remains.
Example (extra practice)
The half-life of radium is 1600 years. What percentage of a sample remains after 400 years?
  1. Use A=A_0\left(\tfrac12\right)^{t/T_{1/2}} with \dfrac{t}{T_{1/2}}=\dfrac{400}{1600}=\dfrac14.
    \dfrac{A}{A_0}=\left(\dfrac12\right)^{1/4}=0.841
  2. So about 84.1\% remains. Check: less than one half-life has passed, so more than 50\% must remain.
\approx84.1\%\ \text{remains}

4.7Newton's law of cooling and warmingOften examined

Newton's law of cooling: the rate at which a body cools is proportional to the difference between its temperature T and the temperature T_s of the surrounding medium (air, water, ice). The same law describes the warming of a cold body placed in a warm room.

\dfrac{dT}{dt}\propto\left(T-T_s\right)\qquad\Longrightarrow\qquad \dfrac{dT}{dt}=k\left(T-T_s\right)

Here t is time and \dfrac{dT}{dt} is the rate of change of the temperature. For a cooling body T>T_s and T decreases, so \dfrac{dT}{dt}<0: the constant k is negative. (Many books write \dfrac{dT}{dt}=-k(T-T_s) with k>0; both forms give the same answers.)

Derivation Solution of Newton's law of cooling

Aim. Solve \dfrac{dT}{dt}=k(T-T_s) with T=T_0 at t=0.

Assumptions.
  • The surroundings stay at the constant temperature T_s.
  • T>T_s for cooling (for warming, T_s-T>0 and the same result follows).
  1. Separate the variables.
    \dfrac{dT}{T-T_s}=k\,dt
  2. Integrate both sides; write the constant as \ln c.
    \ln\left(T-T_s\right)=kt+\ln c
  3. Take antilog.
    T-T_s=c\,e^{kt}
  4. At t=0, T=T_0, so c=T_0-T_s.
    T=T_s+\left(T_0-T_s\right)e^{kt}
ResultT=T_s+\left(T_0-T_s\right)e^{kt},\qquad k<0

The temperature difference T-T_s decays exponentially. The body approaches the temperature of the surroundings but never crosses it.

Example 17
According to Newton, the cooling of a hot body is proportional to the temperature difference between its temperature T and the temperature T_0 of its surrounding medium. If a body at 90^\circ\mathrm{C} is allowed to cool in air at 30^\circ\mathrm{C}, and after 5 min the body has cooled to 70^\circ\mathrm{C}, find the temperature of the body as a function of time.
  1. Newton's law with a constant of proportionality k. In this example the book uses T_0 for the air temperature (called T_s in the derivation above), so T_0=30^\circ.
    \dfrac{dT}{dt}=k\left(T-T_0\right)=k\left(T-30\right)
  2. Separate the variables.
    \dfrac{dT}{T-30}=k\,dt
  3. Integrate both sides, with constant \ln c, and take antilog.
    \ln(T-30)=kt+\ln c\quad\Longrightarrow\quad T-30=c\,e^{kt}\qquad(3)
  4. Initial condition T(0)=90^\circ.
    90-30=c\quad\Longrightarrow\quad c=60\quad\Longrightarrow\quad T=60e^{kt}+30\qquad(4)
  5. Second condition T(5)=70^\circ gives k.
    70=60e^{5k}+30\quad\Longrightarrow\quad e^{5k}=\dfrac{40}{60}=\dfrac23\quad\Longrightarrow\quad k=\dfrac15\ln\dfrac23=-0.0811
  6. Substitute k in (4).
    T=60e^{-0.0811t}+30
t in minutes, T in ^\circ\mathrm{C}. The body cools quickly at first, then more slowly, and approaches the air temperature 30^\circ\mathrm{C} (dashed line) but never goes below it.
T=30+60e^{-0.0811\,t}\quad(t\text{ in minutes})
Check point
A thermometer showing a temperature of 20^\circ\mathrm{C} indoors is placed outdoors. After 8 minutes it reads 25^\circ\mathrm{C} and after another 8 minutes it reads 30^\circ\mathrm{C}. Using Newton's law of cooling, find the outdoor temperature.
  1. Let the outdoor temperature be T_s. The solution of Newton's law with T(0)=20 is
    T-T_s=\left(20-T_s\right)e^{kt}
  2. Use the two readings. The second time is double the first, so the second factor is the square of the first.
    25-T_s=\left(20-T_s\right)e^{8k},\qquad 30-T_s=\left(20-T_s\right)\left(e^{8k}\right)^{2}
  3. Eliminate e^{8k}: square the first equation and divide by (20-T_s).
    \left(25-T_s\right)^{2}=\left(20-T_s\right)\left(30-T_s\right)
  4. Expand both sides.
    625-50T_s+T_s^{2}=600-50T_s+T_s^{2}\quad\Longrightarrow\quad 625=600
  5. This is impossible, so no outdoor temperature fits these readings. Under Newton's law the temperature changes by smaller and smaller amounts in equal times, but here it rises by the same 5^\circ\mathrm{C} in each 8 minutes.
\text{No solution: the printed data are inconsistent}

4.7Flow of electricity: series circuits

The SLOs of this unit also mention the flow of electricity in a series circuit. The book gives no worked example, so here is the standard model. It is solved by separating the variables.

Series RL circuit. When the switch S is closed, the e.m.f. E drives a current i through R and L.

By Kirchhoff's voltage law, the sum of the voltage drops round the loop equals the applied e.m.f. E. The drop across the resistance R is Ri, and across the inductance L it is L\dfrac{di}{dt}, where i(t) is the current. So

L\dfrac{di}{dt}+Ri=E

Derivation Growth of current in a series RL circuit

Aim. Find i(t) when the switch is closed at t=0, so i(0)=0.

Assumptions.
  • E, R and L are constants.
  • E-Ri>0 while the current grows.
  1. Rearrange so that the variables can be separated.
    L\dfrac{di}{dt}=E-Ri\quad\Longrightarrow\quad \dfrac{L\,di}{E-Ri}=dt
  2. Integrate. The derivative of E-Ri is -R, so the left side is -\dfrac{L}{R}\ln(E-Ri).
    -\dfrac{L}{R}\ln\left(E-Ri\right)=t+c
  3. At t=0, i=0: c=-\dfrac{L}{R}\ln E. Substitute and multiply by -\dfrac{R}{L}.
    \ln\left(E-Ri\right)-\ln E=-\dfrac{R}{L}t\quad\Longrightarrow\quad \dfrac{E-Ri}{E}=e^{-Rt/L}
  4. Solve for i.
    i=\dfrac{E}{R}\left(1-e^{-Rt/L}\right)
Resulti=\dfrac{E}{R}\left(1-e^{-Rt/L}\right)

The current does not jump to E/R at once: the inductor opposes the change. L/R is called the time constant.

The current rises from 0 and approaches the steady value E/R. After one time constant \tau=L/R it reaches about 63\% of E/R.
Example (extra practice)
A series circuit has E=12\,\mathrm{V}, R=6\,\Omega and L=2\,\mathrm{H}. Find the current 0.5\,\mathrm{s} after the switch is closed.
  1. Steady current and exponent.
    \dfrac{E}{R}=\dfrac{12}{6}=2\,\mathrm{A},\qquad \dfrac{Rt}{L}=\dfrac{6(0.5)}{2}=1.5
  2. Substitute in the formula.
    i=2\left(1-e^{-1.5}\right)=2(1-0.2231)=1.55\,\mathrm{A}
i\approx1.55\,\mathrm{A}

RC circuit. A capacitor of capacitance C with charge q discharging through a resistance R obeys Ri+\dfrac{q}{C}=0 with i=\dfrac{dq}{dt}. So \dfrac{dq}{dt}=-\dfrac{q}{RC}, whose solution is q=q_0e^{-t/(RC)}: exponential decay, exactly like radioactive decay.

4.7Economics and finance

When interest is added continuously, the amount A in an account grows at a rate proportional to itself. If r is the annual interest rate (as a decimal) and t is the time in years,

\dfrac{dA}{dt}=rA\qquad\Longrightarrow\qquad A=A_0e^{rt}

where A_0 is the amount invested. This is the growth equation again. Similarly, a machine whose value V falls at a rate proportional to its value (depreciation) obeys \dfrac{dV}{dt}=-kV, so V=V_0e^{-kt}.

Example (extra practice)
Rs 50,000 is invested at 8\% per year compounded continuously. Find the amount after 5 years and the time for the money to double.
  1. Use A=A_0e^{rt} with A_0=50\,000, r=0.08, t=5.
    A=50\,000\,e^{0.08\times5}=50\,000\,e^{0.4}=50\,000\times1.4918=74\,591
  2. Doubling: e^{0.08t}=2.
    t=\dfrac{\ln2}{0.08}=8.66\ \text{years}
A\approx\text{Rs }74\,591,\qquad t_d\approx8.66\ \text{years}

Summary: choosing the method

Which method to use
TypeHow to recognise itMethodBook example
\dfrac{dy}{dx}=f(x)no y on the right sideintegrate directly\dfrac{dy}{dx}=1+e^{2x}
Separableright side factorises as f(x)\,g(y)separate, integrate both sidesExample 8, 9, 10
HomogeneousM, N of the same degree; \dfrac{dy}{dx}=F\!\left(\dfrac{y}{x}\right)y=ux, then separateExample 14, 15
Growth or decayrate proportional to the amountP=P_0e^{kt}Exercise 4.4 Q1, Q3, Q4
Cooling or warmingrate proportional to T-T_sT=T_s+(T_0-T_s)e^{kt}Example 17

Formula sheet

Simplest differential equation
\dfrac{dy}{dx}=f(x)\ \Longrightarrow\ y=\int f(x)\,dx+c
c: arbitrary constant of integration
Explicit ODE of order n
y^{(n)}=F\left(x,y,y',\dots,y^{(n-1)}\right)
y^{(n)}: the nth derivative of y; F: a known function
Linear DE of order n
a_n(x)y^{(n)}+a_{n-1}(x)y^{(n-1)}+\dots+a_1(x)y'+a_0(x)y=f(x)
y and its derivatives have power 1; the coefficients a_i(x) and f(x) depend on x only
Separable equation
\dfrac{dy}{dx}=f(x)\,g(y)\ \Longrightarrow\ \int\dfrac{dy}{g(y)}=\int f(x)\,dx+c
f: function of x only; g: function of y only, g(y)\neq0
Initial value problem
\dfrac{dy}{dx}=f(x,y),\qquad y(x_0)=y_0
y(x_0)=y_0: initial condition; it fixes the arbitrary constant
Homogeneous function of degree n
f(tx,ty)=t^{n}f(x,y),\qquad t>0
n: degree (any real number, possibly 0 or a fraction)
Homogeneous differential equation
M(x,y)\,dx+N(x,y)\,dy=0
M and N: homogeneous functions of the same degree; equivalently \dfrac{dy}{dx}=F\!\left(\dfrac{y}{x}\right)
Substitution for homogeneous DEs
y=ux,\qquad \dfrac{dy}{dx}=u+x\dfrac{du}{dx}
u: new dependent variable; at the end replace u by \dfrac{y}{x}
Alternative substitution
x=vy,\qquad \dfrac{dx}{dy}=v+y\dfrac{dv}{dy}
v: new dependent variable; replace v by \dfrac{x}{y} at the end
Exponential growth and decay
\dfrac{dP}{dt}=kP\ \Longrightarrow\ P=P_0e^{kt}
P_0: value at t=0; k: growth constant (k>0 growth, k<0 decay)
Growth constant from two readings
k=\dfrac{1}{t_1}\ln\dfrac{P_1}{P_0}
P_1: value at time t_1
k in per unit of time (per year, per day)
Doubling time
t_d=\dfrac{\ln2}{k}
k>0: growth constant
Radioactive decay
\dfrac{dA}{dt}=-\lambda A\ \Longrightarrow\ A=A_0e^{-\lambda t}
A_0: amount at t=0; \lambda>0: decay constant
Half-life
T_{1/2}=\dfrac{\ln2}{\lambda},\qquad A=A_0\left(\tfrac12\right)^{t/T_{1/2}}
T_{1/2}: time for half of the substance to decay
Newton's law of cooling
\dfrac{dT}{dt}=k\left(T-T_s\right)\ \Longrightarrow\ T=T_s+\left(T_0-T_s\right)e^{kt}
T_s: temperature of the surroundings; T_0: temperature at t=0; k<0
Units: ^\circ\mathrm{C} or ^\circ\mathrm{F}; t usually in minutes
Surrounding temperature from equally spaced readings
T_s=\dfrac{T_0T_2-T_1^{2}}{T_0+T_2-2T_1}
T_0,T_1,T_2: readings at t=0,\tau,2\tau
Falling body (from rest)
\dfrac{dv}{dt}=g\ \Longrightarrow\ v=gt,\qquad s=\tfrac12gt^{2}
v: velocity; s: distance fallen; g=9.8\,\mathrm{m\,s^{-2}}
Units: v in \mathrm{m\,s^{-1}}, s in \mathrm{m}
Body thrown upward
\dfrac{dv}{dt}=-g\ \Longrightarrow\ v=v_0-gt,\qquad H=\dfrac{v_0^{2}}{2g}
v_0: initial upward velocity; H: maximum height
Units: \mathrm{m\,s^{-1}}, \mathrm{m}
Series RL circuit
L\dfrac{di}{dt}+Ri=E\ \Longrightarrow\ i=\dfrac{E}{R}\left(1-e^{-Rt/L}\right)
E: e.m.f.; R: resistance; L: inductance; i(0)=0
Units: i in \mathrm{A}, E in \mathrm{V}, R in \Omega, L in \mathrm{H}
RC discharge
\dfrac{dq}{dt}=-\dfrac{q}{RC}\ \Longrightarrow\ q=q_0e^{-t/(RC)}
q: charge; C: capacitance; q_0: charge at t=0
Units: q in \mathrm{C}, RC in \mathrm{s}
Continuous compound interest
\dfrac{dA}{dt}=rA\ \Longrightarrow\ A=A_0e^{rt}
A_0: amount invested; r: annual rate as a decimal; t: years
Integrals used in this unit
\begin{aligned}\int\dfrac{dx}{x}&=\ln|x|+c\\ \int\tan x\,dx&=-\ln|\cos x|+c\\ \int\cot x\,dx&=\ln|\sin x|+c\\ \int\dfrac{dx}{x^{2}+a^{2}}&=\dfrac1a\tan^{-1}\dfrac{x}{a}+c\end{aligned}
c: constant of integration

Key terms

Differential equation (DE)
An equation containing one or more derivatives of an unknown function.
Ordinary differential equation (ODE)
A DE in which the unknown function depends on only one independent variable, so only ordinary derivatives appear.
Partial differential equation (PDE)
A DE whose unknown depends on two or more independent variables, so it contains partial derivatives.
Dependent variable
The unknown function whose derivatives appear in the DE, usually y.
Independent variable
The variable with respect to which we differentiate, usually x or t.
Order
The order of the highest order derivative present in the DE.
Degree
The power of the highest order derivative, after the DE is made free of radicals and fractional powers of derivatives. It is not defined if a derivative is inside a function such as \log or \sin.
Linear DE
A DE in which y and all its derivatives have power 1 and every coefficient depends only on x (or is constant).
Non-linear DE
A DE that is not linear, for example one containing y^{2}, (y')^{3}, y\,y'' or \sin y.
Solution of a DE
A function, defined on an interval, that turns the DE into an identity when substituted.
Arbitrary constant
A constant that may take any value, such as the constant of integration c.
General solution
A solution containing as many arbitrary constants as the order of the DE; it represents a family of curves.
Particular solution
A solution with no arbitrary constants, obtained by giving the constants particular values; it is one curve of the family.
Trivial solution
A solution that is identically zero on an interval, such as y=0.
Family of curves
The set of curves obtained from a general solution by giving its constant different values.
Explicit solution
A solution written as y=f(x).
Implicit solution
A solution given as a relation f(x,y)=0 that is not solved for y.
Initial condition
A condition y(x_0)=y_0 that fixes the value of the arbitrary constant.
Initial value problem (IVP)
A DE together with an initial condition, \dfrac{dy}{dx}=f(x,y), y(x_0)=y_0.
Separable DE
A first order DE that can be written \dfrac{dy}{dx}=f(x)\,g(y), so the variables can be separated and integrated.
Homogeneous function
A function with f(tx,ty)=t^{n}f(x,y) for all t>0; n is its degree.
Homogeneous DE
A DE M\,dx+N\,dy=0 in which M and N are homogeneous functions of the same degree; it is solved by y=ux.
Slope field
A picture of short line segments showing the slope \dfrac{dy}{dx} given by a DE at many points; solution curves follow the segments.
Mathematical model
A mathematical description, such as a DE, that imitates a real situation so that its behaviour can be predicted.
Malthusian law
The rate of increase of a population is proportional to the population: \dfrac{dP}{dt}=kP.
Growth constant
The constant k in \dfrac{dP}{dt}=kP; positive for growth, negative for decay.
Decay constant
The constant \lambda>0 in \dfrac{dA}{dt}=-\lambda A.
Half-life
The time in which half of a radioactive substance decays: T_{1/2}=\dfrac{\ln2}{\lambda}.
Newton's law of cooling
The rate of change of a body's temperature is proportional to the difference between its temperature and that of its surroundings.
Superposition principle
For a linear DE with right side 0, constant multiples and sums of solutions are again solutions.

Exercise 4.1p. 129–130

Order and degree, forming a DE by eliminating constants, and checking a solution by substitution.

1
Find the order and degree of each of the following differential equations.
Solution
(i) (1-x)y''-4xy'+5y=\cos x
  1. The derivatives present are y'' and y'. The highest is y'' (second order).
  2. y'' appears to the power 1.
\text{order }2,\quad\text{degree }1
(ii) yy'+2y=1+x^{2}
  1. The only derivative is y' (first order).
  2. y' appears to the power 1. The product yy' makes the equation non-linear, but it does not change the degree.
\text{order }1,\quad\text{degree }1
(iii) (y'')^{3}-3y'+2y=x
  1. The highest derivative is y'' (second order).
  2. It appears cubed, so its power is 3.
\text{order }2,\quad\text{degree }3
(iv) (y')^{2}-yy''+2=0
  1. The highest derivative is y'' (second order).
  2. y'' appears to the power 1. The square belongs to y', which is not the highest derivative, so it does not decide the degree.
\text{order }2,\quad\text{degree }1
(v) y\dfrac{d^{2}y}{dx^{2}}+\left(\dfrac{dy}{dx}\right)^{3}=0
  1. The highest derivative is \dfrac{d^{2}y}{dx^{2}} (second order).
  2. It appears to the power 1; the cube is on \dfrac{dy}{dx}.
\text{order }2,\quad\text{degree }1
(vi) \dfrac{dy}{dx}=\sqrt{1+\left(\dfrac{d^{2}y}{dx^{2}}\right)^{2}}
  1. A derivative is under a square root, so first square both sides to make the equation a polynomial in the derivatives.
    \left(\dfrac{dy}{dx}\right)^{2}=1+\left(\dfrac{d^{2}y}{dx^{2}}\right)^{2}
  2. The highest derivative is \dfrac{d^{2}y}{dx^{2}}, and it now appears to the power 2.
\text{order }2,\quad\text{degree }2
(vii) x^{3}\dfrac{d^{4}y}{dx^{4}}-x^{2}\dfrac{d^{2}y}{dx^{2}}+4\left(\dfrac{dy}{dx}\right)^{5}+4xy-3y=0
  1. The highest derivative is \dfrac{d^{4}y}{dx^{4}} (fourth order).
  2. It appears to the power 1. The power 5 on \dfrac{dy}{dx} does not count, because \dfrac{dy}{dx} is not the highest derivative.
\text{order }4,\quad\text{degree }1
2
Eliminate the arbitrary constants from the equations.
Solution
(i) y=ae^{x}+be^{-x}+c
  1. There are three arbitrary constants a, b, c, so differentiate three times. The first derivative removes c.
    y'=ae^{x}-be^{-x}
  2. Differentiate again.
    y''=ae^{x}+be^{-x}
  3. Differentiate a third time.
    y'''=ae^{x}-be^{-x}
  4. The right sides of y''' and y' are the same, so a and b are eliminated together.
    y'''=y'\quad\Longrightarrow\quad \dfrac{d^{3}y}{dx^{3}}-\dfrac{dy}{dx}=0
\dfrac{d^{3}y}{dx^{3}}-\dfrac{dy}{dx}=0
(ii) y=\cos(x+b)
  1. One arbitrary constant b, so differentiate once.
    y'=-\sin(x+b)
  2. Square y and y' and add. Since \cos^{2}\theta+\sin^{2}\theta=1, the constant b disappears.
    y^{2}+(y')^{2}=\cos^{2}(x+b)+\sin^{2}(x+b)=1
\left(\dfrac{dy}{dx}\right)^{2}+y^{2}=1
(iii) y=mx+c
  1. Two constants m and c, so differentiate twice. The first derivative removes c.
    \dfrac{dy}{dx}=m
  2. Differentiate again to remove m.
    \dfrac{d^{2}y}{dx^{2}}=0

So every straight line y=mx+c satisfies y''=0: a straight line has zero curvature.

\dfrac{d^{2}y}{dx^{2}}=0
(iv) y=bx^{2}+2ax
  1. Two constants a and b, so differentiate twice.
    y'=2bx+2a\qquad(1)
  2. Differentiate again and solve for b.
    y''=2b\quad\Longrightarrow\quad b=\dfrac{y''}{2}
  3. Use b in (1) to find 2a.
    2a=y'-2bx=y'-xy''
  4. Substitute b and 2a into y=bx^{2}+(2a)x.
    y=\dfrac{y''}{2}x^{2}+x\left(y'-xy''\right)=xy'-\dfrac{x^{2}}{2}y''
  5. Multiply by 2 and rearrange.
    x^{2}y''-2xy'+2y=0

Check. With y=bx^{2}+2ax: x^{2}(2b)-2x(2bx+2a)+2(bx^{2}+2ax)=2bx^{2}-4bx^{2}-4ax+2bx^{2}+4ax=0.

x^{2}\dfrac{d^{2}y}{dx^{2}}-2x\dfrac{dy}{dx}+2y=0
3
Verify that the indicated function is a solution of the given differential equation.
Solution
(i) 2y'+y=0\ ;\quad y=e^{-x/2}
  1. Differentiate the given function with the chain rule.
    y'=-\dfrac12e^{-x/2}
  2. Substitute in the left side.
    2y'+y=2\left(-\dfrac12e^{-x/2}\right)+e^{-x/2}=-e^{-x/2}+e^{-x/2}=0
  3. The left side equals the right side (0) for every x, so y=e^{-x/2} is a solution.
2y'+y=0\ \checkmark
(ii) \dfrac{dy}{dx}-2y=e^{3x}\ ;\quad y=e^{3x}+10e^{2x}
  1. Differentiate.
    \dfrac{dy}{dx}=3e^{3x}+20e^{2x}
  2. Substitute in the left side.
    \dfrac{dy}{dx}-2y=3e^{3x}+20e^{2x}-2e^{3x}-20e^{2x}=e^{3x}
  3. This equals the right side, so the function is a solution.
\dfrac{dy}{dx}-2y=e^{3x}\ \checkmark
(iii) y'=25+y^{2}\ ;\quad y=5\tan5x
  1. Differentiate: \dfrac{d}{dx}\tan5x=5\sec^{2}5x.
    y'=25\sec^{2}5x
  2. Use \sec^{2}\theta=1+\tan^{2}\theta.
    y'=25\left(1+\tan^{2}5x\right)=25+25\tan^{2}5x=25+(5\tan5x)^{2}=25+y^{2}
  3. So y'=25+y^{2} on any interval where \tan5x is defined, for example -\dfrac{\pi}{10}<x<\dfrac{\pi}{10}.
y'=25+y^{2}\ \checkmark
(iv) y'+y=\sin x\ ;\quad y=\dfrac12\sin x-\dfrac12\cos x+10e^{-x}
  1. Differentiate term by term.
    y'=\dfrac12\cos x+\dfrac12\sin x-10e^{-x}
  2. Add y. The cosine terms and the exponential terms cancel.
    y'+y=\dfrac12\cos x+\dfrac12\sin x-10e^{-x}+\dfrac12\sin x-\dfrac12\cos x+10e^{-x}=\sin x
  3. This equals the right side, so the function is a solution.
y'+y=\sin x\ \checkmark
(v) x^{3}\,dy-2\,dx=0\ ;\quad y=-\dfrac{1}{x^{2}}+6
  1. Write the DE with a derivative: divide by dx.
    x^{3}\dfrac{dy}{dx}-2=0
  2. Differentiate y=-x^{-2}+6.
    \dfrac{dy}{dx}=2x^{-3}=\dfrac{2}{x^{3}}
  3. Substitute in the left side.
    x^{3}\cdot\dfrac{2}{x^{3}}-2=2-2=0
  4. This holds for every x\neq0, so the function is a solution.
x^{3}\,dy-2\,dx=0\ \checkmark
(vi) y'-\dfrac1xy=1\ ;\quad y=x\ln x,\ x>0
  1. Differentiate with the product rule.
    y'=x\cdot\dfrac1x+\ln x=1+\ln x
  2. Substitute in the left side.
    y'-\dfrac{y}{x}=1+\ln x-\dfrac{x\ln x}{x}=1+\ln x-\ln x=1
  3. This equals the right side for every x>0, so the function is a solution.
y'-\dfrac{y}{x}=1\ \checkmark
4
Find the order and degree, if defined, for the differential equation dy-\sin x\,dx=0.
Solution
  1. Divide by dx to see the derivative.
    \dfrac{dy}{dx}-\sin x=0
  2. The only derivative is \dfrac{dy}{dx}, of first order, and it appears to the power 1. The equation is a polynomial in \dfrac{dy}{dx}, so the degree is defined.
  3. As a check, its general solution y=-\cos x+c has one arbitrary constant, as expected for a first order DE.
\text{order }1,\quad\text{degree }1
5
Verify that the function y=a\cos x+b\sin x, where a,b\in\mathbb{R}, is a solution of the differential equation \dfrac{d^{2}y}{dx^{2}}+y=0.
Solution
  1. Differentiate once.
    \dfrac{dy}{dx}=-a\sin x+b\cos x
  2. Differentiate again.
    \dfrac{d^{2}y}{dx^{2}}=-a\cos x-b\sin x=-\left(a\cos x+b\sin x\right)=-y
  3. Substitute in the left side of the DE.
    \dfrac{d^{2}y}{dx^{2}}+y=-y+y=0
  4. This is true for all x and for all real a, b. Since there are two arbitrary constants and the DE is second order, y=a\cos x+b\sin x is in fact its general solution.
\dfrac{d^{2}y}{dx^{2}}+y=0\ \checkmark
6
Show that y_1=x^{2} and y_2=x^{3} are both solutions of x^{2}y''-4xy'+6y=0. (i) Are c_1y_1 and c_2y_2 also solutions? (ii) Check if y_1+y_2 is also a solution.
Solution
Showing that y1 and y2 are solutions
  1. For y_1=x^{2}: y_1'=2x, y_1''=2. Substitute.
    x^{2}(2)-4x(2x)+6x^{2}=2x^{2}-8x^{2}+6x^{2}=0
  2. For y_2=x^{3}: y_2'=3x^{2}, y_2''=6x. Substitute.
    x^{2}(6x)-4x(3x^{2})+6x^{3}=6x^{3}-12x^{3}+6x^{3}=0
  3. Both left sides are 0, so y_1=x^{2} and y_2=x^{3} are solutions.
(i) Are c_1y_1 and c_2y_2 also solutions?
  1. For y=c_1x^{2}: y'=2c_1x, y''=2c_1. Every term contains the factor c_1.
    x^{2}(2c_1)-4x(2c_1x)+6c_1x^{2}=c_1\left(2x^{2}-8x^{2}+6x^{2}\right)=c_1\cdot0=0
  2. For y=c_2x^{3}: y'=3c_2x^{2}, y''=6c_2x.
    x^{2}(6c_2x)-4x(3c_2x^{2})+6c_2x^{3}=c_2\left(6x^{3}-12x^{3}+6x^{3}\right)=0
  3. So yes, both are solutions for any constants c_1, c_2. Reason: the DE is linear and its right side is 0, so a constant factor passes through every term.
\text{Yes: } c_1x^{2}\text{ and } c_2x^{3}\text{ are solutions}
(ii) Check if y_1+y_2 is also a solution.
  1. Let y=x^{2}+x^{3}. Then y'=2x+3x^{2} and y''=2+6x.
    x^{2}(2+6x)-4x\left(2x+3x^{2}\right)+6\left(x^{2}+x^{3}\right)
  2. Expand and collect.
    =2x^{2}+6x^{3}-8x^{2}-12x^{3}+6x^{2}+6x^{3}=0
  3. So y_1+y_2 is also a solution. In fact y=c_1x^{2}+c_2x^{3} is a solution for all c_1,c_2: with two arbitrary constants, it is the general solution of this second order DE.
\text{Yes: } y=x^{2}+x^{3}\text{ is a solution}

Exercise 4.2p. 133

Separate the variables, integrate both sides with one constant, then (for Q9 to Q15) use the initial condition to find the constant.

1
Solve by separating the variables: \dfrac{dy}{dx}=-\dfrac{1}{e^{3x}}
Solution
  1. Write \dfrac{1}{e^{3x}}=e^{-3x} and separate: dy on the left, everything in x on the right.
    dy=-e^{-3x}\,dx
  2. Integrate both sides. Use \int e^{ax}\,dx=\dfrac{e^{ax}}{a} with a=-3.
    y=-\dfrac{e^{-3x}}{-3}+c=\dfrac13e^{-3x}+c
y=\dfrac{1}{3}e^{-3x}+c\qquad\left(\text{or } 3y=e^{-3x}+C\right)
22026 Annual I, Old syllabus (Curriculum 2000), Q.2 (iv) OR, old syllabus
Solve by separating the variables: x\dfrac{dy}{dx}=4y
Solution
  1. Divide by xy and multiply by dx to separate.
    \dfrac{dy}{y}=\dfrac{4\,dx}{x}
  2. Integrate; write the constant as \ln c.
    \ln y=4\ln x+\ln c=\ln x^{4}+\ln c=\ln\left(cx^{4}\right)
  3. Take antilog.
    y=cx^{4}

Dividing by y assumed y\neq0; the solution y=0 is included by c=0.

y=cx^{4}
3
Solve by separating the variables: \dfrac{dy}{dx}=\dfrac{y^{3}}{x^{2}}
Solution
  1. Separate: divide by y^{3} and multiply by dx.
    y^{-3}\,dy=x^{-2}\,dx
  2. Integrate with the power rule \int u^{n}\,du=\dfrac{u^{n+1}}{n+1}.
    \dfrac{y^{-2}}{-2}=\dfrac{x^{-1}}{-1}+c\quad\Longrightarrow\quad -\dfrac{1}{2y^{2}}=-\dfrac{1}{x}+c
  3. Multiply by -2 and rename C=-2c.
    \dfrac{1}{y^{2}}=\dfrac{2}{x}+C

Multiplying by xy^{2} gives the equivalent form x=y^{2}(2+Cx). The constant solution y=0 was lost when dividing by y^{3}, and it also satisfies the DE.

\dfrac{1}{y^{2}}=\dfrac{2}{x}+C
4
Solve by separating the variables: \dfrac{dy}{dx}=e^{2x+3y}
Solution
  1. Use e^{2x+3y}=e^{2x}\cdot e^{3y}; the equation is then separable.
    \dfrac{dy}{dx}=e^{2x}e^{3y}\quad\Longrightarrow\quad e^{-3y}\,dy=e^{2x}\,dx
  2. Integrate both sides.
    -\dfrac13e^{-3y}=\dfrac12e^{2x}+c
  3. Multiply by -6 to clear fractions, and rename C=-6c.
    2e^{-3y}=-3e^{2x}-6c\quad\Longrightarrow\quad 3e^{2x}+2e^{-3y}=C
3e^{2x}+2e^{-3y}=C
5
Solve by separating the variables: \dfrac{dy}{dx}=\dfrac{x^{2}y^{2}}{1+x}
Solution
  1. Separate: divide by y^{2} and multiply by dx.
    \dfrac{dy}{y^{2}}=\dfrac{x^{2}}{1+x}\,dx
  2. The right side is an improper fraction; divide x^{2} by x+1 first.
    \dfrac{x^{2}}{x+1}=x-1+\dfrac{1}{x+1}
  3. Integrate both sides.
    \int y^{-2}\,dy=\int\left(x-1+\dfrac{1}{x+1}\right)dx\quad\Longrightarrow\quad -\dfrac1y=\dfrac{x^{2}}{2}-x+\ln|1+x|+c

Explicitly, y=-\dfrac{1}{\tfrac12x^{2}-x+\ln|1+x|+c}. The constant solution y=0 also satisfies the DE.

-\dfrac{1}{y}=\dfrac{x^{2}}{2}-x+\ln|1+x|+c
6
Solve by separating the variables: 2y(x+1)\,dy=x\,dx
Solution
  1. Divide by (x+1) to separate.
    2y\,dy=\dfrac{x}{x+1}\,dx
  2. Rewrite the fraction: \dfrac{x}{x+1}=\dfrac{(x+1)-1}{x+1}=1-\dfrac{1}{x+1}.
    2y\,dy=\left(1-\dfrac{1}{x+1}\right)dx
  3. Integrate both sides.
    y^{2}=x-\ln|x+1|+c
y^{2}=x-\ln|x+1|+c
7
Solve by separating the variables: \dfrac{dy}{dx}+y^{2}\sin x=0
Solution
  1. Move the second term to the right and separate.
    \dfrac{dy}{dx}=-y^{2}\sin x\quad\Longrightarrow\quad \dfrac{dy}{y^{2}}=-\sin x\,dx
  2. Integrate: \int y^{-2}\,dy=-\dfrac1y and \int(-\sin x)\,dx=\cos x.
    -\dfrac1y=\cos x+c
  3. Solve for y and rename C=-c.
    y=-\dfrac{1}{\cos x+c}=\dfrac{1}{C-\cos x}

The constant solution y=0 (lost when dividing by y^{2}) also satisfies the DE.

y=\dfrac{1}{C-\cos x}
8
Solve by separating the variables: (\sin x+\cos x)\,dx=\cot y\cos x\,dy
Solution
  1. Divide both sides by \cos x to separate the variables.
    \dfrac{\sin x+\cos x}{\cos x}\,dx=\cot y\,dy\quad\Longrightarrow\quad(\tan x+1)\,dx=\cot y\,dy
  2. Integrate. Use \int\cot y\,dy=\ln|\sin y| and \int\tan x\,dx=-\ln|\cos x|.
    \ln|\sin y|=-\ln|\cos x|+x+c
  3. Collect the logarithms on the left.
    \ln|\sin y|+\ln|\cos x|=x+c\quad\Longrightarrow\quad \ln|\sin y\cos x|=x+c
  4. Take antilog and write C=\pm e^{c}.
    \sin y\cos x=Ce^{x}
\sin y\cos x=Ce^{x}
9
Solve the initial value problem: \dfrac{dy}{dx}=\cos x\ ;\ \ y(0)=1
Solution
  1. Separate and integrate.
    dy=\cos x\,dx\quad\Longrightarrow\quad y=\sin x+c
  2. Use y(0)=1: 1=\sin0+c.
    c=1
  3. Substitute c back.
    y=\sin x+1
The particular solution passes through (0,1).
y=\sin x+1
10
Solve the initial value problem: 2\dfrac{dy}{dx}=4xe^{-x}\ ;\ \ y(0)=2
Solution
  1. Divide by 2 and separate.
    dy=2xe^{-x}\,dx
  2. Integrate the right side by parts with u=x, dv=e^{-x}dx: \int xe^{-x}dx=-xe^{-x}-\int(-e^{-x})dx=-xe^{-x}-e^{-x}.
    y=2\left(-xe^{-x}-e^{-x}\right)+c=-2(x+1)e^{-x}+c
  3. Use y(0)=2: 2=-2(1)(1)+c.
    c=4
  4. Substitute c back.
    y=4-2(x+1)e^{-x}
The solution has its lowest point at (0,2) (where \dfrac{dy}{dx}=2xe^{-x}=0) and approaches y=4 as x grows.
y=4-2(x+1)e^{-x}
11
Solve the initial value problem: \dfrac{dy}{dx}+\left(\dfrac{1+x}{x}\right)y=0\ ;\ \ y(1)=1
Solution
  1. Move the y term to the right and separate.
    \dfrac{dy}{y}=-\dfrac{1+x}{x}\,dx=-\left(\dfrac1x+1\right)dx
  2. Integrate both sides.
    \ln|y|=-\ln|x|-x+c
  3. Take antilog; write A=\pm e^{c}.
    y=\dfrac{A\,e^{-x}}{x}
  4. Use y(1)=1: 1=Ae^{-1}.
    A=e
  5. Substitute A back.
    y=\dfrac{e\cdot e^{-x}}{x}=\dfrac{e^{1-x}}{x}
The solution exists for x>0 and passes through (1,1).
y=\dfrac{e^{1-x}}{x}\qquad(\text{or } xy=e^{1-x})
12
Solve the initial value problem: \dfrac{dy}{dx}+y\tan2x=0\ ;\ \ y(0)=2
Solution
  1. Separate the variables.
    \dfrac{dy}{y}=-\tan2x\,dx
  2. Integrate. Since \int\tan2x\,dx=-\dfrac12\ln|\cos2x|, the right side is +\dfrac12\ln|\cos2x|.
    \ln|y|=\dfrac12\ln|\cos2x|+c
  3. Take antilog; write A=\pm e^{c}.
    y=A\sqrt{\cos2x}
  4. Use y(0)=2: 2=A\sqrt{\cos0}=A.
    A=2
  5. Substitute A back.
    y=2\sqrt{\cos2x}\qquad\left(\text{or } y^{2}=4\cos2x\right)
The solution is defined for -\dfrac{\pi}{4}<x<\dfrac{\pi}{4}, where \cos2x>0.
y=2\sqrt{\cos2x}
13
Solve the initial value problem: \dfrac{dy}{dx}=y^{2}+4\ ;\ \ y(0)=-2
Solution
  1. Separate the variables.
    \dfrac{dy}{y^{2}+4}=dx
  2. Integrate with \int\dfrac{dy}{y^{2}+a^{2}}=\dfrac1a\tan^{-1}\dfrac{y}{a}, a=2.
    \dfrac12\tan^{-1}\dfrac{y}{2}=x+c
  3. Use y(0)=-2: \dfrac12\tan^{-1}(-1)=c, and \tan^{-1}(-1)=-\dfrac{\pi}{4}.
    c=-\dfrac{\pi}{8}
  4. Substitute c and multiply by 2.
    \tan^{-1}\dfrac{y}{2}=2x-\dfrac{\pi}{4}
  5. Take \tan of both sides.
    y=2\tan\left(2x-\dfrac{\pi}{4}\right)
The asymptotes x=-\dfrac{\pi}{8} and x=\dfrac{3\pi}{8} bound the solution through (0,-2).
y=2\tan\left(2x-\dfrac{\pi}{4}\right)
14
Solve the initial value problem: (1-x)\,dy+y^{-1}\,dx=0\ ;\ \ y(0)=2
Solution
  1. Move the dx term to the right, then multiply by y and divide by (1-x).
    (1-x)\,dy=-\dfrac{dx}{y}\quad\Longrightarrow\quad y\,dy=-\dfrac{dx}{1-x}
  2. Integrate. Since \dfrac{d}{dx}\ln|1-x|=-\dfrac{1}{1-x}, the right side integrates to \ln|1-x|.
    \dfrac{y^{2}}{2}=\ln|1-x|+c
  3. Use y(0)=2: \dfrac{4}{2}=\ln1+c.
    c=2
  4. Substitute c and multiply by 2.
    y^{2}=2\ln|1-x|+4
  5. Take the positive root, because y(0)=2>0.
    y=\sqrt{4+2\ln(1-x)},\qquad x<1
The solution through (0,2) exists for x<1-e^{-2}\approx0.865, where 4+2\ln(1-x)>0.
y^{2}=4+2\ln|1-x|\qquad\left(y=\sqrt{4+2\ln(1-x)}\right)
15
Solve the initial value problem: 2(y-1)\,dy=\left(3x^{2}+4x+2\right)dx\ ;\ \ y(0)=-1
Solution
  1. The variables are already separated. Integrate both sides; \int2(y-1)\,dy=(y-1)^{2}.
    (y-1)^{2}=x^{3}+2x^{2}+2x+c
  2. Use y(0)=-1: (-1-1)^{2}=0+c.
    c=4
  3. Substitute c back.
    (y-1)^{2}=x^{3}+2x^{2}+2x+4
  4. Take the square root. At x=0, y-1=-2<0, so choose the negative root.
    y-1=-\sqrt{x^{3}+2x^{2}+2x+4}\quad\Longrightarrow\quad y=1-\sqrt{x^{3}+2x^{2}+2x+4}

Since x^{3}+2x^{2}+2x+4=(x+2)(x^{2}+2), the expression under the root is positive exactly when x>-2, so the solution is valid for x>-2.

The solution through (0,-1) is the lower branch, defined for x>-2.
y=1-\sqrt{x^{3}+2x^{2}+2x+4}\qquad\left(\text{or } y^{2}-2y=x^{3}+2x^{2}+2x+3\right)

Exercise 4.3p. 136

Q1 to Q3: test f(tx,ty)=t^{n}f(x,y). Q4 to Q15: substitute y=ux, \dfrac{dy}{dx}=u+x\dfrac{du}{dx}, separate, integrate, and replace u by \dfrac{y}{x}.

1
Check whether the function is homogeneous or not. If homogeneous, find its degree: f(x,y)=6xy^{3}-x^{2}y^{2}
Solution
  1. Replace x by tx and y by ty.
    f(tx,ty)=6(tx)(ty)^{3}-(tx)^{2}(ty)^{2}=6t^{4}xy^{3}-t^{4}x^{2}y^{2}
  2. Take out the common factor t^{4}.
    f(tx,ty)=t^{4}\left(6xy^{3}-x^{2}y^{2}\right)=t^{4}f(x,y)
  3. So f is homogeneous of degree 4 (each term has total degree 1+3=2+2=4).
\text{homogeneous, degree }4
2
Check whether the function is homogeneous or not. If homogeneous, find its degree: f(x,y)=x^{2}-y
Solution
  1. Replace x by tx and y by ty.
    f(tx,ty)=t^{2}x^{2}-ty
  2. The two terms carry different powers of t (t^{2} and t^{1}), so no single power t^{n} can be taken out.
    f(tx,ty)\neq t^{n}f(x,y)
  3. So f is not homogeneous: its terms have total degrees 2 and 1.
\text{not homogeneous}
3
Check whether the function is homogeneous or not. If homogeneous, find its degree: f(x,y)=\dfrac{2y^{3}}{x^{2}y}-7
Solution
  1. Replace x by tx and y by ty.
    f(tx,ty)=\dfrac{2t^{3}y^{3}}{t^{2}x^{2}\cdot ty}-7=\dfrac{2t^{3}y^{3}}{t^{3}x^{2}y}-7
  2. The t^{3} cancels, so the function is unchanged.
    f(tx,ty)=\dfrac{2y^{3}}{x^{2}y}-7=t^{0}f(x,y)
  3. So f is homogeneous of degree 0. The constant -7 is allowed here because the fraction also has degree 3-3=0.
\text{homogeneous, degree }0
4
Solve the homogeneous differential equation: (x-y)\,dx+x\,dy=0
Solution
  1. M=x-y and N=x are both of degree 1, so the equation is homogeneous. Write it as \dfrac{dy}{dx}.
    \dfrac{dy}{dx}=-\dfrac{x-y}{x}=\dfrac{y-x}{x}
  2. Put y=ux, \dfrac{dy}{dx}=u+x\dfrac{du}{dx}.
    u+x\dfrac{du}{dx}=\dfrac{ux-x}{x}=u-1
  3. The u cancels.
    x\dfrac{du}{dx}=-1\quad\Longrightarrow\quad du=-\dfrac{dx}{x}
  4. Integrate.
    u=-\ln|x|+c
  5. Replace u by \dfrac{y}{x} and multiply by x.
    \dfrac{y}{x}=c-\ln|x|\quad\Longrightarrow\quad y=x\left(c-\ln|x|\right)
y=cx-x\ln|x|
5
Solve the homogeneous differential equation: y\,dx-(y-x)\,dy=0
Solution
  1. M=y and N=-(y-x) are both of degree 1, so the equation is homogeneous.
    \dfrac{dy}{dx}=\dfrac{y}{y-x}
  2. Put y=ux.
    u+x\dfrac{du}{dx}=\dfrac{ux}{ux-x}=\dfrac{u}{u-1}
  3. Subtract u.
    x\dfrac{du}{dx}=\dfrac{u}{u-1}-u=\dfrac{u-u^{2}+u}{u-1}=\dfrac{u(2-u)}{u-1}
  4. Separate the variables.
    \dfrac{u-1}{u(2-u)}\,du=\dfrac{dx}{x}
  5. Partial fractions: u-1=A(2-u)+Bu. Put u=0: A=-\tfrac12. Put u=2: B=\tfrac12.
    \dfrac{u-1}{u(2-u)}=-\dfrac{1}{2u}+\dfrac{1}{2(2-u)}
  6. Integrate. Note \int\dfrac{du}{2-u}=-\ln|2-u|.
    -\dfrac12\ln|u|-\dfrac12\ln|2-u|=\ln|x|+c
  7. Multiply by -2 and combine the logarithms.
    \ln\left|u(2-u)\right|+\ln x^{2}=-2c\quad\Longrightarrow\quad u(2-u)\,x^{2}=C_1
  8. Replace u by \dfrac{y}{x}: \dfrac{y}{x}\left(2-\dfrac{y}{x}\right)x^{2}=y(2x-y).
    2xy-y^{2}=C_1\quad\Longrightarrow\quad y^{2}-2xy=C
y^{2}-2xy=C
6
Solve the homogeneous differential equation: \dfrac{dy}{dx}=\dfrac{y-x}{x+y}
Solution
  1. Numerator and denominator are both of degree 1, so the equation is homogeneous. Put y=ux.
    u+x\dfrac{du}{dx}=\dfrac{ux-x}{x+ux}=\dfrac{u-1}{1+u}
  2. Subtract u.
    x\dfrac{du}{dx}=\dfrac{u-1-u-u^{2}}{1+u}=-\dfrac{1+u^{2}}{1+u}
  3. Separate the variables.
    \dfrac{1+u}{1+u^{2}}\,du=-\dfrac{dx}{x}
  4. Split the fraction and integrate: \int\dfrac{du}{1+u^{2}}=\tan^{-1}u and \int\dfrac{u\,du}{1+u^{2}}=\dfrac12\ln(1+u^{2}).
    \tan^{-1}u+\dfrac12\ln\left(1+u^{2}\right)=-\ln|x|+c
  5. Replace u by \dfrac{y}{x}. Note \dfrac12\ln\dfrac{x^{2}+y^{2}}{x^{2}}=\dfrac12\ln(x^{2}+y^{2})-\ln|x|, so the \ln|x| terms cancel.
    \tan^{-1}\dfrac{y}{x}+\dfrac12\ln\left(x^{2}+y^{2}\right)=c
  6. Multiply by 2 and rename C=2c.
    \ln\left(x^{2}+y^{2}\right)+2\tan^{-1}\dfrac{y}{x}=C
\ln\left(x^{2}+y^{2}\right)+2\tan^{-1}\dfrac{y}{x}=C
7
Solve the homogeneous differential equation: \dfrac{dy}{dx}=\dfrac{y^{2}+yx}{x^{2}}
Solution
  1. Numerator and denominator are of degree 2, so the equation is homogeneous. Put y=ux.
    u+x\dfrac{du}{dx}=\dfrac{u^{2}x^{2}+ux^{2}}{x^{2}}=u^{2}+u
  2. Subtract u and separate.
    x\dfrac{du}{dx}=u^{2}\quad\Longrightarrow\quad \dfrac{du}{u^{2}}=\dfrac{dx}{x}
  3. Integrate.
    -\dfrac1u=\ln|x|+c
  4. Replace u by \dfrac{y}{x} and solve for y.
    -\dfrac{x}{y}=\ln|x|+c\quad\Longrightarrow\quad y=-\dfrac{x}{\ln|x|+c}

The solution y=0 (u=0, lost when dividing by u^{2}) also satisfies the DE.

y=-\dfrac{x}{\ln|x|+c}
8
Solve the homogeneous differential equation: \dfrac{dy}{dx}=\dfrac{y}{x+\sqrt{xy}}
Solution
  1. Take x>0, y>0 so that \sqrt{xy} is real. Numerator and denominator are of degree 1. Put y=ux; then \sqrt{xy}=\sqrt{ux^{2}}=x\sqrt u.
    u+x\dfrac{du}{dx}=\dfrac{ux}{x+x\sqrt u}=\dfrac{u}{1+\sqrt u}
  2. Subtract u.
    x\dfrac{du}{dx}=\dfrac{u-u-u\sqrt u}{1+\sqrt u}=-\dfrac{u^{3/2}}{1+\sqrt u}
  3. Separate the variables and split the fraction.
    \dfrac{1+\sqrt u}{u^{3/2}}\,du=-\dfrac{dx}{x}\quad\Longrightarrow\quad\left(u^{-3/2}+u^{-1}\right)du=-\dfrac{dx}{x}
  4. Integrate.
    -2u^{-1/2}+\ln u=-\ln x+c
  5. Replace u by \dfrac{y}{x}: u^{-1/2}=\sqrt{\dfrac{x}{y}} and \ln u=\ln y-\ln x. The \ln x terms cancel.
    -2\sqrt{\dfrac{x}{y}}+\ln y-\ln x=-\ln x+c\quad\Longrightarrow\quad \ln y=2\sqrt{\dfrac{x}{y}}+c

Taking antilog gives the equivalent form y=Ce^{2\sqrt{x/y}}.

\ln y=2\sqrt{\dfrac{x}{y}}+c\qquad\left(\text{or } y=Ce^{2\sqrt{x/y}}\right)
9
Solve the homogeneous differential equation: \dfrac{dy}{dx}=\dfrac{3x^{3}+y^{3}}{xy^{2}}
Solution
  1. Numerator and denominator are of degree 3, so the equation is homogeneous. Put y=ux.
    u+x\dfrac{du}{dx}=\dfrac{3x^{3}+u^{3}x^{3}}{x\cdot u^{2}x^{2}}=\dfrac{3+u^{3}}{u^{2}}=\dfrac{3}{u^{2}}+u
  2. Subtract u and separate.
    x\dfrac{du}{dx}=\dfrac{3}{u^{2}}\quad\Longrightarrow\quad u^{2}\,du=\dfrac{3\,dx}{x}
  3. Integrate.
    \dfrac{u^{3}}{3}=3\ln|x|+c
  4. Replace u by \dfrac{y}{x}, multiply by 3x^{3} and rename C=3c.
    \dfrac{y^{3}}{x^{3}}=9\ln|x|+C\quad\Longrightarrow\quad y^{3}=9x^{3}\ln|x|+Cx^{3}
y^{3}=x^{3}\left(9\ln|x|+C\right)
10
Solve the homogeneous differential equation: \dfrac{dy}{dx}=\dfrac{x+3y}{3x+y}
Solution
  1. Numerator and denominator are of degree 1. Put y=ux.
    u+x\dfrac{du}{dx}=\dfrac{x+3ux}{3x+ux}=\dfrac{1+3u}{3+u}
  2. Subtract u.
    x\dfrac{du}{dx}=\dfrac{1+3u-3u-u^{2}}{3+u}=\dfrac{1-u^{2}}{3+u}
  3. Separate the variables.
    \dfrac{3+u}{(1-u)(1+u)}\,du=\dfrac{dx}{x}
  4. Partial fractions: 3+u=A(1+u)+B(1-u). Put u=1: A=2. Put u=-1: B=1.
    \dfrac{3+u}{(1-u)(1+u)}=\dfrac{2}{1-u}+\dfrac{1}{1+u}
  5. Integrate; \int\dfrac{2\,du}{1-u}=-2\ln|1-u|. Write the constant as \ln c.
    -2\ln|1-u|+\ln|1+u|=\ln|x|+\ln c\quad\Longrightarrow\quad \dfrac{1+u}{(1-u)^{2}}=cx
  6. Replace u by \dfrac{y}{x}: \dfrac{(x+y)/x}{(x-y)^{2}/x^{2}}=\dfrac{x(x+y)}{(x-y)^{2}}.
    \dfrac{x(x+y)}{(x-y)^{2}}=cx\quad\Longrightarrow\quad x+y=c(x-y)^{2}
x+y=c\,(x-y)^{2}
11
Solve the homogeneous differential equation: \left[y+x\cot\left(\dfrac{y}{x}\right)\right]dx-x\,dy=0
Solution
  1. Divide by x\,dx. The right side depends only on \dfrac{y}{x}, so the equation is homogeneous.
    \dfrac{dy}{dx}=\dfrac{y}{x}+\cot\dfrac{y}{x}
  2. Put y=ux. The u cancels.
    u+x\dfrac{du}{dx}=u+\cot u\quad\Longrightarrow\quad x\dfrac{du}{dx}=\cot u
  3. Separate: \dfrac{1}{\cot u}=\tan u.
    \tan u\,du=\dfrac{dx}{x}
  4. Integrate; \int\tan u\,du=-\ln|\cos u|. Write the constant as -\ln C.
    -\ln|\cos u|=\ln|x|-\ln C\quad\Longrightarrow\quad \ln\left|x\cos u\right|=\ln C
  5. Take antilog and replace u by \dfrac{y}{x}.
    x\cos\dfrac{y}{x}=C
x\cos\left(\dfrac{y}{x}\right)=C
12
Solve the initial value problem: xy^{2}\dfrac{dy}{dx}=y^{3}-x^{3}\ ;\ \ y(1)=2
Solution
  1. Divide by xy^{2}. Numerator and denominator are of degree 3, so the equation is homogeneous.
    \dfrac{dy}{dx}=\dfrac{y^{3}-x^{3}}{xy^{2}}
  2. Put y=ux.
    u+x\dfrac{du}{dx}=\dfrac{u^{3}x^{3}-x^{3}}{x\cdot u^{2}x^{2}}=\dfrac{u^{3}-1}{u^{2}}=u-\dfrac{1}{u^{2}}
  3. Subtract u and separate.
    x\dfrac{du}{dx}=-\dfrac{1}{u^{2}}\quad\Longrightarrow\quad u^{2}\,du=-\dfrac{dx}{x}
  4. Integrate.
    \dfrac{u^{3}}{3}=-\ln|x|+c
  5. Replace u by \dfrac{y}{x}, multiply by 3x^{3}, and rename C=3c.
    y^{3}=x^{3}\left(C-3\ln|x|\right)
  6. Use y(1)=2: 8=1\cdot(C-0).
    C=8
  7. Substitute C back.
    y^{3}=x^{3}\left(8-3\ln x\right)
y^{3}=x^{3}\left(8-3\ln x\right)
132026 Annual I, Set D, Q.6
Solve the initial value problem: \left(x^{2}+2y^{2}\right)dx=xy\,dy\ ;\ \ y(1)=1
Solution
  1. Both sides have degree 2, so the equation is homogeneous.
    \dfrac{dy}{dx}=\dfrac{x^{2}+2y^{2}}{xy}
  2. Put y=ux.
    u+x\dfrac{du}{dx}=\dfrac{x^{2}+2u^{2}x^{2}}{x\cdot ux}=\dfrac{1+2u^{2}}{u}
  3. Subtract u.
    x\dfrac{du}{dx}=\dfrac{1+2u^{2}-u^{2}}{u}=\dfrac{1+u^{2}}{u}
  4. Separate the variables.
    \dfrac{u\,du}{1+u^{2}}=\dfrac{dx}{x}
  5. Integrate: the numerator is half the derivative of 1+u^{2}. Write the constant as \ln\sqrt{C}.
    \dfrac12\ln\left(1+u^{2}\right)=\ln|x|+\dfrac12\ln C\quad\Longrightarrow\quad 1+u^{2}=Cx^{2}
  6. Replace u by \dfrac{y}{x} and multiply by x^{2}.
    x^{2}+y^{2}=Cx^{4}
  7. Use y(1)=1: 1+1=C.
    C=2
  8. Substitute C back.
    x^{2}+y^{2}=2x^{4}
Explicit form of x^{2}+y^{2}=2x^{4} for x>0, y\geq0: it passes through (1,1) and starts at x=\tfrac{1}{\sqrt2}.
x^{2}+y^{2}=2x^{4}
14
Solve the initial value problem: 2x^{2}\dfrac{dy}{dx}=3xy+y^{2}\ ;\ \ y(1)=-2
Solution
  1. Divide by 2x^{2}. Every term has degree 2, so the equation is homogeneous.
    \dfrac{dy}{dx}=\dfrac{3xy+y^{2}}{2x^{2}}
  2. Put y=ux.
    u+x\dfrac{du}{dx}=\dfrac{3ux^{2}+u^{2}x^{2}}{2x^{2}}=\dfrac{3u+u^{2}}{2}
  3. Subtract u.
    x\dfrac{du}{dx}=\dfrac{3u+u^{2}-2u}{2}=\dfrac{u(1+u)}{2}
  4. Separate the variables.
    \dfrac{2\,du}{u(1+u)}=\dfrac{dx}{x}
  5. Partial fractions: \dfrac{1}{u(1+u)}=\dfrac1u-\dfrac{1}{1+u}. Integrate.
    2\ln|u|-2\ln|1+u|=\ln|x|+c\quad\Longrightarrow\quad \ln\left(\dfrac{u}{1+u}\right)^{2}=\ln|x|+c
  6. Take antilog (C=e^{c}) and replace u by \dfrac{y}{x}: \dfrac{u}{1+u}=\dfrac{y}{x+y}.
    \left(\dfrac{y}{x+y}\right)^{2}=Cx
  7. Use y(1)=-2: \left(\dfrac{-2}{1-2}\right)^{2}=C.
    C=4\quad\Longrightarrow\quad y^{2}=4x\,(x+y)^{2}
  8. Solve for y. At x=1, \dfrac{y}{x+y}=\dfrac{-2}{-1}=2>0, so take \dfrac{y}{x+y}=+2\sqrt x.
    y=2\sqrt x\,(x+y)\quad\Longrightarrow\quad y\left(1-2\sqrt x\right)=2x\sqrt x\quad\Longrightarrow\quad y=\dfrac{2x^{3/2}}{1-2\sqrt x}
The solution through (1,-2) is defined for x>\tfrac14; the line x=\tfrac14 is a vertical asymptote.
y=\dfrac{2x^{3/2}}{1-2\sqrt{x}}\qquad\left(\text{or } y^{2}=4x(x+y)^{2}\right)
15
Solve the initial value problem: \left(x+ye^{y/x}\right)dx-xe^{y/x}\,dy=0\ ;\ \ y(1)=0
Solution
  1. Write the equation as \dfrac{dy}{dx}. Dividing numerator and denominator by x shows it depends only on \dfrac{y}{x}, so it is homogeneous.
    \dfrac{dy}{dx}=\dfrac{x+ye^{y/x}}{xe^{y/x}}
  2. Put y=ux.
    u+x\dfrac{du}{dx}=\dfrac{x+uxe^{u}}{xe^{u}}=\dfrac{1+ue^{u}}{e^{u}}=e^{-u}+u
  3. The u cancels. Separate the variables.
    x\dfrac{du}{dx}=e^{-u}\quad\Longrightarrow\quad e^{u}\,du=\dfrac{dx}{x}
  4. Integrate.
    e^{u}=\ln|x|+c\quad\Longrightarrow\quad e^{y/x}=\ln|x|+c
  5. Use y(1)=0: e^{0}=\ln1+c.
    c=1
  6. Substitute c and take \ln of both sides.
    e^{y/x}=1+\ln x\quad\Longrightarrow\quad y=x\ln\left(1+\ln x\right)
The solution through (1,0) needs 1+\ln x>0, that is x>\tfrac1e.
e^{y/x}=1+\ln x\qquad\left(y=x\ln(1+\ln x)\right)

Exercise 4.4p. 138

Write the DE from the words, solve it by separating the variables, then use the two pieces of data to find the constants.

12026 Annual I, Set B, Q.2 (i) OR2026 Annual I, Set D, Q.2 (i) OR
Thomas Malthus in 1798 proved that the increase in population of a country or a city at a certain time is proportional to the total population of the country at that time \left(\dfrac{dP}{dt}\propto P\right). If at present the population of city A is 20 million and after 4 years it is expected to be 25 million, what would be the population of that city after 12 years?
Solution
Given
  • P(0) = 20\ \text{million} present population
  • P(4) = 25\ \text{million}
To find
  • P(12)
  1. Model. Let P(t) be the population in millions after t years. Introduce the constant of proportionality k.
    \dfrac{dP}{dt}=kP
  2. Separate the variables and integrate.
    \dfrac{dP}{P}=k\,dt\quad\Longrightarrow\quad \ln P=kt+\ln c\quad\Longrightarrow\quad P=c\,e^{kt}
  3. Use P(0)=20: c=20.
    P=20e^{kt}
  4. Use P(4)=25 to find k.
    25=20e^{4k}\quad\Longrightarrow\quad e^{4k}=\dfrac54\quad\Longrightarrow\quad k=\dfrac14\ln1.25=0.0558\ \text{per year}
  5. Population after 12 years: e^{12k}=\left(e^{4k}\right)^{3}.
    P(12)=20\left(\dfrac54\right)^{3}=20\times\dfrac{125}{64}=39.06\ \text{million}
t in years, P in millions. The population is multiplied by 1.25 every 4 years.
P(12)\approx39.06\ \text{million}
2
Ayesha was preparing a pizza in a baking oven. She observed that the temperature of the cooked pizza was 150^\circ\mathrm{C}. Four minutes after removing it from the oven, the temperature of the pizza was 90^\circ\mathrm{C}. How long will it take to cool off to a temperature of 40^\circ\mathrm{C} if the room temperature is 20^\circ\mathrm{C}?
Solution
Given
  • T(0) = 150^\circ\mathrm{C}
  • T(4) = 90^\circ\mathrm{C}
  • T_s = 20^\circ\mathrm{C} room temperature
To find
  • t when T=40^\circ\mathrm{C}
  1. Model. Newton's law of cooling, with t in minutes.
    \dfrac{dT}{dt}=k\left(T-20\right)
  2. Separate, integrate and take antilog.
    \dfrac{dT}{T-20}=k\,dt\quad\Longrightarrow\quad \ln(T-20)=kt+\ln c\quad\Longrightarrow\quad T-20=c\,e^{kt}
  3. Use T(0)=150: c=130.
    T=20+130e^{kt}
  4. Use T(4)=90 to find k.
    90=20+130e^{4k}\quad\Longrightarrow\quad e^{4k}=\dfrac{70}{130}=\dfrac{7}{13}\quad\Longrightarrow\quad k=\dfrac14\ln\dfrac{7}{13}=-0.1548
  5. Put T=40 and solve for t.
    40=20+130e^{kt}\quad\Longrightarrow\quad e^{kt}=\dfrac{2}{13}\quad\Longrightarrow\quad t=\dfrac{\ln(2/13)}{k}=\dfrac{4\ln(2/13)}{\ln(7/13)}
  6. Evaluate.
    t=\dfrac{4(-1.8718)}{-0.6190}=12.1\ \text{min}
t in minutes, T in ^\circ\mathrm{C}. The pizza cools towards room temperature 20^\circ\mathrm{C} (dashed line).

So the pizza reaches 40^\circ\mathrm{C} about 12.1 minutes after it is taken out of the oven, that is about 8.1 minutes after the 90^\circ\mathrm{C} reading.

t\approx12.1\ \text{minutes}
3
In a culture, the rate of growth of bacteria is proportional to the population present. If the population of bacteria becomes four times in two days, how much would the population be after ten days at the same rate, if the initial population was 20?
Solution
Given
  • P(0) = 20
  • P(2) = 4\times20=80 four times in two days
To find
  • P(10)
  1. Model. Let P(t) be the number of bacteria after t days.
    \dfrac{dP}{dt}=kP\quad\Longrightarrow\quad P=P(0)\,e^{kt}=20e^{kt}
  2. Use P(2)=80.
    80=20e^{2k}\quad\Longrightarrow\quad e^{2k}=4\quad\Longrightarrow\quad k=\dfrac12\ln4=\ln2=0.693\ \text{per day}
  3. After 10 days: e^{10k}=\left(e^{2k}\right)^{5}=4^{5}.
    P(10)=20\times4^{5}=20\times1024=20\,480
P(10)=20\,480\ \text{bacteria}
4
Most radioactive substances disintegrate at a rate proportional to the amount present. If the amount of a radioactive substance is 50 grams and its half-life is 1000 years, find the amount of substance present after 800 years.
Solution
Given
  • A(0) = 50\,\mathrm{g}
  • T_{1/2} = 1000\ \text{years} so A(1000)=25\,\mathrm{g}
To find
  • A(800)
  1. Model. Let A(t) be the amount in grams after t years.
    \dfrac{dA}{dt}=kA\quad\Longrightarrow\quad A=50e^{kt}
  2. Half-life: A(1000)=25.
    \begin{aligned}25&=50e^{1000k}\\ \Longrightarrow\ e^{1000k}&=\dfrac12\\ \Longrightarrow\ k&=-\dfrac{\ln2}{1000}\\ &=-0.000693\ \text{per year}\end{aligned}
  3. After 800 years: e^{800k}=\left(e^{1000k}\right)^{0.8}=\left(\tfrac12\right)^{0.8}.
    A(800)=50\left(\dfrac12\right)^{0.8}=50\times0.5743=28.72\,\mathrm{g}
t in years, A in grams. Half of the substance is left after 1000 years and a quarter after 2000 years.
A(800)\approx28.7\,\mathrm{g}
5
A thermometer showing a room temperature of 80^\circ\mathrm{F} is placed on a block of ice with a temperature of 30^\circ\mathrm{F}. After one minute the temperature of the thermometer is 40^\circ\mathrm{F}. How long will it take for the thermometer to have a temperature of 70^\circ\mathrm{F}?
Solution
Given
  • T(0) = 80^\circ\mathrm{F}
  • T_s = 30^\circ\mathrm{F} block of ice
  • T(1) = 40^\circ\mathrm{F}
To find
  • t when T=70^\circ\mathrm{F}
  1. Model. Newton's law of cooling, t in minutes. As in Q2 the solution is
    T=30+(80-30)e^{kt}=30+50e^{kt}
  2. Use T(1)=40.
    40=30+50e^{k}\quad\Longrightarrow\quad e^{k}=\dfrac15\quad\Longrightarrow\quad k=-\ln5=-1.609\ \text{per minute}
  3. Put T=70.
    70=30+50e^{kt}\quad\Longrightarrow\quad e^{kt}=\dfrac45
  4. Take logarithms and solve for t.
    t=\dfrac{\ln(4/5)}{-\ln5}=\dfrac{\ln1.25}{\ln5}=\dfrac{0.2231}{1.6094}=0.139\ \text{min}
  5. Convert to seconds.
    t=0.139\times60\approx8.3\,\mathrm{s}
t in minutes, T in ^\circ\mathrm{F}. The reading falls quickly from 80^\circ\mathrm{F} towards the ice temperature 30^\circ\mathrm{F} (dashed line); it passes 70^\circ\mathrm{F} after about 8 seconds.
t\approx0.139\ \text{min}\approx8.3\ \text{s}

Note. 70^\circ\mathrm{F} lies between the starting reading 80^\circ\mathrm{F} and the one-minute reading 40^\circ\mathrm{F}, so the thermometer passes 70^\circ\mathrm{F} during the first minute. The small answer is therefore expected.

6
A ball is thrown upward with a velocity of 40\,\mathrm{m\,s^{-1}}. Develop a differential equation representing the flow phenomenon and find the velocity of the ball after 1 second. Also find the maximum height attained by the ball. Neglect the air resistance.
Solution
Given
  • v(0) = 40\,\mathrm{m\,s^{-1}} upward
  • g = 9.8\,\mathrm{m\,s^{-2}} downward
To find
  • v(1)
  • maximum height H
  1. DE. Take upward as positive. Gravity reduces the velocity at the constant rate g.
    \dfrac{dv}{dt}=-g\qquad\text{or}\qquad dv=-g\,dt
  2. Integrate and use v(0)=40.
    v=-gt+c_1,\quad c_1=40\quad\Longrightarrow\quad v=40-9.8t\qquad(\text{i})
  3. Velocity after 1 s.
    v(1)=40-9.8=30.2\,\mathrm{m\,s^{-1}}\ \ (\text{upward})
  4. Height. \dfrac{ds}{dt}=40-9.8t. Integrate, with s(0)=0.
    s=40t-4.9t^{2}\qquad(\text{ii})
  5. At the maximum height the ball stops for an instant: v=0 in (i).
    0=40-9.8t\quad\Longrightarrow\quad t=\dfrac{40}{9.8}=4.08\,\mathrm{s}
  6. Substitute t=4.08\,\mathrm{s} in (ii).
    H=40(4.082)-4.9(4.082)^{2}=163.3-81.6=81.6\,\mathrm{m}
t in seconds, s (height) in metres. The ball rises for 4.08 s, reaches 81.6 m, and falls back.
v(1)=30.2\,\mathrm{m\,s^{-1}},\qquad H\approx81.6\,\mathrm{m}

Review exercisep. 139

Mixed revision of the whole unit: MCQs, order and degree, verifying solutions, and solving separable and homogeneous equations.

1 (i)2026 Annual I, Old syllabus (Curriculum 2006), MCQ 16, old syllabus
The order of the differential equation x\dfrac{d^{3}y}{dx^{3}}-2\left(\dfrac{dy}{dx}\right)^{4}+y=0 is:
  1. A1
  2. B2
  3. C3Answer
  4. D4
Why

The highest derivative present is \dfrac{d^{3}y}{dx^{3}}, so the order is 3.

Option (d) 4 is the power of \dfrac{dy}{dx}. That power decides neither the order nor the degree (the degree here is 1, the power of \dfrac{d^{3}y}{dx^{3}}).

1 (ii)
The degree of the differential equation \dfrac{d^{2}y}{dx^{2}}+9y^{3}=\sin x is:
  1. A0
  2. B1Answer
  3. C2
  4. D3
Why

The highest derivative \dfrac{d^{2}y}{dx^{2}} appears to the power 1, so the degree is 1.

The cube in 9y^{3} is on y, not on a derivative, so (d) is wrong; (c) 2 is the order, not the degree.

1 (iii)
y=8 is a solution of the differential equation:
  1. A\dfrac{dy}{dx}+8y=32
  2. B\dfrac{dy}{dx}+6y=32
  3. C\dfrac{dy}{dx}+5y=32
  4. D\dfrac{dy}{dx}+4y=32Answer
Why

For the constant function y=8, \dfrac{dy}{dx}=0. The left side becomes k(8), which must equal 32, so k=4.

Check (d): 0+4(8)=32. The other options give 64, 48 and 40.

1 (iv)
f(x,y)=\dfrac{x^{3}-y^{3}}{x-y} is a homogeneous function of degree:
  1. A1
  2. B2Answer
  3. C3
  4. D4
Why

f(tx,ty)=\dfrac{t^{3}\left(x^{3}-y^{3}\right)}{t(x-y)}=t^{2}f(x,y), so the degree is 2.

Quick rule: degree of numerator minus degree of denominator, 3-1=2. Indeed f=x^{2}+xy+y^{2} for x\neq y. Option (c) forgets the denominator.

1 (v)
The solution of the differential equation dy=dx is:
  1. Ay=x+cAnswer
  2. By=x^{2}+c
  3. Cy^{2}=x^{2}+c
  4. Dy^{2}=x+c
Why

Integrate both sides: \int dy=\int dx gives y=x+c.

Check: \dfrac{d}{dx}(x+c)=1, so dy=dx.

1 (vi)
The number of arbitrary constants present in the general solution of a differential equation of first order is:
  1. A1Answer
  2. B2
  3. C3
  4. D0
Why

The general solution contains as many arbitrary constants as the order of the DE. A first order DE is solved by one integration, which gives exactly one constant.

1 (vii)
The differential equation \dfrac{dy}{dx}=e^{x+y} has solution:
  1. Ae^{-x-y}=c
  2. Be^{-x}+e^{y}=c
  3. Ce^{x}+e^{y}=c
  4. De^{x}+e^{-y}=cAnswer
Why

Write e^{x+y}=e^{x}e^{y} and separate: e^{-y}\,dy=e^{x}\,dx.

Integrate: -e^{-y}=e^{x}+c_1, so e^{x}+e^{-y}=c with c=-c_1.

1 (viii)
The general solution of y^{2}\,dy-x^{2}\,dx=0 is:
  1. Ax^{2}-y^{2}=c
  2. Bx^{3}+y^{3}=c
  3. Cx^{3}-y^{3}=cAnswer
  4. Dx^{2}+y^{2}=c
Why

The variables are already separated: y^{2}\,dy=x^{2}\,dx. Integrate: \dfrac{y^{3}}{3}=\dfrac{x^{3}}{3}+c_1.

Multiply by 3: x^{3}-y^{3}=c with c=-3c_1. Options (a) and (d) forget that \int x^{2}dx raises the power to 3.

1 (ix)
The solution of \cos x\sin y\,dx+\sin x\cos y\,dy=0 is:
  1. A\sin x\cos y=c
  2. B\cos x\cos y=c
  3. C\cos x\sin y=c
  4. D\sin x\sin y=cAnswer
Why

Divide by \sin x\sin y: \cot x\,dx+\cot y\,dy=0. Integrate: \ln\sin x+\ln\sin y=\ln c, so \sin x\sin y=c.

Check: the left side of the DE is exactly d(\sin x\sin y) by the product rule.

1 (x)
Which of the following cannot be the order of a differential equation?
  1. A-1Answer
  2. B1
  3. C10
  4. D100
Why

The order is the order of a derivative that appears, so it is always a positive integer (1, 2, 3, ...). -1 is impossible; 10 and 100 are unusual but possible.

2
Find the order and degree of the differential equations.
Solution
(i) x\left(\dfrac{dy}{dx}\right)^{2}+2\sqrt{xy}\,\dfrac{dy}{dx}+y=0
  1. The only derivative is \dfrac{dy}{dx} (first order).
  2. The equation is already a polynomial in \dfrac{dy}{dx}. The root \sqrt{xy} contains no derivative, so it is just a coefficient and need not be removed.
  3. The highest power of \dfrac{dy}{dx} is 2.
\text{order }1,\quad\text{degree }2
(ii) \dfrac{dy}{dx}=\sqrt{1+\left(\dfrac{d^{2}y}{dx^{2}}\right)^{4}}
  1. A derivative is under the square root, so square both sides first.
    \left(\dfrac{dy}{dx}\right)^{2}=1+\left(\dfrac{d^{2}y}{dx^{2}}\right)^{4}
  2. The highest derivative is \dfrac{d^{2}y}{dx^{2}} (second order), and its power is 4.
\text{order }2,\quad\text{degree }4

Note. In part (ii) the book prints the second derivative as \dfrac{d^{2}y}{d^{2}x}; it should be \dfrac{d^{2}y}{dx^{2}}.

3
Verify that the indicated function is a solution of the given differential equation.
Solution
(i) 2\dfrac{dy}{dx}+y=0\ ;\quad y=e^{-x/2}
  1. Differentiate.
    \dfrac{dy}{dx}=-\dfrac12e^{-x/2}
  2. Substitute in the left side.
    2\left(-\dfrac12e^{-x/2}\right)+e^{-x/2}=-e^{-x/2}+e^{-x/2}=0
  3. The left side equals the right side, so y=e^{-x/2} is a solution.
2\dfrac{dy}{dx}+y=0\ \checkmark
(ii) \dfrac{dy}{dx}+20y=24\ ;\quad y=\dfrac65-\dfrac65e^{-20x}
  1. Differentiate: \dfrac{d}{dx}e^{-20x}=-20e^{-20x}.
    \dfrac{dy}{dx}=-\dfrac65\left(-20e^{-20x}\right)=24e^{-20x}
  2. Substitute in the left side.
    24e^{-20x}+20\left(\dfrac65-\dfrac65e^{-20x}\right)=24e^{-20x}+24-24e^{-20x}=24
  3. The left side equals 24, so the function is a solution.
\dfrac{dy}{dx}+20y=24\ \checkmark
(iii) \dfrac{dy}{dx}=25+y^{2}\ ;\quad y=5\tan5x
  1. Differentiate with the chain rule.
    \dfrac{dy}{dx}=25\sec^{2}5x
  2. Use \sec^{2}\theta=1+\tan^{2}\theta.
    \dfrac{dy}{dx}=25+25\tan^{2}5x=25+(5\tan5x)^{2}=25+y^{2}
  3. So the function satisfies the DE wherever \tan5x is defined.
\dfrac{dy}{dx}=25+y^{2}\ \checkmark
(iv) x^{2}\,dy+2xy\,dx=0\ ;\quad y=\dfrac{1}{x^{2}}
  1. Divide by dx.
    x^{2}\dfrac{dy}{dx}+2xy=0
  2. Differentiate y=x^{-2}.
    \dfrac{dy}{dx}=-2x^{-3}=-\dfrac{2}{x^{3}}
  3. Substitute in the left side.
    x^{2}\left(-\dfrac{2}{x^{3}}\right)+2x\cdot\dfrac{1}{x^{2}}=-\dfrac2x+\dfrac2x=0
  4. True for every x\neq0, so the function is a solution.
x^{2}\,dy+2xy\,dx=0\ \checkmark
42026 Annual I, Set B, Q.6
Solve the differential equations.
Solution
(i) \dfrac{dy}{dx}=x\ln x
  1. Separate: dy=x\ln x\,dx, then integrate by parts with u=\ln x, dv=x\,dx (so du=\dfrac{dx}{x}, v=\dfrac{x^{2}}{2}).
    y=\dfrac{x^{2}}{2}\ln x-\int\dfrac{x^{2}}{2}\cdot\dfrac1x\,dx
  2. Finish the integral.
    y=\dfrac{x^{2}}{2}\ln x-\dfrac{x^{2}}{4}+c
y=\dfrac{x^{2}}{2}\ln x-\dfrac{x^{2}}{4}+c
(ii) (y+1)\dfrac{dy}{dx}+x\sin x
  1. Take the equation as (y+1)\dfrac{dy}{dx}+x\sin x=0 and separate the variables.
    (y+1)\,dy=-x\sin x\,dx
  2. Integrate the right side by parts: \int x\sin x\,dx=-x\cos x+\int\cos x\,dx=-x\cos x+\sin x.
    \int(y+1)\,dy=-\left(-x\cos x+\sin x\right)+c
  3. Write the result.
    \dfrac{(y+1)^{2}}{2}=x\cos x-\sin x+c
\dfrac{(y+1)^{2}}{2}=x\cos x-\sin x+c
(iii) \dfrac{dy}{dx}=\dfrac{x^{2}+y^{2}}{xy}\ ;\quad y(1)=2
  1. Numerator and denominator have degree 2, so the equation is homogeneous. Put y=ux.
    u+x\dfrac{du}{dx}=\dfrac{x^{2}+u^{2}x^{2}}{x\cdot ux}=\dfrac{1+u^{2}}{u}=\dfrac1u+u
  2. Subtract u and separate.
    x\dfrac{du}{dx}=\dfrac1u\quad\Longrightarrow\quad u\,du=\dfrac{dx}{x}
  3. Integrate.
    \dfrac{u^{2}}{2}=\ln|x|+c
  4. Replace u by \dfrac{y}{x} and multiply by 2x^{2}.
    y^{2}=2x^{2}\left(\ln|x|+c\right)
  5. Use y(1)=2: 4=2(0+c).
    c=2
  6. Substitute c. Since y(1)=2>0, take the positive root for the explicit form.
    y^{2}=2x^{2}\left(\ln x+2\right)\quad\Longrightarrow\quad y=x\sqrt{2\ln x+4}
The solution through (1,2) exists for x>e^{-2}, where 2\ln x+4>0.
y^{2}=2x^{2}\left(\ln x+2\right)

FBISE 2026 questions

Questions from the FBISE HSSC-II 2026 (1st Annual) papers that belong to this unit, answered the way an examiner expects.

2026 Annual I, Set B, MCQ 41 mark
The general solution of \dfrac{dy}{dx}=\dfrac{x}{y} is:
  1. Ay^{2}=x^{2}+CAnswer
  2. By^{2}=-x^{2}+C
  3. Cy^{2}=2x^{2}+C
  4. Dy=x+C
Why

Separate the variables: y\,dy=x\,dx. Integrate: \dfrac{y^{2}}{2}=\dfrac{x^{2}}{2}+c, so y^{2}=x^{2}+C with C=2c.

Option (B) comes from a sign slip (it solves \dfrac{dy}{dx}=-\dfrac{x}{y}), and (C) from integrating x as x^{2} without the factor \tfrac12.

2026 Annual I, Set B, Q.2 (i) OR4 marks
The population of a city grows at a rate proportional to its current population. In 2020, the population was 50,000. In 2025, it became 60,000. (a) Find the growth constant k. (b) Predict the population in 2030.
Solution
Given
  • P(0) = 50\,000 in 2020, so t=0
  • P(5) = 60\,000 in 2025
To find
  • (a) k
  • (b) P(10), the population in 2030
Model and general solution
  1. "Rate proportional to the current population": with t in years after 2020,
    \dfrac{dP}{dt}=kP
  2. Separate the variables and integrate.
    \dfrac{dP}{P}=k\,dt\quad\Longrightarrow\quad \ln P=kt+\ln c\quad\Longrightarrow\quad P=c\,e^{kt}
  3. At t=0, P=50\,000, so c=50\,000.
    P=50\,000\,e^{kt}
(a) Growth constant
  1. Use P(5)=60\,000.
    60\,000=50\,000\,e^{5k}\quad\Longrightarrow\quad e^{5k}=\dfrac{6}{5}=1.2
  2. Take natural logarithms.
    k=\dfrac{1}{5}\ln1.2=\dfrac{0.18232}{5}=0.0365\ \text{per year}
(b) Population in 2030
  1. In 2030, t=10. Use e^{10k}=\left(e^{5k}\right)^{2}=1.2^{2}.
    P(10)=50\,000\times(1.2)^{2}=50\,000\times1.44=72\,000
t in years after 2020, P in thousands. Equal time steps multiply the population by the same factor: 1.2 every 5 years.
k=\tfrac15\ln1.2\approx0.0365\ \text{per year};\qquad P(2030)=72\,000

Note. Same method as Exercise 4.4 Q1 (Malthus model) with different data.

2026 Annual I, Set B, Q.2 (iv)4 marks
Solve the separable differential equation: \dfrac{dy}{dx}=\dfrac{y+1}{x-2}
Solution
  1. Separate the variables: divide by (y+1) and multiply by dx.
    \dfrac{dy}{y+1}=\dfrac{dx}{x-2}
  2. Integrate both sides; write the constant as \ln|c|.
    \ln|y+1|=\ln|x-2|+\ln|c|
  3. Combine the logarithms and take antilog.
    \ln|y+1|=\ln\left|c(x-2)\right|\quad\Longrightarrow\quad y+1=c(x-2)
  4. Check. From y=c(x-2)-1: \dfrac{dy}{dx}=c, and \dfrac{y+1}{x-2}=\dfrac{c(x-2)}{x-2}=c. Both sides agree.

Dividing by y+1 assumed y\neq-1; the constant solution y=-1 is included by c=0.

The solutions y+1=c(x-2) are straight lines through (2,-1).
y+1=c\,(x-2)
2026 Annual I, Set B, Q.68 marks
Find the general solution of the homogeneous differential equation \dfrac{dy}{dx}=\dfrac{x^{2}+y^{2}}{xy}.
Solution
Step 1: Show that the equation is homogeneous

Let f(x,y)=\dfrac{x^{2}+y^{2}}{xy}. Then f(tx,ty)=\dfrac{t^{2}\left(x^{2}+y^{2}\right)}{t^{2}xy}=t^{0}f(x,y). Numerator and denominator are both of degree 2, so f is homogeneous of degree 0 and the DE is homogeneous. It can be solved with y=ux.

Step 2: Substitute y = ux
  1. Put y=ux, where u is a function of x. By the product rule:
    \dfrac{dy}{dx}=u+x\dfrac{du}{dx}
  2. Substitute in the DE and cancel x^{2}.
    u+x\dfrac{du}{dx}=\dfrac{x^{2}+u^{2}x^{2}}{x\cdot ux}=\dfrac{1+u^{2}}{u}=\dfrac{1}{u}+u
Step 3: Separate the variables
  1. Subtract u from both sides. The u terms cancel.
    x\dfrac{du}{dx}=\dfrac{1}{u}
  2. Multiply by u\,dx and divide by x.
    u\,du=\dfrac{dx}{x}
Step 4: Integrate
  1. Integrate both sides, with one constant c.
    \int u\,du=\int\dfrac{dx}{x}\quad\Longrightarrow\quad \dfrac{u^{2}}{2}=\ln|x|+c
Step 5: Return to x and y
  1. Replace u by \dfrac{y}{x}.
    \dfrac{y^{2}}{2x^{2}}=\ln|x|+c
  2. Multiply by 2x^{2} and write C=2c.
    y^{2}=2x^{2}\ln|x|+Cx^{2}
Verification
  1. Differentiate y^{2}=2x^{2}\ln|x|+Cx^{2} implicitly.
    2y\dfrac{dy}{dx}=4x\ln|x|+2x+2Cx\quad\Longrightarrow\quad \dfrac{dy}{dx}=\dfrac{2x\ln|x|+x+Cx}{y}
  2. The right side of the DE gives the same value, because y^{2}=2x^{2}\ln|x|+Cx^{2}.
    \dfrac{x^{2}+y^{2}}{xy}=\dfrac{x^{2}+2x^{2}\ln|x|+Cx^{2}}{xy}=\dfrac{x+2x\ln|x|+Cx}{y}\ \checkmark
y^{2}=2x^{2}\ln|x|+Cx^{2}\qquad\left(\text{or } y^{2}=x^{2}\left(2\ln|x|+C\right)\right)

Note. Nearly the same as Review Exercise Q4 (iii), which adds the initial condition y(1) = 2.

2026 Annual I, Set D, MCQ 51 mark
The general solution of \dfrac{dy}{dx}=\dfrac{3x^{2}}{y} is:
  1. Ay^{2}=2x^{3}+cAnswer
  2. By=6x^{3}+c
  3. Cy^{2}=x^{3}+c
  4. Dy^{2}=3x^{3}+c
Why

Separate: y\,dy=3x^{2}\,dx. Integrate: \dfrac{y^{2}}{2}=x^{3}+c. Multiply by 2: y^{2}=2x^{3}+C.

Option (C) forgets to multiply x^{3} by 2, and (D) integrates 3x^{2} wrongly.

2026 Annual I, Set D, MCQ 91 mark
A population grows as P(t)=1000e^{0.03}. The population after 10 years is closest to:
  1. A2000
  2. B1000
  3. C1350Answer
  4. D1500
Why

The intended model is P(t)=1000e^{0.03t}. At t=10: P=1000e^{0.3}=1000\times1.3499\approx1350.

Option (A) 2000 would need e^{0.3}=2, and (B) 1000 is the starting population.

Note. The paper prints the exponent as e^{0.03} without t. Read literally, P=1000e^{0.03}\approx1030 for all t, which would make (B) closest; the intended answer is clearly (C) for P=1000e^{0.03t}.

2026 Annual I, Set D, Q.2 (i) OR4 marks
The population of a city grows at a rate proportional to its current population. In 2020, the population was 50,000. In 2025, it became 60,000. (a) Find the growth constant k. (b) Predict the population in 2030.
Solution
Given
  • P(0) = 50\,000 in 2020, so t=0
  • P(5) = 60\,000 in 2025
To find
  • (a) k
  • (b) P(10), the population in 2030
Model and general solution
  1. "Rate proportional to the current population": with t in years after 2020,
    \dfrac{dP}{dt}=kP
  2. Separate the variables and integrate.
    \dfrac{dP}{P}=k\,dt\quad\Longrightarrow\quad \ln P=kt+\ln c\quad\Longrightarrow\quad P=c\,e^{kt}
  3. At t=0, P=50\,000, so c=50\,000.
    P=50\,000\,e^{kt}
(a) Growth constant
  1. Use P(5)=60\,000.
    60\,000=50\,000\,e^{5k}\quad\Longrightarrow\quad e^{5k}=\dfrac{6}{5}=1.2
  2. Take natural logarithms.
    k=\dfrac{1}{5}\ln1.2=\dfrac{0.18232}{5}=0.0365\ \text{per year}
(b) Population in 2030
  1. In 2030, t=10. Use e^{10k}=\left(e^{5k}\right)^{2}=1.2^{2}.
    P(10)=50\,000\times(1.2)^{2}=50\,000\times1.44=72\,000
t in years after 2020, P in thousands. Equal time steps multiply the population by the same factor: 1.2 every 5 years.
k=\tfrac15\ln1.2\approx0.0365\ \text{per year};\qquad P(2030)=72\,000

Note. The same question appeared in Set B. Same method as Exercise 4.4 Q1 with different data.

2026 Annual I, Set D, Q.2 (iv)4 marks
Solve the separable differential equation: \dfrac{dy}{dx}=\dfrac{y-2}{x+6}
Solution
  1. Separate the variables: divide by (y-2) and multiply by dx.
    \dfrac{dy}{y-2}=\dfrac{dx}{x+6}
  2. Integrate both sides; write the constant as \ln|c|.
    \ln|y-2|=\ln|x+6|+\ln|c|
  3. Combine and take antilog.
    y-2=c\,(x+6)
  4. Check. From y=c(x+6)+2: \dfrac{dy}{dx}=c and \dfrac{y-2}{x+6}=\dfrac{c(x+6)}{x+6}=c. Both sides agree.

The solutions are straight lines through the point (-6,2). The constant solution y=2 (lost when dividing by y-2) is included by c=0.

y-2=c\,(x+6)
2026 Annual I, Set D, Q.68 marks
Find the general solution of the homogeneous differential equation: \dfrac{dy}{dx}=\dfrac{x^{2}+3y^{2}}{xy}
Solution
Step 1: Show that the equation is homogeneous

f(x,y)=\dfrac{x^{2}+3y^{2}}{xy} gives f(tx,ty)=\dfrac{t^{2}\left(x^{2}+3y^{2}\right)}{t^{2}xy}=t^{0}f(x,y). Numerator and denominator are both of degree 2, so the DE is homogeneous.

Step 2: Substitute y = ux
  1. Put y=ux and \dfrac{dy}{dx}=u+x\dfrac{du}{dx}. Cancel x^{2}.
    u+x\dfrac{du}{dx}=\dfrac{x^{2}+3u^{2}x^{2}}{x\cdot ux}=\dfrac{1+3u^{2}}{u}
Step 3: Separate the variables
  1. Subtract u.
    x\dfrac{du}{dx}=\dfrac{1+3u^{2}-u^{2}}{u}=\dfrac{1+2u^{2}}{u}
  2. Separate.
    \dfrac{u\,du}{1+2u^{2}}=\dfrac{dx}{x}
Step 4: Integrate
  1. The derivative of 1+2u^{2} is 4u, so the left side is \dfrac14\ln\left(1+2u^{2}\right). Write the constant as \dfrac14\ln C.
    \dfrac14\ln\left(1+2u^{2}\right)=\ln|x|+\dfrac14\ln C
  2. Multiply by 4 and take antilog.
    \ln\left(1+2u^{2}\right)=\ln\left(Cx^{4}\right)\quad\Longrightarrow\quad 1+2u^{2}=Cx^{4}
Step 5: Return to x and y
  1. Replace u by \dfrac{y}{x} and multiply by x^{2}.
    1+\dfrac{2y^{2}}{x^{2}}=Cx^{4}\quad\Longrightarrow\quad x^{2}+2y^{2}=Cx^{6}
Verification
  1. Differentiate x^{2}+2y^{2}=Cx^{6} implicitly and use C=\dfrac{x^{2}+2y^{2}}{x^{6}}.
    2x+4y\dfrac{dy}{dx}=6Cx^{5}=\dfrac{6\left(x^{2}+2y^{2}\right)}{x}
  2. Solve for \dfrac{dy}{dx}: this is the given DE.
    4y\dfrac{dy}{dx}=\dfrac{6x^{2}+12y^{2}-2x^{2}}{x}=\dfrac{4\left(x^{2}+3y^{2}\right)}{x}\quad\Longrightarrow\quad\dfrac{dy}{dx}=\dfrac{x^{2}+3y^{2}}{xy}\ \checkmark
x^{2}+2y^{2}=Cx^{6}

Note. Nearly the same as Exercise 4.3 Q13, (x^{2}+2y^{2})dx=xy\,dy, which has coefficient 2 instead of 3 and an initial condition.

2026 Annual I, Old syllabus (Curriculum 2000), MCQ 17, old syllabus1 mark
The solution of the differential equation \dfrac{dy}{dx}=xy is:
  1. Ay=e^{x^{2}/2}+C
  2. By=Ce^{x}
  3. Cy=\dfrac{x^{2}}{2}+C
  4. Dy=Ce^{x^{2}/2}Answer
Why

Separate: \dfrac{dy}{y}=x\,dx. Integrate: \ln y=\dfrac{x^{2}}{2}+c. Take antilog: y=e^{c}e^{x^{2}/2}=Ce^{x^{2}/2}.

Option (A) is tempting, but the constant multiplies the exponential; it is not added to it.

2026 Annual I, Old syllabus (Curriculum 2000), Q.2 (iv) OR, old syllabus4 marks
Consider the differential equation \dfrac{dy}{dx}=\dfrac{y}{x}. Solve it using separation of variables.
Solution
  1. Separate the variables: divide by y and multiply by dx.
    \dfrac{dy}{y}=\dfrac{dx}{x}
  2. Integrate both sides; write the constant as \ln|c|.
    \ln|y|=\ln|x|+\ln|c|=\ln|cx|
  3. Take antilog.
    y=cx
  4. Check. \dfrac{dy}{dx}=c=\dfrac{cx}{x}=\dfrac{y}{x}.

The general solution is the family of straight lines through the origin (for x\neq0). The equation is also homogeneous, but separation is quicker.

Solutions y=cx: lines through the origin.
y=cx

Note. Nearly the same as Exercise 4.2 Q2, x\dfrac{dy}{dx}=4y (answer y=cx^{4}).

2026 Annual I, Old syllabus (Curriculum 2006), MCQ 16, old syllabus1 mark
Order of differential equation \dfrac{d^{3}y}{dx^{3}}+2\left(\dfrac{dy}{dx}\right)^{4}=y^{2} is:
  1. A1
  2. B2
  3. C3Answer
  4. D4
Why

The highest derivative is \dfrac{d^{3}y}{dx^{3}}, so the order is 3.

Option (D) 4 is the power of \dfrac{dy}{dx}, which is not the highest derivative. (The degree of this DE is 1.)

2026 Annual I, Old syllabus (Curriculum 2006), MCQ 17, old syllabus1 mark
A radioactive material decays at a rate proportional to its remaining mass M. The differential equation is:
  1. A\dfrac{dM}{dt}=-kMAnswer
  2. B\dfrac{dM}{dt}=kM
  3. C\dfrac{dM}{dt}=kt
  4. D\dfrac{dM}{dt}=-kt
Why

"Rate" is \dfrac{dM}{dt}, "proportional to M" means k\times M, and "decays" means the rate is negative: \dfrac{dM}{dt}=-kM with k>0.

Option (B) describes growth, and (C), (D) make the rate depend on time instead of on the mass.

2026 Annual I, Old syllabus (Curriculum 2006), Q.2 (xi), old syllabus4 marks
Solve the separable differential equation \dfrac{dy}{dx}=\dfrac{y}{x+1}.
Solution
  1. Separate the variables: divide by y and multiply by dx.
    \dfrac{dy}{y}=\dfrac{dx}{x+1}
  2. Integrate both sides; write the constant as \ln|c|.
    \ln|y|=\ln|x+1|+\ln|c|
  3. Combine and take antilog.
    \ln|y|=\ln\left|c(x+1)\right|\quad\Longrightarrow\quad y=c(x+1)
  4. Check. \dfrac{dy}{dx}=c and \dfrac{y}{x+1}=\dfrac{c(x+1)}{x+1}=c.

This is the book's Example 9, (1+x)\,dy-y\,dx=0, written in another form. The solutions are straight lines through (-1,0).

y=c\,(x+1)
2026 Annual I, Old syllabus (Curriculum 2006), Q.5, old syllabus8 marks
Solve the homogeneous differential equation \dfrac{dy}{dx}=\dfrac{x+y}{x-y}.
Solution
Step 1: Show that the equation is homogeneous

f(x,y)=\dfrac{x+y}{x-y} gives f(tx,ty)=\dfrac{t(x+y)}{t(x-y)}=t^{0}f(x,y): numerator and denominator are both of degree 1, so the DE is homogeneous.

Step 2: Substitute y = ux
  1. Put y=ux, \dfrac{dy}{dx}=u+x\dfrac{du}{dx}, and cancel x.
    u+x\dfrac{du}{dx}=\dfrac{x+ux}{x-ux}=\dfrac{1+u}{1-u}
Step 3: Separate the variables
  1. Subtract u.
    x\dfrac{du}{dx}=\dfrac{1+u-u(1-u)}{1-u}=\dfrac{1+u^{2}}{1-u}
  2. Separate.
    \dfrac{1-u}{1+u^{2}}\,du=\dfrac{dx}{x}
Step 4: Integrate
  1. Split the fraction: \dfrac{1-u}{1+u^{2}}=\dfrac{1}{1+u^{2}}-\dfrac{u}{1+u^{2}}. Use \int\dfrac{du}{1+u^{2}}=\tan^{-1}u and \int\dfrac{u\,du}{1+u^{2}}=\dfrac12\ln\left(1+u^{2}\right).
    \tan^{-1}u-\dfrac12\ln\left(1+u^{2}\right)=\ln|x|+c
Step 5: Return to x and y
  1. Replace u by \dfrac{y}{x}. Since \dfrac12\ln\dfrac{x^{2}+y^{2}}{x^{2}}=\dfrac12\ln\left(x^{2}+y^{2}\right)-\ln|x|, the \ln|x| terms cancel.
    \tan^{-1}\dfrac{y}{x}-\dfrac12\ln\left(x^{2}+y^{2}\right)+\ln|x|=\ln|x|+c
  2. Simplify.
    \tan^{-1}\dfrac{y}{x}=\ln\sqrt{x^{2}+y^{2}}+c
Verification
  1. Differentiate both sides implicitly: \dfrac{d}{dx}\tan^{-1}\dfrac{y}{x}=\dfrac{xy'-y}{x^{2}+y^{2}} and \dfrac{d}{dx}\ln\sqrt{x^{2}+y^{2}}=\dfrac{x+yy'}{x^{2}+y^{2}}.
    xy'-y=x+yy'\quad\Longrightarrow\quad y'(x-y)=x+y\quad\Longrightarrow\quad y'=\dfrac{x+y}{x-y}\ \checkmark
In polar coordinates (x=r\cos\theta, y=r\sin\theta) the solutions are the spirals r=Ae^{\theta}; the curve shown has A=1.
\tan^{-1}\dfrac{y}{x}=\ln\sqrt{x^{2}+y^{2}}+c

Exam focus

  1. Find the order and degree of a DE, including one with a derivative under a root such as \dfrac{dy}{dx}=\sqrt{1+\left(\dfrac{d^{2}y}{dx^{2}}\right)^{2}}. (short)
    A regular MCQ and short question; the 2026 Old syllabus paper asked for the order in MCQ 16.
    Go to answer
  2. Form the DE by eliminating the arbitrary constants from y=a\cos x+b\sin x or y=bx^{2}+2ax. (problem)
    Tests differentiation and elimination; Examples 6 and Exercise 4.1 Q2.
    Go to answer
  3. Verify that a given function is a solution of a DE, for example y=5\tan5x for y'=25+y^{2}. (problem)
    Quick marks by substitution; Exercise 4.1 Q3 and Review Q3.
    Go to answer
  4. Solve a separable DE such as \dfrac{dy}{dx}=\dfrac{y+1}{x-2}. (problem)
    Asked for 4 marks in both 2026 new-syllabus sets (Q.2 (iv)) and in both old-syllabus papers (Q.2 (iv) OR, Q.2 (xi)).
    Go to answer
  5. Solve an initial value problem such as \dfrac{dy}{dx}=y^{2}+4, y(0)=-2. (problem)
    Combines separation with finding the constant and choosing the correct branch; Exercise 4.2 Q9 to Q15.
    Go to answer
  6. Test whether f(x,y) is homogeneous and find its degree. (short)
    Exercise 4.3 Q1 to Q3, Example 13 and Review MCQ (iv).
    Go to answer
  7. Solve a homogeneous DE by y=ux, for example \dfrac{dy}{dx}=\dfrac{x^{2}+y^{2}}{xy}. (problem)
    The 8 mark Section C question in 2026 Set B, Set D and the Old syllabus paper.
    Go to answer
  8. Population growth: find k from two readings and predict a future value. (problem)
    2026 Sets B and D Q.2 (i) OR; Exercise 4.4 Q1 and Q3; Set D MCQ 9.
    Go to answer
  9. Newton's law of cooling: find the temperature as a function of time, or the time to reach a given temperature. (problem)
    Example 17 and Exercise 4.4 Q2 and Q5.
    Go to answer
  10. Radioactive decay: write the DE and use the half-life. (problem)
    Exercise 4.4 Q4; the 2026 Old syllabus MCQ 17 asked for \dfrac{dM}{dt}=-kM.
    Go to answer
  11. Define general and particular solutions and state how many arbitrary constants a general solution has. (short)
    Definitions for short questions and MCQs (Review MCQ (vi)).
    Go to answer

Mark this unit as done to fill the Class 12 Mathematics ring.

Next, unit 5
Kinematics of motion in a straight line

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