Differential Equations: Class 12 Maths · Chapter 9

Equations about rates of change, and the functions that satisfy them

Mathematics Part 2 · Companion notes, simulations and practice · Official NCERT chapter PDF ↗

Published 12 Oct 2026 · Updated 12 Oct 2026

100 min
1 simulation
47 practice questions
23 MCQs

What you will learn in Differential Equations

Recognise a differential equation and find its order and, when it is defined, its degree
Check by substitution whether a given function, explicit or implicit, solves a differential equation
Tell a general solution from a particular one, and use given conditions to fix the arbitrary constants
Solve first order, first degree equations whose variables separate
Recognise homogeneous equations and solve them with the substitution y = vx or x = vy
Solve first order linear equations with an integrating factor, in x or in y
Turn a slope condition, a growth law or a cooling law into a differential equation and solve it

Original companion text aligned to NCERT Mathematics Class 12, Chapter 9, pp. 300–337. Not affiliated with or endorsed by NCERT.

9.1 Introduction

A glass of chai poured at 90 °C does not cool at a steady pace. In the first minute it loses about six degrees; ten minutes later, at around 50 °C, it loses only two degrees a minute. What we can observe directly is the rate: the further the chai is above room temperature, the faster it cools (Fig. 9.1).

Write $T$ for the temperature after $t$ minutes in a room at 30 °C. The observation becomes $\frac{d T}{d t} = - k (T - 30),$ where $k$ is a positive constant. The unknown here is not a number but a whole function, $T (t),$ and the equation says only how fast that function changes. An equation like this is a differential equation.

3/3The slope is proportional to the gap above the room: $\frac{d T}{d t} = - k (T - 30) .$

Fig. 9.1The hotter the chai, the steeper the graph: the rate of cooling depends on the temperature itself.

A graph of temperature T in degrees Celsius against time in minutes, from 0 to 30 minutes. An orange curve starts at 90 degrees and bends down towards a dashed horizontal line at 30 degrees, labelled room. In the first step a steep tangent is drawn where the curve starts, labelled 90 degrees: steep, cools fast. In the second step a second, dashed tangent touches the curve at 50 degrees, near 11 minutes; it is much flatter and labelled gentler slope. In the third step both tangents stay and the rule dT by dt equals minus k times T minus 30 appears.

Equations of this kind turn up wherever something changes at a rate we can describe: carbon-14 decaying in old wood (which lets archaeologists date it), a rumour or an infection spreading through a town, a capacitor charging, a loan whose interest is compounded continuously. You have already solved the simplest case. In the chapter on integrals, finding $y$ from $\frac{d y}{d x} = g (x)$ was just antidifferentiation. The new feature is that the rate may depend on the unknown itself, as it does for the chai.

This chapter first fixes the vocabulary: order, degree, and general and particular solutions. It then develops three exact methods for first order equations: separating the variables, a substitution for homogeneous equations, and an integrating factor for linear equations. At every stage we check an answer by putting it back into the equation it came from.

Quick check

1 of 2

Which of these are differential equations?

\frac{d y}{d x} = 3 x + y

x^{2} + y^{2} = 25

\frac{d^{2} s}{d t^{2}} = - 9.8

y = e^{2 x} + 1

Select all that apply

Show answer

Answer

(A) $\frac{d y}{d x} = 3 x + y$
(C) $\frac{d^{2} s}{d t^{2}} = - 9.8$

A and C contain derivatives of the unknown functions

y (x)

and

s (t) .

The circle

x^{2} + y^{2} = 25

and the formula

y = e^{2 x} + 1

contain no derivative; the second is in fact a solution of

\frac{d y}{d x} = 2 y - 2,

not an equation of that kind.

9.2 Basic Concepts

In school algebra the unknown is a number: $x^{2} - 5 x + 6 = 0$ holds for $x = 2$ and $x = 3$ and for nothing else. In $x \frac{d y}{d x} + 2 y = 0$ the unknown is a function $y (x),$ and the equation involves its derivative. The variable $x$ is the independent variable and $y,$ which depends on it, is the dependent variable.

Definition

Differential equation

An equation in which an unknown function appears together with at least one of its derivatives.

Example $\frac{d y}{d x} = x^{2} - y,$ where $y$ is an unknown function of $x .$

When only one independent variable is involved, as in every equation of this chapter, we have an ordinary differential equation. The temperature along a heated iron rod depends on both the position along the rod and the time, so the equation for it contains partial derivatives with respect to each; such partial differential equations belong to later courses. From here on, "differential equation" means an ordinary one.

For derivatives we use the short forms $y^{'} = \frac{d y}{d x},$ $y^{''} = \frac{d^{2} y}{d x^{2}}$ and $y^{'''} = \frac{d^{3} y}{d x^{3}} .$ Beyond the third derivative the dashes pile up, so the $n$ th derivative $\frac{d^{n} y}{d x^{n}}$ is written $y_{n}$ or $y^{(n)} .$

9.2.1 Order of a differential equation

Definition

Order

Count how many times the dependent variable has been differentiated in each term; the largest count anywhere in the equation is its order.

Example $y^{''} + 4 y = 0$ has order 2.

Only the number of times $y$ has been differentiated matters here. Powers and coefficients play no part:

$\frac{d y}{d x} = 3 x^{2} - y$ has order 1;
$\frac{d^{2} y}{d x^{2}} + 4 y = \sin x$ has order 2;
${(\frac{d y}{d x})}^{3} + \frac{d^{3} y}{d x^{3}} = x$ has order 3, even though the cube sits on the first derivative.

9.2.2 Degree of a differential equation

The degree comes with a condition. Each derivative ( $y^{'},$ $y^{''},$ …) has to enter the equation the way a variable enters a polynomial: through whole-number powers, sums and products, never inside a sine, a logarithm, an exponential or a root. Their coefficients, on the other hand, may be any functions of $x$ and $y .$

Definition

Degree

Look at the highest order derivative. If the equation is a polynomial in all its derivatives, the degree is the largest whole-number exponent carried by that derivative; if it is not, the equation has no degree.

Example $(y^{''})^{2} + y^{'} = x$ has order 2 and degree 2.

Read an equation in two moves (Fig. 9.2). First find the highest derivative: that gives the order. Then look at the power of that derivative alone: that gives the degree. Powers of lower derivatives are irrelevant, however large.

4/4The power 5 belongs to a lower derivative, so it does not decide the degree.

Fig. 9.2Order from the highest derivative; degree from the power of that same derivative.

Three boxes in a row hold the terms of an equation: the second derivative of y cubed, plus x times the first derivative to the fifth power, minus y, equals 0. In the first step a note says to find the highest derivative first. In the second step the first box is highlighted and an arrow points to the words order 2, second derivative. In the third step the cube on that box is circled and the words degree 3, its power, appear. In the fourth step the middle box gets a dashed outline and an arrow labelled power 5: ignore, lower derivative.

$(y^{'})^{2} + y^{'} \cos y = x ⁣ :$ order 1, degree 2. The $\cos y$ is harmless, because $y$ itself is not a derivative.
$y^{''} + e^{y^{'}} = 0 ⁣ :$ order 2, but $e^{y^{'}}$ is not a polynomial in $y^{'},$ so the degree is not defined.
$\sqrt{x} y^{'''} + x^{2} y = 0 ⁣ :$ order 3, degree 1. A root of $x$ is fine; a root of a derivative would not be.

Example 1NCERT Example 1 · p. 303

Order and degree of three equations

Find the order and the degree, if it is defined, of: (i) ${(\frac{d y}{d x})}^{2} - 4 x y = 0;$ (ii) $\frac{d^{2} y}{d x^{2}} + x^{2} {(\frac{d y}{d x})}^{3} + y = 0;$ (iii) $y^{'''} + x \sin (y^{''}) = 0.$

Try it yourself first. Then reveal the solution one step at a time.

Practice · 4 questions

Practice 9.1: questions with answers

1
Short answerEasyNCERT Exercise 9.1, Q1–Q5
Find the order and the degree of each equation. Type a number, or not defined when the degree does not exist.
1. $y^{'''} - e^{y^{''}} = 0 ⁣ :$ order , degree
  Show solution
  The highest derivative is $y^{'''},$ so the order is 3. Because $e^{y^{''}}$ is not a polynomial in $y^{''},$ the degree is not defined.
2. $3 y^{'} = 2 y ⁣ :$ order , degree
  Show solution
  Only $y^{'}$ occurs, to the first power: order 1, degree 1.
3. ${(\frac{d r}{d t})}^{3} - 2 r \frac{d^{2} r}{d t^{2}} = 0 ⁣ :$ order , degree
  Show solution
  Here $r$ is a function of $t .$ The highest derivative, $\frac{d^{2} r}{d t^{2}},$ appears to the first power: order 2, degree 1. The cube is on a lower derivative.
4. ${(\frac{d^{3} y}{d x^{3}})}^{2} + \sin (\frac{d^{2} y}{d x^{2}}) = 0 ⁣ :$ order , degree
  Show solution
  The order is 3. The sine of $y^{''}$ is not a polynomial in a derivative, so the degree is not defined, even though the highest derivative itself is only squared.
5. $\frac{d^{3} y}{d x^{3}} = e^{2 x} + \sin x ⁣ :$ order , degree
  Show solution
  Functions of $x$ alone do not affect the degree. Order 3, degree 1.
2
Short answerEasyNCERT Exercise 9.1, Q6–Q10
Find the order and the degree of each equation, as in Question 1.
1. $(y^{'''})^{2} + (y^{''})^{4} - (y^{'})^{5} + y^{3} = 0 ⁣ :$ order , degree
  Show solution
  The highest derivative $y^{'''}$ is squared: order 3, degree 2. The powers 4 and 5 belong to lower derivatives.
2. $y^{(4)} - 3 y^{''} + 5 y = 0 ⁣ :$ order , degree
  Show solution
  The fourth derivative appears to the first power: order 4, degree 1.
3. $y^{'} - 2 y = \cos x ⁣ :$ order , degree
  Show solution
  Order 1, degree 1; $\cos x$ involves no derivative.
4. $2 y^{''} + (y^{'})^{3} = x y ⁣ :$ order , degree
  Show solution
  The highest derivative $y^{''}$ appears to the first power: order 2, degree 1.
5. $y^{''} + e^{y} y^{'} = 0 ⁣ :$ order , degree
  Show solution
  The factor $e^{y}$ contains no derivative, so the equation is a polynomial in $y^{''}$ and $y^{'} ⁣ :$ order 2, degree 1.
Quick answerEasyNCERT Exercise 9.1, Q11
The degree of ${(\frac{d^{3} y}{d x^{3}})}^{2} - {(\frac{d y}{d x})}^{4} + e^{d y / d x} = 0$ is
2not defined43
Show answer
Answer (B) not defined
The term $e^{d y / d x}$ is not a polynomial in a derivative, so no degree exists, even though the highest derivative is only squared.
Quick answerEasyNCERT Exercise 9.1, Q12
The order of ${(\frac{d y}{d x})}^{4} + x^{2} \frac{d^{2} y}{d x^{2}} = 3 y$ is
412not defined
Show answer
Answer (C) 2
The highest derivative is $\frac{d^{2} y}{d x^{2}},$ so the order is 2. The power 4 is a power, not an order, and an order is always defined.

9.3 General and Particular Solutions of a Differential Equation

A solution of $x^{2} - 9 = 0$ is a number that makes the two sides equal. A solution of a differential equation is a function that does the same job: put it and its derivatives into the equation, and the equation turns into an identity, true for every $x$ in some interval.

Take the plain equation $\frac{d y}{d x} = x .$ Integrating, every function $y = \frac{x^{2}}{2} + C$ works, whatever constant $C$ we choose. Their graphs are one parabola slid up and down, and each of them is a solution curve (also called an integral curve) of the equation (Fig. 9.3).

4/4Only the member with $C = - 1$ passes through $(2, 1) .$

Fig. 9.3Every member of the family

y = \frac{x^{2}}{2} + C

follows the slope marks; a given point picks out one member.

A square grid of short line segments, the slope marks: at each point the segment has slope equal to the x-coordinate, flat on the y-axis and steeper further left or right. In the first step only the marks are shown. In the second step a blue parabola, y equals x squared over 2, runs along the marks. In the third step four more parabolas appear, shifted up and down, labelled C equals minus 2 to 2 at their lowest points. In the fourth step the parabola with C equals minus 1 is picked out with a dashed orange line through the marked point (2, 1).

The little segments show the slope that the equation demands at each point; the whole pattern is a slope field (or direction field). A solution curve is a curve that runs along these segments everywhere. Through a chosen point passes just one member of the family: for $(2, 1)$ we need $1 = 2 + C,$ so $C = - 1.$

Definition

General solution

The whole family of solutions, written with free constants, one for each order of the equation. It is also called the primitive.

Example $y = \frac{x^{2}}{2} + C$ for $\frac{d y}{d x} = x .$

Definition

Particular solution

One member of the family: every constant in the general solution has been given a definite value, usually to meet stated conditions.

Example $y = \frac{x^{2}}{2} - 1,$ the member through $(2, 1) .$

Why as many constants as the order? A second order equation such as $\frac{d^{2} y}{d x^{2}} = 0$ must be integrated twice: first $\frac{d y}{d x} = A,$ then $y = A x + B .$ Two integrations bring two arbitrary constants, and the family is every non-vertical straight line. To single out one line we need two facts, such as a point and a slope, or two points.

InteractiveSlope field explorer

A square graph covered with short slope marks for a chosen differential equation, with five tabs: dy by dx equals x; dy by dx equals y over 2; dy by dx equals 3 x squared over 2y; the homogeneous equation (2x minus y) dx plus (2y minus x) dy equals 0; and the linear equation dy by dx plus y equals sin x plus cos x. A draggable point P, also set by two sliders, fixes the arbitrary constant; the panel shows the slope at P, the value of C and the particular solution, drawn as a thick orange curve. A Show family button adds other members of the general solution in blue, and a Trace button moves a green dot with its tangent along the curve. Nothing moves until the learner acts.

Fig. 9.4Drag

P

(or use the sliders) and watch the solution through it change. Turn on the family, or press Trace to send a point along the curve with its tangent.

Try this

Walk along a slope field

You’ll need

a calculator
graph paper (optional)

Predict first

Start at

(0, 1)

and take steps of

0.25

x

along the slope marks of

\frac{d y}{d x} = y .

When you reach

x = 1,

will your

y

be below, at or above the true value

e \approx 2.72

belowexactly atabove

Start at $x = 0,$ $y = 1.$
The slope there is $\frac{d y}{d x} = y .$ Move right by $0.25$ and up by $0.25 \times slope .$
Repeat from the new point until $x = 1,$ writing each new $y$ in the table.
Halve the step to $0.125$ and compare your final $y$ with $e .$

Table 9.1Euler steps for

\frac{d y}{d x} = y

$x$	$y$ from the walk	true value $e^{x}$
0	1	1
0.25		1.284
0.5		1.649
0.75		2.117
1		2.718

Your entries are saved automatically.

Example 2NCERT Example 2 · p. 305

Checking a solution

Show that $y = 3 e^{2 x} - 1$ is a solution of $\frac{d y}{d x} = 2 y + 2.$

Try it yourself first. Then reveal the solution one step at a time.

Example 3NCERT Example 3 · p. 305

A two-constant family

Verify that $y = (A + B x) e^{x},$ where $A$ and $B$ are constants, solves $\frac{d^{2} y}{d x^{2}} - 2 \frac{d y}{d x} + y = 0.$ Then find the solution with $y = 2$ and $\frac{d y}{d x} = 1$ at $x = 0.$

Try it yourself first. Then reveal the solution one step at a time.

Practice · 4 questions

Practice 9.2: questions with answers

1
Short answerEasyNCERT Exercise 9.2, Q1–Q5
Check by substitution that each function satisfies the equation beside it.
1. $y = 2 e^{- x} + 3;$ $y^{''} + y^{'} = 0$
  Show solution
  $y^{'} = - 2 e^{- x}$ and $y^{''} = 2 e^{- x},$ so $y^{''} + y^{'} = 0.$
2. $y = x^{3} - 4 x + C;$ $y^{'} = 3 x^{2} - 4$
  Show solution
  Differentiating gives $y^{'} = 3 x^{2} - 4,$ whatever the value of $C .$
3. $y = \sin 2 x + C;$ $y^{'} - 2 \cos 2 x = 0$
  Show solution
  $y^{'} = 2 \cos 2 x,$ so $y^{'} - 2 \cos 2 x = 0.$
4. $y = (1 + x^{2})^{3 / 2};$ $(1 + x^{2}) y^{'} = 3 x y$
  Show solution
  By the chain rule $y^{'} = 3 x (1 + x^{2})^{1 / 2} .$ Multiplying by $1 + x^{2}$ gives $3 x (1 + x^{2})^{3 / 2},$ which is $3 x y .$
5. $y = B x^{2};$ $x y^{'} = 2 y$
  Show solution
  $y^{'} = 2 B x,$ so $x y^{'} = 2 B x^{2} = 2 y .$
2
Short answerMediumNCERT Exercise 9.2, Q6–Q10
Each relation below defines $y,$ some only implicitly. Check that it satisfies the equation beside it.
1. $y = x \cos x,$ $0 < x < π;$ $x y^{'} = y - x \sqrt{x^{2} - y^{2}}$
  Show solution
  $y^{'} = \cos x - x \sin x,$ so $x y^{'} = y - x^{2} \sin x .$
  Also $x^{2} - y^{2} = x^{2} \sin^{2} x .$ For $0 < x < π,$ $x \sin x > 0,$ so $x \sqrt{x^{2} - y^{2}} = x^{2} \sin x,$ as required.
2. $x e^{y} = y + C;$ $y^{'} = \frac{e^{y}}{1 - x e^{y}}$ (where $x e^{y} \neq 1$ )
  Show solution
  Differentiate: $e^{y} + x e^{y} y^{'} = y^{'} .$ Collect the $y^{'}$ terms: $y^{'} (1 - x e^{y}) = e^{y} .$
3. $y + \sin y = x;$ $(y \cos y - \sin y + x) y^{'} = y$
  Show solution
  Differentiating, $y^{'} (1 + \cos y) = 1.$
  On the curve, $x - \sin y = y,$ so the bracket equals $y \cos y + y = y (1 + \cos y) .$ The left side becomes $y (1 + \cos y) y^{'} = y .$
4. $x - y = \tan^{- 1} y;$ $(y^{2} + 2) y^{'} = 1 + y^{2}$
  Show solution
  Differentiate: $1 - y^{'} = \frac{y^{'}}{1 + y^{2}} .$ Multiply by $1 + y^{2} ⁣ :$ $1 + y^{2} = y^{'} (2 + y^{2}) .$
5. $y = \frac{b}{a} \sqrt{a^{2} - x^{2}},$ $- a < x < a;$ $a^{2} y y^{'} + b^{2} x = 0$
  Show solution
  $y^{'} = - \frac{b}{a} \cdot \frac{x}{\sqrt{a^{2} - x^{2}}},$ so $y y^{'} = - \frac{b^{2}}{a^{2}} x .$ Multiply by $a^{2}$ and rearrange.
Quick answerEasyNCERT Exercise 9.2, Q11
How many arbitrary constants does the general solution of a third order differential equation contain?
1302
Show answer
Answer (B) 3
One constant for each integration, and a third order equation needs three.
Quick answerEasyNCERT Exercise 9.2, Q12
How many arbitrary constants does a particular solution of a second order differential equation contain?
1230
Show answer
Answer (D) 0
A particular solution has all its constants fixed, so it contains none.

9.4 Methods of Solving First Order, First Degree Differential Equations

From now on every equation has order 1 and degree 1, so it can be written as $\frac{d y}{d x} = F (x, y),$ or, with the roles of the variables swapped, as $\frac{d x}{d y} = G (x, y) .$ No single trick solves all of these. We study three shapes that can be solved exactly, and the first job with any equation is to recognise its shape.

Compare

Three solvable shapes

Variables separable

$F (x, y) = g (x) h (y)$
move $y$ to $d y$ and $x$ to $d x$
integrate both sides

Homogeneous

$F (λ x, λ y) = F (x, y)$
put $y = v x$
the new equation separates

Linear

$y^{'} + P y = Q,$ with $P,$ $Q$ in $x$
multiply by $e^{\int P d x}$
integrate an exact derivative

9.4.1 Differential equations with variables separable

The easiest case is when $F (x, y)$ splits into a factor containing only $x$ times a factor containing only $y .$ Such an equation has its variables separable:

\frac{d y}{d x} = g (x) h (y)

(9.1)

where

$g (x)$: a function of $x$ only
$h (y)$: a function of $y$ only

Where $h (y) \neq 0,$ divide by $h (y)$ and multiply by $d x,$ so that everything in $y$ stands with $d y$ and everything in $x$ with $d x$ (Fig. 9.5). Then integrate each side:

\begin{aligned} \int \frac{d y}{h (y)} & = \int g (x) d x \\ H (y) & = G (x) + C \end{aligned}

(9.2)

where

$H$: an antiderivative of $\frac{1}{h (y)}$
$G$: an antiderivative of $g (x)$
$C$: one arbitrary constant, enough for both integrals

3/3Integrate each side; one constant $C$ is enough.

Fig. 9.5Sort the variables, then integrate: the orange

y

-part and the blue

x

-part end up on opposite sides.

Three rows of colour-coded boxes. The first row shows dy by dx equals e to the x, in a blue box labelled x-part, times 1 plus y squared, in an orange box labelled y-part. In the second step a second row appears, marked sort: dy over 1 plus y squared in an orange box equals e to the x dx in a blue box. In the third step a third row appears, marked integrate: inverse tan y in an orange box equals e to the x in a blue box, plus C.

In this chapter $\log$ means the natural logarithm, to base $e,$ as in the chapter on integrals.

Example 4NCERT Example 4 · p. 307

A family of hyperbolas

Find the general solution of $\frac{d y}{d x} = \frac{x - 2}{y + 3},$ where $y \neq - 3.$

Try it yourself first. Then reveal the solution one step at a time.

Example 5NCERT Example 5 · p. 308

An inverse tangent appears

Find the general solution of $\frac{d y}{d x} = e^{x} (1 + y^{2}) .$

Try it yourself first. Then reveal the solution one step at a time.

Example 6NCERT Example 6 · p. 308

Fixing the constant

Find the particular solution of $\frac{d y}{d x} = y^{2} \cos x$ for which $y = \frac{1}{2}$ when $x = 0.$

Try it yourself first. Then reveal the solution one step at a time.

Example 7NCERT Example 7 · p. 308

A curve through a given point

Find the curve through $(0, 1)$ that satisfies $(x^{2} + 1) d y = 2 x d x .$

Try it yourself first. Then reveal the solution one step at a time.

Example 8NCERT Example 8 · p. 309

From a slope condition to a curve

At every point $(x, y)$ of a curve the tangent has slope $\frac{3 x^{2}}{2 y},$ and the curve passes through $(2, 3) .$ Find its equation.

Try it yourself first. Then reveal the solution one step at a time.

Example 9NCERT Example 9 · p. 310

How long until the chai is drinkable?

Chai at 90 °C is left in a room at 30 °C, and its temperature obeys $\frac{d T}{d t} = - k (T - 30)$ with $t$ in minutes. After 4 minutes it has cooled to 70 °C. When will it reach 50 °C?

Try it yourself first. Then reveal the solution one step at a time.

Quick check

1 of 3

Which equation has its variables separable?

\frac{d y}{d x} = x + y

\frac{d y}{d x} = e^{x + y}

\frac{d y}{d x} = \sin (x y)

\frac{d y}{d x} = x^{2} + y^{2}

Show answer

Answer (B) $\frac{d y}{d x} = e^{x + y}$

e^{x + y} = e^{x} e^{y}

is a product of a function of

x

and a function of

y .

A sum such as

x + y

x^{2} + y^{2}

cannot be split into such a product, and neither can

\sin (x y) .

Practice · 12 questions

Practice 9.3: questions with answers

1
Short answerEasyNCERT Exercise 9.3, Q1–Q5
Find the general solution.
1. $\frac{d y}{d x} = \frac{1 + \cos 2 x}{1 - \cos 2 x}$
  Show solution
  Answer $y = - \cot x - x + C$
  The fraction is $\frac{2 \cos^{2} x}{2 \sin^{2} x} = \cot^{2} x = {cosec}^{2} x - 1.$ Integrating gives $y = - \cot x - x + C .$
2. $\frac{d y}{d x} = x \sqrt{1 - y^{2}},$ $- 1 < y < 1$
  Show solution
  Answer $\sin^{- 1} y = \frac{x^{2}}{2} + C$
  $\frac{d y}{\sqrt{1 - y^{2}}} = x d x,$ so $\sin^{- 1} y = \frac{x^{2}}{2} + C .$
3. $\frac{d y}{d x} = 3 - 2 y$
  Show solution
  Answer $y = \frac{3}{2} + A e^{- 2 x}$
  $\frac{d y}{3 - 2 y} = d x$ gives $- \frac{1}{2} \log ∣ 3 - 2 y ∣ = x + C_{1} .$ So $3 - 2 y = B e^{- 2 x},$ and $y = \frac{3}{2} + A e^{- 2 x}$ with $A = - \frac{B}{2} .$
4. $(1 + y^{2}) x d x + (1 + x^{2}) y d y = 0$
  Show solution
  Answer $(1 + x^{2}) (1 + y^{2}) = A$
  Divide by $(1 + x^{2}) (1 + y^{2}) ⁣ :$ $\frac{x d x}{1 + x^{2}} + \frac{y d y}{1 + y^{2}} = 0.$ Integrating, $\frac{1}{2} \log (1 + x^{2}) + \frac{1}{2} \log (1 + y^{2}) = C,$ which gives $(1 + x^{2}) (1 + y^{2}) = A .$
5. $(2 + \sin x) d y - \cos x d x = 0$
  Show solution
  Answer $y = \log (2 + \sin x) + C$
  $d y = \frac{\cos x}{2 + \sin x} d x;$ the numerator is the derivative of the denominator, which is always positive.
2
Short answerMediumNCERT Exercise 9.3, Q6–Q10
Find the general solution.
1. $\frac{d y}{d x} = (4 + y^{2}) (1 + 2 x)$
  Show solution
  Answer $\frac{1}{2} \tan^{- 1} \frac{y}{2} = x + x^{2} + C$
  $\frac{d y}{4 + y^{2}} = (1 + 2 x) d x,$ and $\int \frac{d y}{4 + y^{2}} = \frac{1}{2} \tan^{- 1} \frac{y}{2} .$
2. $x \log x d y - y d x = 0,$ $x > 1$
  Show solution
  Answer $y = C \log x$
  $\frac{d y}{y} = \frac{d x}{x \log x} .$ With $u = \log x$ the right side is $\frac{d u}{u},$ so $\log ∣ y ∣ = \log (\log x) + \log ∣ C ∣ .$
3. $x^{3} \frac{d y}{d x} = - y^{3}$
  Show solution
  Answer $\frac{1}{x^{2}} + \frac{1}{y^{2}} = C$
  $\frac{d y}{y^{3}} = - \frac{d x}{x^{3}} .$ Integrating, $- \frac{1}{2 y^{2}} = \frac{1}{2 x^{2}} + C_{1};$ multiply by $- 2.$
4. $\frac{d y}{d x} = \tan^{- 1} x$
  Show solution
  Answer $y = x \tan^{- 1} x - \frac{1}{2} \log (1 + x^{2}) + C$
  By parts with $u = \tan^{- 1} x ⁣ :$ $\int \tan^{- 1} x d x = x \tan^{- 1} x - \int \frac{x d x}{1 + x^{2}} .$
5. $3 e^{x} \tan y d x + (2 - e^{x}) \sec^{2} y d y = 0$
  Show solution
  Answer $\tan y = C (e^{x} - 2)^{3}$
  Divide by $(2 - e^{x}) \tan y ⁣ :$ $\frac{\sec^{2} y}{\tan y} d y = \frac{3 e^{x}}{e^{x} - 2} d x .$ Both sides have the form $\frac{f^{'}}{f},$ so $\log ∣ \tan y ∣ = 3 \log ∣ e^{x} - 2 ∣ + \log ∣ C ∣ .$
3
Short answerMediumNCERT Exercise 9.3, Q11–Q14
Use the condition given to pick out one solution of each equation.
1. $(x^{3} + 2 x^{2} + x + 2) \frac{d y}{d x} = 3 x^{2} + 5 x + 3;$ $y = 0$ when $x = 0$
  Show solution
  Answer $y = \tan^{- 1} x$ $+ \log \frac{(x + 2) (x^{2} + 1)}{2}$
  The bracket factors as $(x + 2) (x^{2} + 1),$ and partial fractions give $\frac{1}{x + 2} + \frac{2 x + 1}{x^{2} + 1} .$
  Integrating, $y = \log ∣ x + 2 ∣ + \log (x^{2} + 1) + \tan^{- 1} x + C .$ At $x = 0,$ $0 = \log 2 + C .$
2. $x (x^{2} - 4) \frac{d y}{d x} = 8;$ $y = 0$ when $x = 4$
  Show solution
  Answer $y = \log \frac{4 (x^{2} - 4)}{3 x^{2}}$
  Partial fractions: $\frac{8}{x (x - 2) (x + 2)} = - \frac{2}{x} + \frac{1}{x - 2} + \frac{1}{x + 2} .$
  So $y = \log ∣ \frac{x^{2} - 4}{x^{2}} ∣ + C .$ At $x = 4 ⁣ :$ $0 = \log \frac{3}{4} + C,$ so $C = \log \frac{4}{3} .$
3. $e^{d y / d x} = x + 1;$ $y = 2$ when $x = 0$
  Show solution
  Answer $y = (x + 1) \log (x + 1) - x + 2$
  Taking logs, $\frac{d y}{d x} = \log (x + 1) .$ By parts, $\int \log (x + 1) d x = (x + 1) \log (x + 1) - x + C,$ and $x = 0$ gives $C = 2.$
4. $\frac{d y}{d x} = y \cot x;$ $y = 2$ when $x = \frac{π}{2}$
  Show solution
  Answer $y = 2 \sin x$
  $\frac{d y}{y} = \cot x d x$ gives $\log ∣ y ∣ = \log ∣ \sin x ∣ + \log ∣ C ∣,$ so $y = C \sin x,$ and the condition gives $C = 2.$
Short answerMediumNCERT Exercise 9.3, Q15
Which curve through $(0, 1)$ has $y^{'} = e^{x} \cos x$ at every point?
Show solution
Answer $y = \frac{e^{x} (\sin x + \cos x) + 1}{2}$
$\int e^{x} \cos x d x = \frac{e^{x} (\sin x + \cos x)}{2} + C .$
At $(0, 1) ⁣ :$ $1 = \frac{1}{2} + C,$ so $C = \frac{1}{2} .$
Short answerMediumNCERT Exercise 9.3, Q16
For $(x + 1) \frac{d y}{d x} = x (y - 1),$ find the solution curve through $(0, 3) .$
Show solution
Answer $y = 1 + \frac{2 e^{x}}{x + 1}$
$\frac{d y}{y - 1} = (1 - \frac{1}{x + 1}) d x,$ so $\log ∣ y - 1 ∣ = x - \log ∣ x + 1 ∣ + C_{1} .$
Hence $(y - 1) (x + 1) = A e^{x},$ and $(0, 3)$ gives $A = 2.$
Short answerEasyNCERT Exercise 9.3, Q17
A curve passes through $(0, 3),$ and at each of its points the slope of the tangent equals the product of the coordinates. Find the curve.
Show solution
Answer $y = 3 e^{x^{2} / 2}$
$\frac{d y}{d x} = x y,$ so $\frac{d y}{y} = x d x$ and $y = A e^{x^{2} / 2} .$ At $(0, 3),$ $A = 3.$
Short answerMediumNCERT Exercise 9.3, Q18
A curve through $(2, 3)$ has this property: wherever $P (x, y)$ lies on it, the tangent at $P$ is three times as steep as the segment from $A (1, 2)$ to $P .$ Find the curve.
Show solution
Answer $y = 2 + (x - 1)^{3}$
$\frac{d y}{d x} = \frac{3 (y - 2)}{x - 1},$ so $\frac{d y}{y - 2} = \frac{3 d x}{x - 1}$ and $y - 2 = A (x - 1)^{3} .$ At $(2, 3),$ $A = 1.$
NumericalMediumNCERT Exercise 9.3, Q19
A spherical soap bubble is blown so that its volume grows at a constant rate. Its radius is 2 cm at first and 4 cm after 7 seconds. Express its radius as a function of the time $t,$ in seconds.
Show solution
Answer $r = 2 (t + 1)^{1 / 3}$ cm
Integrating, $\frac{4}{3} π r^{3} = k t + C .$ At $t = 0,$ $C = \frac{4}{3} π \cdot 8.$
At $t = 7 ⁣ :$ $\frac{4}{3} π \cdot 64 = 7 k + \frac{4}{3} π \cdot 8,$ so $k = \frac{4}{3} π \cdot 8.$
Dividing by $\frac{4}{3} π ⁣ :$ $r^{3} = 8 t + 8 = 8 (t + 1) .$
NumericalMediumNCERT Exercise 9.3, Q20
Money in an account grows continuously at $r %$ a year. ₹5000 becomes ₹15,000 in 15 years. Find $r .$ (Use $\log 3 = 1.0986$ .)
%
Show solution
Answerabout $7.32 %$ a year
$15000 = 5000 e^{15 r / 100},$ so $3 = e^{0.15 r}$ and $r = \frac{\log 3}{0.15} = \frac{1.0986}{0.15} \approx 7.32.$
NumericalEasyNCERT Exercise 9.3, Q21
₹20,000 is deposited at 6% a year, compounded continuously. What is it worth after 5 years? (Use $e^{0.3} = 1.3499$ .)
₹
Show solution
Answerabout ₹26,998
At $t = 5,$ $P = 20000 e^{0.3} = 20000 \times 1.3499 \approx 26998.$
NumericalMediumNCERT Exercise 9.3, Q22
A yeast culture starts with 50,000 cells and grows at a rate proportional to its current size. After 3 hours it has grown by 20%. When will it reach 1,00,000 cells? (Use $\log 2 = 0.6931$ and $\log 1.2 = 0.1823$ .)
hours
Show solution
Answerafter about 11.4 hours
$1.2 = e^{3 k}$ gives $k = \frac{\log 1.2}{3} .$ Then $2 = e^{k t}$ gives $t = \frac{3 \log 2}{\log 1.2} = \frac{2.0793}{0.1823} \approx 11.4.$
Quick answerEasyNCERT Exercise 9.3, Q23
The general solution of $\frac{d y}{d x} = e^{x - y}$ is
$e^{x} + e^{y} = C$ $e^{- x} + e^{- y} = C$ $e^{y} - e^{x} = C$ $e^{x} + e^{- y} = C$
Show answer
Answer (C) $e^{y} - e^{x} = C$
$e^{x - y} = e^{x} e^{- y},$ so $e^{y} d y = e^{x} d x$ and $e^{y} = e^{x} + C .$

9.4.2 Homogeneous differential equations

Some equations do not separate as they stand but become separable after a change of variable. The most common family is recognised by a scaling test.

Replace $x$ by $λ x$ and $y$ by $λ y,$ for a non-zero constant $λ .$ The function $3 x^{2} - x y$ becomes $λ^{2} (3 x^{2} - x y) .$ In $\frac{x + 2 y}{x - y}$ the $λ$ cancels completely, and $e^{y / x}$ does not change either. But $x + y^{2}$ turns into $λ x + λ^{2} y^{2},$ which is not a single power of $λ$ times $x + y^{2} .$

Homogeneous function of degree n

\begin{aligned} F (λ x, λ y) \\ = λ^{n} F (x, y) \end{aligned}

(9.3)

Definition

Homogeneous function

Scaling both variables by any non-zero constant $λ$ multiplies the function by $λ^{n} ⁣ :$ $F (λ x, λ y) = λ^{n} F (x, y) .$ The exponent $n$ is its degree.

Example $3 x^{2} - x y$ has degree 2; $\frac{x + 2 y}{x - y}$ and $e^{y / x}$ have degree 0; $x + y^{2}$ is not homogeneous.

Choosing $λ = \frac{1}{x}$ (for $x \neq 0$ ) shows what such a function looks like: $F (x, y) = x^{n} F (1, \frac{y}{x}),$ a power of $x$ times a function of $\frac{y}{x}$ alone. For example, $3 x^{2} - x y = x^{2} (3 - \frac{y}{x}) .$ In the same way, $F (x, y) = y^{n} F (\frac{x}{y}, 1)$ when $y \neq 0.$

Definition

Homogeneous differential equation

Written as $\frac{d y}{d x} = F (x, y),$ its right side has degree zero: scaling $x$ and $y$ leaves $F$ unchanged, so $F$ depends only on $\frac{y}{x} .$ The form $\frac{d x}{d y} = h (\frac{x}{y})$ counts too.

Example $\frac{d y}{d x} = \frac{x + 2 y}{x - y},$ whose right side equals $\frac{1 + 2 (y / x)}{1 - (y / x)} .$

Degree zero has a picture. All points on one line through the origin share the same ratio $\frac{y}{x},$ so a homogeneous equation demands the same slope at every one of them (Fig. 9.6). As a result, enlarging or shrinking a solution curve about the origin produces another solution curve.

4/4Enlarging or shrinking it about the origin gives the other solutions.

Fig. 9.6For

\frac{d y}{d x} = \frac{2 x - y}{x - 2 y}

the slope depends only on

\frac{y}{x},

so the solutions are scaled copies of one another.

Axes with dashed lines through the origin. In the first step one line, y equals x over 3, is drawn in purple with six dots on it, three on each side of the origin, and at every dot the slope mark is the same steep segment. In the second step faint lines every 15 degrees carry slope marks that are parallel along each line but change from line to line. In the third step a blue ellipse, x squared minus xy plus y squared equals 3, runs along the marks. In the fourth step three more dashed ellipses of the same shape, smaller and larger, surround it.

SimulationScaling along a ray

A coordinate plane from minus 8 to 8 on both axes with a fixed origin O and two draggable points, P at (3, 1) and Q at (6, 2), on the same line through O. The distances OP and OQ are drawn and measured, and the mirror image of each point in the origin is shown, so the learner can compare points obtained by scaling with lambda equal to 2 and to minus 1 and check the slope formula of a homogeneous equation at each.

Fig. 9.7

Q

starts at twice

P,

and the mirror point

P^{'}

P

scaled by

- 1.

All three lie on one line through

O .

Try this

Leave $P$ at $(3, 1)$ and $Q$ at $(6, 2) .$ Work out $\frac{2 x - y}{x - 2 y}$ at both: you should get 5 each time. Is $O Q$ twice $O P$ ?
The mirror point $P^{'}$ is $(- 3, - 1),$ the scaling with $λ = - 1.$ Check that the slope there is 5 too.
Drag $Q$ to $(4, 3),$ off the line. Now $\frac{y}{x}$ changes, and the slope becomes $- 2.5.$

DerivationThe substitution

y = v x

Let $\frac{d y}{d x} = g (\frac{y}{x}) .$ Give the ratio its own name, $v = \frac{y}{x},$ so that $y = v x$ with $v$ a function of $x .$ By the product rule,

\frac{d y}{d x} = v + x \frac{d v}{d x}

(9.4)

Substituting into the equation gives $v + x \frac{d v}{d x} = g (v),$ that is $x \frac{d v}{d x} = g (v) - v .$ Now the variables separate:

\begin{aligned} \int \frac{d v}{g (v) - v} \\ = \int \frac{d x}{x} + C \end{aligned}

(9.5)

Integrate, then replace $v$ by $\frac{y}{x}$ to return to the original variables.

Example 10NCERT Example 10 · p. 314

A family of ellipses

Solve $(2 x - y) d x + (2 y - x) d y = 0$ after checking that it is homogeneous.

Try it yourself first. Then reveal the solution one step at a time.

Example 11NCERT Example 11 · p. 316

A tangent inside the equation

Check that $x \frac{d y}{d x} = y + x \tan \frac{y}{x}$ is homogeneous, and find its general solution ( $x \neq 0$ ).

Try it yourself first. Then reveal the solution one step at a time.

Example 12NCERT Example 12 · p. 318

Using $x = v y$

For $x y d x = (x^{2} + y^{2}) d y,$ confirm that the equation is homogeneous and find the solution with $x = 2$ when $y = 1.$

Try it yourself first. Then reveal the solution one step at a time.

Example 13NCERT Example 13 · p. 319

Curves from a slope rule

A family of curves has this property: the tangent at any point $(x, y)$ has slope $\frac{2 x y}{x^{2} - y^{2}} .$ Find the family and say what the curves are.

Try it yourself first. Then reveal the solution one step at a time.

Practice · 5 questions

Practice 9.4: questions with answers

1
Short answerMediumNCERT Exercise 9.4, Q1–Q5
Check that each equation is homogeneous, then find its general solution.
1. $(x y - x^{2}) d y = y^{2} d x$
  Show solution
  Answer $\frac{y}{x} = \log ∣ y ∣ + C$
  $\frac{d y}{d x} = \frac{y^{2}}{x y - x^{2}} = \frac{v^{2}}{v - 1},$ so $x \frac{d v}{d x} = \frac{v^{2}}{v - 1} - v = \frac{v}{v - 1} .$
  Then $(1 - \frac{1}{v}) d v = \frac{d x}{x}$ and $v - \log ∣ v ∣ = \log ∣ x ∣ + C .$ Since $\log ∣ v ∣ + \log ∣ x ∣ = \log ∣ y ∣,$ this is $\frac{y}{x} = \log ∣ y ∣ + C .$
2. $\frac{d y}{d x} = \frac{y + \sqrt{x y}}{x},$ with $x, y > 0$
  Show solution
  Answer $2 \sqrt{\frac{y}{x}} = \log x + C$
  The right side is $v + \sqrt{v},$ so $x \frac{d v}{d x} = \sqrt{v},$ $\frac{d v}{\sqrt{v}} = \frac{d x}{x}$ and $2 \sqrt{v} = \log x + C .$
3. $(x + 2 y) d x - (2 x - y) d y = 0$
  Show solution
  Answer $2 \tan^{- 1} \frac{y}{x} = \log \sqrt{x^{2} + y^{2}} + C$
  $\frac{d y}{d x} = \frac{1 + 2 v}{2 - v},$ so $x \frac{d v}{d x} = \frac{1 + v^{2}}{2 - v}$ and $\frac{(2 - v) d v}{1 + v^{2}} = \frac{d x}{x} .$
  Integrating, $2 \tan^{- 1} v - \frac{1}{2} \log (1 + v^{2}) = \log ∣ x ∣ + C .$ Since $\frac{1}{2} \log (1 + v^{2}) + \log ∣ x ∣ = \log \sqrt{x^{2} + y^{2}},$ the answer follows.
4. $(x^{2} + 3 y^{2}) d x - 2 x y d y = 0$
  Show solution
  Answer $x^{2} + y^{2} = C x^{3}$
  $\frac{d y}{d x} = \frac{1 + 3 v^{2}}{2 v},$ so $x \frac{d v}{d x} = \frac{1 + v^{2}}{2 v} .$ Then $\log (1 + v^{2}) = \log ∣ x ∣ + \log C,$ i.e. $x^{2} + y^{2} = C x^{3} .$
5. $x^{2} \frac{d y}{d x} = x^{2} + x y + y^{2}$
  Show solution
  Answer $\tan^{- 1} \frac{y}{x} = \log ∣ x ∣ + C$
  $v + x \frac{d v}{d x} = 1 + v + v^{2},$ so $\frac{d v}{1 + v^{2}} = \frac{d x}{x} .$
2
Short answerMediumNCERT Exercise 9.4, Q6–Q10
Confirm that each equation is homogeneous and solve it.
1. $x d y - y d x = \sqrt{x^{2} - y^{2}} d x,$ with $x > 0,$ $∣ y ∣ < x$
  Show solution
  Answer $\sin^{- 1} \frac{y}{x} = \log x + C$
  $\frac{d y}{d x} = v + \sqrt{1 - v^{2}},$ so $\frac{d v}{\sqrt{1 - v^{2}}} = \frac{d x}{x} .$
2. $x \tan \frac{y}{x} d y = (x + y \tan \frac{y}{x}) d x$
  Show solution
  Answer $x \cos \frac{y}{x} = C$
  $\frac{d y}{d x} = \frac{y}{x} + \cot \frac{y}{x} .$ With $y = v x ⁣ :$ $x \frac{d v}{d x} = \cot v,$ so $\tan v d v = \frac{d x}{x}$ and $- \log ∣ \cos v ∣ = \log ∣ x ∣ + C_{1} .$
3. $x \frac{d y}{d x} - y = x e^{y / x}$
  Show solution
  Answer $e^{- y / x} + \log ∣ x ∣ = C$
  $x \frac{d v}{d x} = e^{v},$ so $e^{- v} d v = \frac{d x}{x}$ and $- e^{- v} = \log ∣ x ∣ - C .$
4. $x \frac{d y}{d x} = y + y \log \frac{y}{x},$ with $x, y > 0$
  Show solution
  Answer $y = x e^{C x}$
  $x \frac{d v}{d x} = v \log v .$ With $u = \log v,$ $\frac{d v}{v \log v} = \frac{d u}{u},$ so $\log ∣ \log v ∣ = \log ∣ x ∣ + \log ∣ C ∣$ and $\log \frac{y}{x} = C x .$
5. $y \frac{d x}{d y} = x + y \cos^{2} \frac{x}{y}$
  Show solution
  Answer $\tan \frac{x}{y} = \log ∣ y ∣ + C$
  Put $x = v y ⁣ :$ $v + y \frac{d v}{d y} = v + \cos^{2} v,$ so $\sec^{2} v d v = \frac{d y}{y} .$
3
Short answerHardNCERT Exercise 9.4, Q11–Q15
Solve each equation and choose the solution that meets the condition.
1. $(x^{2} - 2 y^{2}) d x + x y d y = 0;$ $y = 2$ when $x = 1$
  Show solution
  Answer $y^{2} = x^{2} + 3 x^{4}$
  $\frac{d y}{d x} = 2 v - \frac{1}{v},$ so $x \frac{d v}{d x} = \frac{v^{2} - 1}{v}$ and $\frac{v d v}{v^{2} - 1} = \frac{d x}{x} .$
  So $v^{2} - 1 = A x^{2},$ that is $y^{2} - x^{2} = A x^{4} .$ At $(1, 2),$ $A = 3.$
2. $x^{2} \frac{d y}{d x} = y^{2} + 2 x y;$ $y = 1$ when $x = 1$
  Show solution
  Answer $y = \frac{x^{2}}{2 - x}$
  $x \frac{d v}{d x} = v^{2} + v .$ Partial fractions, $\frac{1}{v (v + 1)} = \frac{1}{v} - \frac{1}{v + 1},$ give $\frac{v}{v + 1} = A x,$ i.e. $\frac{y}{x + y} = A x .$
  At $(1, 1),$ $A = \frac{1}{2} ⁣ :$ $2 y = x (x + y),$ so $y = \frac{x^{2}}{2 - x} .$
3. $x \frac{d y}{d x} = y + x \cos^{2} \frac{y}{x};$ $y = \frac{π}{4}$ when $x = 1$
  Show solution
  Answer $\tan \frac{y}{x} = 1 + \log ∣ x ∣$
  $x \frac{d v}{d x} = \cos^{2} v,$ so $\sec^{2} v d v = \frac{d x}{x}$ and $\tan v = \log ∣ x ∣ + C .$ At $x = 1,$ $\tan \frac{π}{4} = 1 = C .$
4. $x \frac{d y}{d x} = y + x \cos \frac{y}{x};$ $y = 0$ when $x = 1$
  Show solution
  Answer $\sec \frac{y}{x} + \tan \frac{y}{x} = x$
  $x \frac{d v}{d x} = \cos v,$ so $\sec v d v = \frac{d x}{x}$ and $\log ∣ \sec v + \tan v ∣ = \log ∣ x ∣ + \log A .$ At $(1, 0),$ $1 = A .$
5. $x^{3} \frac{d y}{d x} = x^{2} y + y^{3};$ $y = 1$ when $x = 1$
  Show solution
  Answer $y^{2} = \frac{x^{2}}{1 - 2 \log ∣ x ∣}$
  $x \frac{d v}{d x} = v^{3},$ so $\frac{d v}{v^{3}} = \frac{d x}{x}$ and $- \frac{1}{2 v^{2}} = \log ∣ x ∣ + C .$ At $(1, 1),$ $C = - \frac{1}{2},$ which gives $\frac{x^{2}}{y^{2}} = 1 - 2 \log ∣ x ∣ .$
Quick answerEasyNCERT Exercise 9.4, Q16
With the substitution $y = v x,$ the derivative $\frac{d y}{d x}$ becomes
$x \frac{d v}{d x}$ $v + \frac{d v}{d x}$ $v + x \frac{d v}{d x}$ $v x \frac{d v}{d x}$
Show answer
Answer (C) $v + x \frac{d v}{d x}$
By the product rule, $\frac{d}{d x} (v x) = v + x \frac{d v}{d x} .$
Quick answerEasyNCERT Exercise 9.4, Q17
Which of these is a homogeneous differential equation?
$(x^{2} - x y) d x + (y^{2} + 3 x y) d y = 0$ $(2 x + 3 y + 1) d y - (x - y) d x = 0$ $(x^{2} + y) d x + x y d y = 0$ $x y d x - (x^{3} - y^{2}) d y = 0$
Show answer
Answer (A) $(x^{2} - x y) d x + (y^{2} + 3 x y) d y = 0$
In A both brackets are homogeneous of degree 2, so $\frac{d y}{d x}$ is a ratio of degree 0. The constant 1 spoils B, and C and D mix different degrees.

9.4.3 Linear differential equations

A first order equation is linear when $y$ and $\frac{d y}{d x}$ occur only to the first power and are never multiplied together. Every such equation can be arranged as

First order linear equation

\frac{d y}{d x} + P y = Q

(9.6)

where

$P, Q$: constants or functions of $x$ only

$\frac{d y}{d x} - 2 y = x,$ with $P = - 2$ and $Q = x;$
$\frac{d y}{d x} + 2 x y = e^{- x^{2}},$ with $P = 2 x;$
$x \frac{d y}{d x} + y = \cos x,$ which becomes $\frac{d y}{d x} + \frac{1}{x} y = \frac{\cos x}{x}$ after dividing by $x .$

On the other hand, $y \frac{d y}{d x} + x = 0$ (a product of $y$ and $\frac{d y}{d x}$ ) and $\frac{d y}{d x} + y^{2} = x$ (a square) are not linear. With the roles of the variables swapped, the linear form is $\frac{d x}{d y} + P_{1} x = Q_{1},$ in which $P_{1}$ and $Q_{1}$ depend on $y$ alone (or are constants), as in $\frac{d x}{d y} + 2 x = y e^{- y} .$

DerivationFinding an integrating factor

The left side $\frac{d y}{d x} + P y$ is nearly the derivative of a product. Multiply the whole equation by a positive function $g (x),$ still to be chosen:

g \frac{d y}{d x} + P g y = Q g

By the product rule, $\frac{d}{d x} (y g) = g \frac{d y}{d x} + g^{'} y .$ The left side matches this exactly when $g^{'} = P g,$ that is when $\frac{g^{'}}{g} = P .$ Integrating, $\log g = \int P d x,$ so

Integrating factor

g (x) = e^{\int P d x}

(9.7)

With this choice the equation reads $\frac{d}{d x} (y g) = Q g,$ and one integration finishes the job:

\begin{aligned} y e^{\int P d x} \\ = \int Q e^{\int P d x} d x \\ + C \end{aligned}

(9.8)

Definition

Integrating factor

The function $e^{\int P d x}$ that turns the left side of $\frac{d y}{d x} + P y = Q$ into an exact derivative, $\frac{d}{d x} (y e^{\int P d x}) .$ It is written I.F.

ExampleFor $\frac{d y}{d x} + \frac{2}{x} y = 1$ with $x > 0,$ I.F. $= e^{2 \log x} = x^{2} .$

The method in three steps:

Divide through so that $\frac{d y}{d x}$ has coefficient 1, then read off $P$ and $Q .$
Compute $e^{\int P d x},$ the integrating factor; no constant is needed here.
Integrate $Q$ times the integrating factor, add $C,$ and divide by the integrating factor to get $y,$ as in (9.8).

For an equation $\frac{d x}{d y} + P_{1} x = Q_{1}$ the steps are the same with the letters exchanged:

\begin{aligned} x e^{\int P_{1} d y} \\ = \int Q_{1} e^{\int P_{1} d y} d y \\ + C \end{aligned}

(9.9)

Example 14NCERT Example 14 · p. 324

Every solution approaches a sine wave

Find the general solution of $\frac{d y}{d x} + y = \sin x + \cos x .$

Try it yourself first. Then reveal the solution one step at a time.

3/3The term $C e^{- x}$ dies away, so every solution approaches $y = \sin x .$

Fig. 9.8The general solution is

y = \sin x + C e^{- x} ⁣ :

different starts, the same long-run behaviour.

Axes with x from 0 to about 7.5, ticks at pi and 2 pi, and y from minus 3 to 3. In the first step a thick blue sine wave, y equals sin x, is drawn: the solution with C equals 0. In the second step six dashed orange curves start on the y-axis at heights 3, 2, 1, minus 1, minus 2 and minus 3, labelled C equals 3 and C equals minus 3 at the ends, and bend towards the sine wave. In the third step the region beyond x equals 3.6 is shaded, where all the curves have merged with the wave, with the labels curves merge and C e to the minus x tends to 0.

Example 15NCERT Example 15 · p. 325

An integrating factor that is a power

Find the general solution of $x \frac{d y}{d x} - 2 y = x^{3},$ for $x \neq 0.$

Try it yourself first. Then reveal the solution one step at a time.

Example 16NCERT Example 16 · p. 326

Linear in x instead of y

Solve $y^{2} d x + (3 x y - 1) d y = 0,$ for $y > 0.$

Try it yourself first. Then reveal the solution one step at a time.

Example 17NCERT Example 17 · p. 326

A particular solution

Find the solution of $\frac{d y}{d x} + y \tan x = \sec x$ with $y = 2$ when $x = 0,$ for $- \frac{π}{2} < x < \frac{π}{2} .$

Try it yourself first. Then reveal the solution one step at a time.

Example 18NCERT Example 18 · p. 327

A curve from a slope rule, by the linear method

A curve passes through the origin, and at each point $(x, y)$ its slope equals the square of the abscissa plus the ordinate. Find it.

Try it yourself first. Then reveal the solution one step at a time.

Practice · 7 questions

Practice 9.5: questions with answers

1
Short answerMediumNCERT Exercise 9.5, Q1–Q6
Find the general solution.
1. $\frac{d y}{d x} + 3 y = \cos x$
  Show solution
  Answer $y = \frac{3 \cos x + \sin x}{10} + C e^{- 3 x}$
  I.F. $= e^{3 x},$ and $y e^{3 x} = \int e^{3 x} \cos x d x = \frac{e^{3 x} (3 \cos x + \sin x)}{10} + C$ (by parts twice).
2. $\frac{d y}{d x} - 2 y = e^{3 x}$
  Show solution
  Answer $y = e^{3 x} + C e^{2 x}$
  I.F. $= e^{- 2 x},$ and $y e^{- 2 x} = \int e^{x} d x = e^{x} + C .$
3. $\frac{d y}{d x} + \frac{3 y}{x} = \frac{1}{x^{2}},$ $x > 0$
  Show solution
  Answer $y = \frac{1}{2 x} + \frac{C}{x^{3}}$
  I.F. $= e^{3 \log x} = x^{3},$ and $x^{3} y = \int x d x = \frac{x^{2}}{2} + C .$
4. $\frac{d y}{d x} + y \cot x = \cos x,$ $0 < x < π$
  Show solution
  Answer $y = \frac{\sin x}{2} + C cosec x$
  I.F. $= e^{\log \sin x} = \sin x,$ and $y \sin x = \int \sin x \cos x d x = \frac{\sin^{2} x}{2} + C .$
5. $x^{2} \frac{d y}{d x} + y = e^{1 / x},$ $x \neq 0$
  Show solution
  Answer $y = e^{1 / x} (C - \frac{1}{x})$
  Divide by $x^{2} ⁣ :$ $P = \frac{1}{x^{2}},$ I.F. $= e^{- 1 / x} .$ Then $y e^{- 1 / x} = \int \frac{d x}{x^{2}} = - \frac{1}{x} + C .$
6. $x \frac{d y}{d x} + y = \log x,$ $x > 0$
  Show solution
  Answer $y = \log x - 1 + \frac{C}{x}$
  The left side is already $\frac{d}{d x} (x y) .$ So $x y = \int \log x d x = x \log x - x + C .$
2
Short answerHardNCERT Exercise 9.5, Q7–Q12
Find the general solution.
1. $x \log x \frac{d y}{d x} + y = 2 x \log x,$ $x > 1$
  Show solution
  Answer $y = 2 x - \frac{2 x - C}{\log x}$
  Divide by $x \log x ⁣ :$ $P = \frac{1}{x \log x},$ I.F. $= e^{\log (\log x)} = \log x .$ Then $y \log x = \int 2 \log x d x = 2 x \log x - 2 x + C .$
2. $(4 + x^{2}) d y + 2 x y d x = e^{x} d x$
  Show solution
  Answer $y = \frac{e^{x} + C}{4 + x^{2}}$
  The left side is $d [(4 + x^{2}) y],$ so $(4 + x^{2}) y = e^{x} + C .$ (In standard form the I.F. is $4 + x^{2}$ .)
3. $x \frac{d y}{d x} + y - x y \tan x = x,$ $0 < x < \frac{π}{2}$
  Show solution
  Answer $y = \tan x + \frac{1}{x} + \frac{C}{x \cos x}$
  Divide by $x ⁣ :$ $P = \frac{1}{x} - \tan x,$ so I.F. $= e^{\log x + \log \cos x} = x \cos x .$ Then $x y \cos x = \int x \cos x d x = x \sin x + \cos x + C .$
4. $(2 x + y) \frac{d y}{d x} = 1$
  Show solution
  Answer $x = - \frac{y}{2} - \frac{1}{4} + C e^{2 y}$
  $\frac{d x}{d y} - 2 x = y,$ with I.F. $e^{- 2 y} .$ Then $x e^{- 2 y} = \int y e^{- 2 y} d y = - (\frac{y}{2} + \frac{1}{4}) e^{- 2 y} + C .$
5. $y d x + (2 x - y^{3}) d y = 0,$ $y > 0$
  Show solution
  Answer $x = \frac{y^{3}}{5} + \frac{C}{y^{2}}$
  $\frac{d x}{d y} + \frac{2}{y} x = y^{2},$ I.F. $= y^{2},$ and $x y^{2} = \int y^{4} d y = \frac{y^{5}}{5} + C .$
6. $(x + 2 y^{3}) \frac{d y}{d x} = y,$ $y > 0$
  Show solution
  Answer $x = y^{3} + C y$
  $\frac{d x}{d y} - \frac{1}{y} x = 2 y^{2},$ I.F. $= \frac{1}{y},$ and $\frac{x}{y} = \int 2 y d y = y^{2} + C .$
3
Short answerHardNCERT Exercise 9.5, Q13–Q15
Solve, then use the condition to fix the constant.
1. $\frac{d y}{d x} - y \tan x = \cos x;$ $y = 0$ when $x = 0$
  Show solution
  Answer $y = \frac{x + \sin x \cos x}{2 \cos x}$
  I.F. $= e^{- \int \tan x d x} = \cos x,$ so $y \cos x = \int \cos^{2} x d x = \frac{x}{2} + \frac{\sin 2 x}{4} + C,$ and $x = 0$ gives $C = 0.$
2. $(1 + x^{2}) \frac{d y}{d x} + 4 x y = \frac{1}{1 + x^{2}};$ $y = 1$ when $x = 0$
  Show solution
  Answer $y = \frac{x + 1}{(1 + x^{2})^{2}}$
  Divide by $1 + x^{2} ⁣ :$ $P = \frac{4 x}{1 + x^{2}},$ I.F. $= (1 + x^{2})^{2} .$ Then $y (1 + x^{2})^{2} = \int d x = x + C,$ and $C = 1.$
3. $\frac{d y}{d x} - 2 y \cot x = \cos x;$ $y = 3$ when $x = \frac{π}{2}$
  Show solution
  Answer $y = 4 \sin^{2} x - \sin x$
  I.F. $= e^{- 2 \log \sin x} = \frac{1}{\sin^{2} x},$ and $\frac{y}{\sin^{2} x} = \int \frac{\cos x}{\sin^{2} x} d x = - \frac{1}{\sin x} + C .$ So $y = C \sin^{2} x - \sin x,$ and $x = \frac{π}{2}$ gives $C = 4.$
Short answerMediumNCERT Exercise 9.5, Q16
Find the curve through $(0, 1)$ whose slope at every point $(x, y)$ equals twice the abscissa minus the ordinate.
Show solution
Answer $y = 2 x - 2 + 3 e^{- x}$
$\frac{d y}{d x} + y = 2 x,$ I.F. $e^{x},$ $y e^{x} = \int 2 x e^{x} d x = 2 (x - 1) e^{x} + C .$ So $y = 2 x - 2 + C e^{- x},$ and $(0, 1)$ gives $C = 3.$
Short answerMediumNCERT Exercise 9.5, Q17
Along a certain curve, the tangent slope at $(x, y)$ is always 3 less than $y - x .$ The curve passes through $(0, 5);$ find it.
Show solution
Answer $y = x + 4 + e^{x}$
In standard form, $\frac{d y}{d x} - y = - x - 3,$ with I.F. $e^{- x} .$
$y e^{- x} = \int (- x - 3) e^{- x} d x = (x + 4) e^{- x} + C,$ so $y = x + 4 + C e^{x} .$ At $(0, 5),$ $C = 1.$
Quick answerEasyNCERT Exercise 9.5, Q18
The integrating factor of $x \frac{d y}{d x} + 2 y = x^{3}$ ( $x > 0$ ) is
$x^{2}$ $\frac{1}{x^{2}}$ $2 \log x$ $e^{x^{2}}$
Show answer
Answer (A) $x^{2}$
In standard form $P = \frac{2}{x},$ and $e^{\int P d x} = e^{2 \log x} = x^{2} .$
Quick answerMediumNCERT Exercise 9.5, Q19
The integrating factor of $(4 - y^{2}) \frac{d x}{d y} - y x = y^{2}$ ( $- 2 < y < 2$ ) is
$\frac{1}{\sqrt{4 - y^{2}}}$ $4 - y^{2}$ $\sqrt{4 - y^{2}}$ $e^{- y^{2} / 2}$
Show answer
Answer (C) $\sqrt{4 - y^{2}}$
Here $P_{1} = - \frac{y}{4 - y^{2}},$ and $\int P_{1} d y = \frac{1}{2} \log (4 - y^{2}),$ so I.F. $= \sqrt{4 - y^{2}} .$

Miscellaneous Examples

The last four examples mix the skills of the chapter. Before starting any of them, decide which kind of equation you have (Fig. 9.9).

1/4Write the equation as $\frac{d y}{d x} = F (x, y),$ then ask three questions.

Fig. 9.9Three questions to ask of any first order, first degree equation.

At the top a box holds dy by dx equals F of x and y. Below it three question boxes are stacked, joined by arrows labelled no: F equals g of x times h of y?, F of lambda x and lambda y equals F of x and y?, and y dash plus P y equals Q?. From each, an arrow labelled yes leads to an answer box on the right: separate, put y equals v x, and use the IF. In the first step all rows look alike. In the second, third and fourth steps the separable, homogeneous and linear rows in turn are highlighted in colour while the others fade.

Example 19NCERT Example 19 · p. 329

A second order check

Verify that $y = A x + \frac{B}{x},$ with arbitrary constants $A$ and $B,$ satisfies $x^{2} \frac{d^{2} y}{d x^{2}} + x \frac{d y}{d x} - y = 0$ for $x \neq 0.$

Try it yourself first. Then reveal the solution one step at a time.

Example 20NCERT Example 20 · p. 330

Unwrapping a logarithm

A curve passes through the origin, and at every point $(x, y)$ on it the natural logarithm of the slope equals $x - 2 y .$ Find the curve.

Try it yourself first. Then reveal the solution one step at a time.

Example 21NCERT Example 21 · p. 331

A perfect square after substitution

Solve $x^{2} \frac{d y}{d x} = x^{2} + 3 x y + y^{2},$ for $x \neq 0.$

Try it yourself first. Then reveal the solution one step at a time.

Example 22NCERT Example 22 · p. 333

Linear in x, with a substitution inside

Find the solution of $d x = (\sin y - x) \cos y d y$ with $x = 0$ when $y = 0.$

Try it yourself first. Then reveal the solution one step at a time.

Quick check

1 of 3

Match each equation with the method that solves it.

Tap an item on the left, then its match on the right.

Show answer

Answer

$\frac{d y}{d x} = \frac{e^{x}}{y}$ : separate the variables
$\frac{d y}{d x} = \frac{x^{2} + y^{2}}{x y}$ : put $y = v x$
$\frac{d y}{d x} + \frac{y}{x} = e^{x}$ : integrating factor
$\frac{d x}{d y} = \frac{x}{y} + \cos \frac{x}{y}$ : put $x = v y$

\frac{e^{x}}{y}

is a product of a function of

x

and one of

y;

the second right side has degree 0; the third is linear in

y;

the fourth is a function of

\frac{x}{y}

and comes as

\frac{d x}{d y} .

From history

Who solved the first ones?

Differential equations were born together with calculus. In November 1675 Gottfried Wilhelm Leibniz wrote $\int y d y = \frac{1}{2} y^{2}$ in his notes, introducing both the integral sign and the $d y$ notation we still use. Within twenty years he had found all three methods of this chapter: separation of variables in 1691, the reduction of homogeneous equations soon afterwards, and then the method for linear equations.

The words came later. Jakob Bernoulli called the answer an integral of the equation in 1690, and his younger brother Johann gave the separation method its name. Joseph-Louis Lagrange used the word solution in 1774, and Henri Poincaré, about a century later, made it standard. Poincaré also started the qualitative study of differential equations: describing the shapes of solution curves, as a slope field does, even when no formula for them exists.

Recap

Differential Equations: key points and summary

A differential equation involves derivatives of an unknown function; in an ordinary one there is a single independent variable (Section 9.2).
Order: how many times the unknown is differentiated in its most differentiated term (Section 9.2.1).
Degree: the exponent on the highest derivative, which exists only when the equation is polynomial in all its derivatives. Both numbers are positive integers (Section 9.2.2).
A solution is a function that turns the equation into an identity; its graph is a solution curve (Section 9.3).
An order- $n$ equation’s general solution carries $n$ arbitrary constants; fixing them all with given conditions yields a particular solution (Section 9.3).
If $\frac{d y}{d x} = g (x) h (y),$ separate and integrate: $\int \frac{d y}{h (y)} = \int g (x) d x + C$ (Section 9.4.1).
$F (x, y)$ is homogeneous of degree $n$ if $F (λ x, λ y) = λ^{n} F (x, y);$ $\frac{d y}{d x} = F (x, y)$ is homogeneous when $F$ has degree 0 (Section 9.4.2).
For a homogeneous equation put $y = v x,$ or $x = v y$ for $\frac{d x}{d y} = h (\frac{x}{y});$ the new equation separates (Section 9.4.2).
$\frac{d y}{d x} + P y = Q,$ with $P$ and $Q$ functions of $x,$ is linear: $y \cdot I.F. = \int Q \cdot I.F. d x + C$ with $I.F. = e^{\int P d x}$ (Section 9.4.3).
For $\frac{d x}{d y} + P_{1} x = Q_{1}$ use $I.F. = e^{\int P_{1} d y}$ and $x \cdot I.F. = \int Q_{1} \cdot I.F. d y + C$ (Section 9.4.3).
Classify an equation before solving it, and check the answer by substituting it back (Miscellaneous Examples).

Flash cards · 8

Key terms

1 / 8

Miscellaneous Exercise on Chapter 9

Find your NCERT exercise question

Our practice questions are written for ncert.ai and follow the NCERT exercises. Find the one that matches the question in your textbook; its answer is with it.

In your NCERT textbook	Practice question here
NCERT Exercise 9.1, Q1–Q5	Practice 9.1 · Q1
NCERT Exercise 9.1, Q6–Q10	Practice 9.1 · Q2
NCERT Exercise 9.1, Q11	Practice 9.1 · Q3
NCERT Exercise 9.1, Q12	Practice 9.1 · Q4
NCERT Exercise 9.2, Q1–Q5	Practice 9.2 · Q1
NCERT Exercise 9.2, Q6–Q10	Practice 9.2 · Q2
NCERT Exercise 9.2, Q11	Practice 9.2 · Q3
NCERT Exercise 9.2, Q12	Practice 9.2 · Q4
NCERT Exercise 9.3, Q1–Q5	Practice 9.3 · Q1
NCERT Exercise 9.3, Q6–Q10	Practice 9.3 · Q2
NCERT Exercise 9.3, Q11–Q14	Practice 9.3 · Q3
NCERT Exercise 9.3, Q15	Practice 9.3 · Q4
NCERT Exercise 9.3, Q16	Practice 9.3 · Q5
NCERT Exercise 9.3, Q17	Practice 9.3 · Q6
NCERT Exercise 9.3, Q18	Practice 9.3 · Q7
NCERT Exercise 9.3, Q19	Practice 9.3 · Q8
NCERT Exercise 9.3, Q20	Practice 9.3 · Q9
NCERT Exercise 9.3, Q21	Practice 9.3 · Q10
NCERT Exercise 9.3, Q22	Practice 9.3 · Q11
NCERT Exercise 9.3, Q23	Practice 9.3 · Q12
NCERT Exercise 9.4, Q1–Q5	Practice 9.4 · Q1
NCERT Exercise 9.4, Q6–Q10	Practice 9.4 · Q2
NCERT Exercise 9.4, Q11–Q15	Practice 9.4 · Q3
NCERT Exercise 9.4, Q16	Practice 9.4 · Q4
NCERT Exercise 9.4, Q17	Practice 9.4 · Q5
NCERT Exercise 9.5, Q1–Q6	Practice 9.5 · Q1
NCERT Exercise 9.5, Q7–Q12	Practice 9.5 · Q2
NCERT Exercise 9.5, Q13–Q15	Practice 9.5 · Q3
NCERT Exercise 9.5, Q16	Practice 9.5 · Q4
NCERT Exercise 9.5, Q17	Practice 9.5 · Q5
NCERT Exercise 9.5, Q18	Practice 9.5 · Q6
NCERT Exercise 9.5, Q19	Practice 9.5 · Q7
NCERT Miscellaneous Exercise on Chapter 9, Q1	Miscellaneous exercise · Q1
NCERT Miscellaneous Exercise on Chapter 9, Q2	Miscellaneous exercise · Q2
NCERT Miscellaneous Exercise on Chapter 9, Q3	Miscellaneous exercise · Q3
NCERT Miscellaneous Exercise on Chapter 9, Q4	Miscellaneous exercise · Q4
NCERT Miscellaneous Exercise on Chapter 9, Q5	Miscellaneous exercise · Q5
NCERT Miscellaneous Exercise on Chapter 9, Q6	Miscellaneous exercise · Q6
NCERT Miscellaneous Exercise on Chapter 9, Q7	Miscellaneous exercise · Q7
NCERT Miscellaneous Exercise on Chapter 9, Q8	Miscellaneous exercise · Q8
NCERT Miscellaneous Exercise on Chapter 9, Q9	Miscellaneous exercise · Q9
NCERT Miscellaneous Exercise on Chapter 9, Q10	Miscellaneous exercise · Q10
NCERT Miscellaneous Exercise on Chapter 9, Q11	Miscellaneous exercise · Q11
NCERT Miscellaneous Exercise on Chapter 9, Q12	Miscellaneous exercise · Q12
NCERT Miscellaneous Exercise on Chapter 9, Q13	Miscellaneous exercise · Q13
NCERT Miscellaneous Exercise on Chapter 9, Q14	Miscellaneous exercise · Q14
NCERT Miscellaneous Exercise on Chapter 9, Q15	Miscellaneous exercise · Q15

Quick check: MCQs on Differential Equations

9.1 Introduction2 questions
9.2.2 Degree of a differential equation4 questions
9.3 General and Particular Solutions of a Differential Equation4 questions
9.4.1 Differential equations with variables separable3 questions
9.4.2 Homogeneous differential equations3 questions
9.4.3 Linear differential equations4 questions
Miscellaneous Examples3 questions

Glossary · 14

Key terms (English–Hindi)

arbitrary constantस्वेच्छ अचरsvechchh achar: A constant, such as $C,$ that may take any value in a general solution.First seen in the chapter →

degreeघातghaat: The exponent on the highest derivative, once the equation is a polynomial in its derivatives; otherwise not defined.
differential equationअवकल समीकरणavakal sameekaran: An equation involving derivatives of an unknown function.First seen in the chapter →

general solutionव्यापक हलvyaapak hal: A solution with free constants, one for each integration: $n$ of them when the equation has order $n .$

homogeneous differential equationसमघातीय अवकल समीकरणsamaghaateey avakal sameekaran: $\frac{d y}{d x} = F (x, y)$ with $F$ homogeneous of degree zero.
homogeneous functionसमघातीय फलनsamaghaateey phalan: $F$ with $F (λ x, λ y) = λ^{n} F (x, y)$ for every non-zero $λ;$ $n$ is its degree.

integrating factorसमाकलन गुणकsamaakalan gunak: $e^{\int P d x},$ which makes the left side of a linear equation an exact derivative.

linear differential equationरैखिक अवकल समीकरणraikhik avakal sameekaran: A first order equation in which $y$ and $y^{'}$ appear only to the first power, so it reads $y^{'} + P y = Q$ with $P,$ $Q$ depending on $x$ alone.First seen in the chapter →

orderकोटिkoti: The highest number of times the unknown function is differentiated anywhere in the equation.
ordinary differential equationसाधारण अवकल समीकरणsaadhaaran avakal sameekaran: A differential equation with only one independent variable.First seen in the chapter →

particular solutionविशिष्ट हलvishisht hal: A solution free of arbitrary constants, fixed by given conditions.

solutionहलhal: A function that turns the differential equation into an identity when substituted.First seen in the chapter →
solution curveहल वक्रhal vakra: The graph of a solution; it follows the slope the equation demands at each point.First seen in the chapter →

variables separableचर पृथक्करणीयchar prithakkaraneey: Of the form $\frac{d y}{d x} = g (x) h (y),$ so each variable can be moved to its own side.First seen in the chapter →

Differential Equations: Class 12 Maths · Chapter 9

What you will learn in Differential Equations

9.1 Introduction

9.2 Basic Concepts

9.2.1 Order of a differential equation

9.2.2 Degree of a differential equation

Practice 9.1: questions with answers

9.3 General and Particular Solutions of a Differential Equation

Practice 9.2: questions with answers

9.4 Methods of Solving First Order, First Degree Differential Equations

9.4.1 Differential equations with variables separable

Practice 9.3: questions with answers

9.4.2 Homogeneous differential equations

Practice 9.4: questions with answers

9.4.3 Linear differential equations

Practice 9.5: questions with answers

Miscellaneous Examples

Differential Equations: key points and summary

Miscellaneous Exercise on Chapter 9

Practice questions with answers (NCERT “Miscellaneous exercise”)

Find your NCERT exercise question

Quick check: MCQs on Differential Equations

Key terms (English–Hindi)

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