Class 12 Maths NCERT Solutions Chapter 9 Ex 9.3 – Differential Equations with Variables Separable | Boundless Maths

Ex 9.3 Class 12 Maths NCERT Solutions

Class 12 Maths NCERT Solutions Chapter 9 Ex 9.3 – Differential Equations with Variables Separable

This Class 12 Maths NCERT Solutions Chapter 9 Ex 9.3 page covers all 23 questions, solved step-by-step — rearranging a differential equation so every y-term sits with dy and every x-term sits with dx, then integrating both sides independently to build the general or particular solution.

Questions 1–10 build the core skill on a range of forms — trigonometric identities, exponential functions, log y, and inverse trig integrals. Questions 11–14 add an initial condition to pin down the arbitrary constant and get a particular solution, including the classic partial-fractions setup in Q11 and Q12. Questions 15–18 turn the method into curve-fitting problems: given a slope condition and a point the curve passes through, find its equation. The exercise closes with four real-world rate problems — a balloon inflating at a constant rate, continuous compound interest (twice, in Q20 and Q21), and exponential bacterial growth in Q22 — all of which reduce to the same separable-variables technique, plus a quick MCQ.

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23Questions

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2026-27CBSE Syllabus

Class 12 Maths NCERT Solutions Chapter 9 Ex 9.3 — All 23 Questions

Find the general solution: \dfrac{dy}{dx}=\dfrac{1-\cos x}{1+\cos x}

Medium +

Solution

Using half-angle identities, 1-\cos x=2\sin^2\frac{x}{2} and 1+\cos x=2\cos^2\frac{x}{2}, so \dfrac{dy}{dx}=\tan^2\dfrac{x}{2}=\sec^2\dfrac{x}{2}-1.

Integrating: y=\displaystyle\int\left(\sec^2\dfrac{x}{2}-1\right)dx=2\tan\dfrac{x}{2}-x+C.

Answer: y = 2tan(x/2) − x + C

Find the general solution: \dfrac{dy}{dx}=\sqrt{4-y^2} (-2 \lt y \lt 2)

Easy +

Solution

Separating variables: \dfrac{dy}{\sqrt{4-y^2}}=dx.

Integrating: \sin^{-1}\dfrac{y}{2}=x+C.

Answer: sin⁻¹(y/2) = x + C

Find the general solution: \dfrac{dy}{dx}+y=1 (y\neq 1)

Easy +

Solution

Separating variables: \dfrac{dy}{1-y}=dx.

Integrating: -\log|1-y|=x+C_1, so 1-y=Ce^{-x}.

Answer: y = 1 − Ce⁻ˣ

Find the general solution: \sec^2x\tan y\,dx+\sec^2y\tan x\,dy=0

Medium +

Solution

Separating variables: \dfrac{\sec^2x}{\tan x}dx=-\dfrac{\sec^2y}{\tan y}dy.

Answer: tan x · tan y = C

Find the general solution: (e^x+e^{-x})dy-(e^x-e^{-x})dx=0

Medium +

Solution

Separating variables: dy=\dfrac{e^x-e^{-x}}{e^x+e^{-x}}dx.

The right side is the derivative of \log(e^x+e^{-x}), so integrating gives y=\log(e^x+e^{-x})+C.

Answer: y = log(eˣ + e⁻ˣ) + C

Find the general solution: \dfrac{dy}{dx}=(1+x^2)(1+y^2)

Easy +

Solution

Separating variables: \dfrac{dy}{1+y^2}=(1+x^2)dx.

Integrating: \tan^{-1}y=x+\dfrac{x^3}{3}+C.

Answer: tan⁻¹y = x + x³/3 + C

Find the general solution: y\log y\,dx-x\,dy=0

Medium +

Solution

Separating variables: \dfrac{dy}{y\log y}=\dfrac{dx}{x}.

Integrating (using t=\log y on the left): \log|\log y|=\log|x|+C, so \log y=Cx.

Answer: log y = Cx

Find the general solution: x^5\dfrac{dy}{dx}=-y^5

Medium +

Solution

Separating variables: \dfrac{dy}{y^5}=-\dfrac{dx}{x^5}.

Integrating: -\dfrac{1}{4y^4}=\dfrac{1}{4x^4}+C_1.

Answer: 1/x⁴ + 1/y⁴ = C

Find the general solution: \dfrac{dy}{dx}=\sin^{-1}x

Medium +

Solution

dy=\sin^{-1}x\,dx. Integrating by parts (taking \sin^{-1}x as the first function): y=\displaystyle\int\sin^{-1}x\,dx=x\sin^{-1}x+\sqrt{1-x^2}+C.

Answer: y = x sin⁻¹x + √(1−x²) + C

Find the general solution: e^x\tan y\,dx+(1-e^x)\sec^2y\,dy=0

Medium +

Solution

Separating variables: \dfrac{\sec^2y}{\tan y}dy=-\dfrac{e^x}{1-e^x}dx.

Integrating: \log|\tan y|=\log|1-e^x|+C_1, so \tan y=C(1-e^x).

Answer: tan y = C(1 − eˣ)

Find a particular solution: (x^3+x^2+x+1)\dfrac{dy}{dx}=2x^2+x; y=1 when x=0.

Hard +

Solution

Factor: x^3+x^2+x+1=x^2(x+1)+(x+1)=(x+1)(x^2+1), so dy=\dfrac{2x^2+x}{(x+1)(x^2+1)}dx.

By partial fractions, \dfrac{2x^2+x}{(x+1)(x^2+1)}=\dfrac{1/2}{x+1}+\dfrac{\frac{3}{2}x-\frac{1}{2}}{x^2+1}.

Integrating: y=\dfrac{1}{2}\log|x+1|+\dfrac{3}{4}\log(x^2+1)-\dfrac{1}{2}\tan^{-1}x+C.

At x=0,y=1: all the log and arctan terms vanish, so C=1.

Answer: y = (1/2)log|x+1| + (3/4)log(x²+1) − (1/2)tan⁻¹x + 1

Find a particular solution: x(x^2-1)\dfrac{dy}{dx}=1; y=0 when x=2.

Hard +

Solution

dy=\dfrac{dx}{x(x-1)(x+1)}. By partial fractions, \dfrac{1}{x(x-1)(x+1)}=-\dfrac{1}{x}+\dfrac{1/2}{x-1}+\dfrac{1/2}{x+1}.

Integrating: y=-\log|x|+\dfrac{1}{2}\log|x^2-1|+C.

At x=2,y=0: 0=-\log 2+\dfrac{1}{2}\log 3+C, so C=\log 2-\dfrac{1}{2}\log 3.

Answer: y = (1/2)log|x²−1| − log|x| + log 2 − (1/2)log 3

Find a particular solution: \cos\left(\dfrac{dy}{dx}\right)=a (a\in\mathbb{R}); y=1 when x=0.

Medium +

Solution

Since a is a constant, \dfrac{dy}{dx}=\cos^{-1}a, itself a constant.

Integrating: y=x\cos^{-1}a+C. At x=0,y=1: C=1.

Answer: y = x cos⁻¹a + 1

Find a particular solution: \dfrac{dy}{dx}=y\tan x; y=1 when x=0.

Easy +

Solution

Separating variables: \dfrac{dy}{y}=\tan x\,dx.

Integrating: \log|y|=\log|\sec x|+C_1, so y=C\sec x. At x=0,y=1: C=1.

Answer: y = sec x

Find the equation of a curve passing through the point (0,0) and whose differential equation is y'=e^x\sin x.

Hard +

Solution

y=\displaystyle\int e^x\sin x\,dx. Using the standard reduction (integrating by parts twice), \displaystyle\int e^x\sin x\,dx=\dfrac{e^x(\sin x-\cos x)}{2}+C.

At x=0,y=0: 0=\dfrac{1(0-1)}{2}+C, so C=\dfrac{1}{2}.

Answer: y = [eˣ(sin x − cos x) + 1] / 2

For the differential equation xy\dfrac{dy}{dx}=(x+2)(y+2), find the solution curve passing through the point (1,-1).

Hard +

Solution

Separating variables: \dfrac{y}{y+2}dy=\dfrac{x+2}{x}dx, i.e. \left(1-\dfrac{2}{y+2}\right)dy=\left(1+\dfrac{2}{x}\right)dx.

Integrating: y-2\log|y+2|=x+2\log|x|+C, i.e. y-x=2\log|x(y+2)|+C.

At (1,-1): y+2=1, so -1-1=2\log(1)+C\Rightarrow C=-2.

Answer: y − x + 2 = 2 log|x(y + 2)|

Find the equation of a curve passing through the point (0,-2) given that at any point (x,y) on the curve, the product of the slope of its tangent and y coordinate of the point is equal to the x coordinate of the point.

Medium +

Solution

The condition translates to y\dfrac{dy}{dx}=x, i.e. y\,dy=x\,dx.

Integrating: \dfrac{y^2}{2}=\dfrac{x^2}{2}+C_1, so y^2-x^2=C.

At (0,-2): 4-0=C\Rightarrow C=4.

Answer: y² − x² = 4

At any point (x,y) of a curve, the slope of the tangent is twice the slope of the line segment joining the point of contact to the point (-4,-3). Find the equation of the curve given that it passes through (-2,1).

Hard +

Solution

The slope of the segment joining (x,y) to (-4,-3) is \dfrac{y+3}{x+4}, so \dfrac{dy}{dx}=\dfrac{2(y+3)}{x+4}.

Separating variables: \dfrac{dy}{y+3}=\dfrac{2\,dx}{x+4}. Integrating: \log|y+3|=2\log|x+4|+C_1, so y+3=C(x+4)^2.

At (-2,1): 1+3=C(2)^2\Rightarrow C=1.

Answer: y = (x + 4)² − 3

The volume of a spherical balloon being inflated changes at a constant rate. If initially its radius is 3 units and after 3 seconds it is 6 units, find the radius of the balloon after t seconds.

Hard +

Solution

Let \dfrac{dV}{dt}=k (constant), where V=\dfrac{4}{3}\pi r^3. Differentiating, 4\pi r^2\dfrac{dr}{dt}=k, so r^2\,dr=\dfrac{k}{4\pi}dt.

Integrating: \dfrac{r^3}{3}=\dfrac{k}{4\pi}t+C. At t=0,r=3: C=9.

At t=3,r=6: 72=\dfrac{3k}{4\pi}+9\Rightarrow \dfrac{k}{4\pi}=21.

So \dfrac{r^3}{3}=21t+9, i.e. r^3=63t+27.

Answer: r = (63t + 27)^(1/3)

In a bank, principal increases continuously at the rate of r\% per year. Find the value of r if Rs 100 double itself in 10 years (\log_e 2=0.6931).

Medium +

Solution

\dfrac{dP}{dt}=\dfrac{r}{100}P. Separating and integrating: \log P=\dfrac{r}{100}t+C, i.e. P=P_0e^{(r/100)t}.

Doubling means 2P_0=P_0e^{(r/100)(10)}, so \log 2=\dfrac{r}{10}, giving r=10\log_e 2=10(0.6931).

Answer: r = 6.93% (approximately)

In a bank, principal increases continuously at the rate of 5\% per year. An amount of Rs 1000 is deposited with this bank, how much will it be worth after 10 years (e^{0.5}=1.648)?

Medium +

Solution

As in the previous question, P=P_0e^{(5/100)t}=1000e^{0.05t}.

At t=10: P=1000e^{0.5}=1000(1.648).

Answer: Rs 1648

In a culture, the bacteria count is 1{,}00{,}000. The number is increased by 10\% in 2 hours. In how many hours will the count reach 2{,}00{,}000, if the rate of growth of bacteria is proportional to the number present?

Hard +

Solution

\dfrac{dN}{dt}=kN\Rightarrow N=N_0e^{kt}.

At t=2, N=1.1N_0: 1.1N_0=N_0e^{2k}\Rightarrow k=\dfrac{1}{2}\log(1.1).

For N=2N_0: 2N_0=N_0e^{kt}\Rightarrow kt=\log 2\Rightarrow t=\dfrac{\log 2}{k}=\dfrac{2\log 2}{\log 1.1}.

Answer: t = (2 log 2) / (log 1.1) hours

The general solution of the differential equation \dfrac{dy}{dx}=e^{x+y} is

Easy +

Solution

(A) eˣ + e⁻ʸ = C (B) eˣ + eʸ = C (C) e⁻ˣ + eʸ = C (D) e⁻ˣ + e⁻ʸ = C

Since e^{x+y}=e^x\cdot e^y, separating variables gives e^{-y}dy=e^x\,dx.

Integrating: -e^{-y}=e^x+C_1, i.e. e^x+e^{-y}=C.

Answer: (A) eˣ + e⁻ʸ = C

Common Questions

FAQs — Class 12 Maths NCERT Solutions Chapter 9 Ex 9.3

How many questions are there in Exercise 9.3?

Exercise 9.3 has 23 questions — 10 on finding the general solution, 4 on finding a particular solution given an initial condition, 8 applied problems (curves, balloon inflation, compound interest and bacterial growth), and a final MCQ.

What does "variables separable" mean?

A differential equation is of variable separable type when it can be rearranged so that every term involving y (and dy) is on one side and every term involving x (and dx) is on the other. Once separated, both sides can be integrated independently to get the general solution.

Where can I find the official NCERT textbook for this exercise?

Exercise 9.3 is from Chapter 9, Differential Equations, in the NCERT Class 12 Mathematics textbook (Part II), published by the National Council of Educational Research and Training (NCERT) and prescribed by CBSE. You can download the official textbook PDF directly from ncert.nic.in, NCERT's official website — the solutions on this page follow the questions exactly as they appear there.

Class 12 Maths NCERT Solutions Chapter 9 Ex 9.3 – Differential Equations with Variables Separable

Class 12 Maths NCERT Solutions Chapter 9 Ex 9.3 — All 23 Questions

Find the general solution: \dfrac{dy}{dx}=\dfrac{1-\cos x}{1+\cos x}

Find the general solution: \dfrac{dy}{dx}=\sqrt{4-y^2} (-2 \lt y \lt 2)

Find the general solution: \dfrac{dy}{dx}+y=1 (y\neq 1)

Find the general solution: \sec^2x\tan y\,dx+\sec^2y\tan x\,dy=0

Find the general solution: (e^x+e^{-x})dy-(e^x-e^{-x})dx=0

Find the general solution: \dfrac{dy}{dx}=(1+x^2)(1+y^2)

Find the general solution: y\log y\,dx-x\,dy=0

Find the general solution: x^5\dfrac{dy}{dx}=-y^5

Find the general solution: \dfrac{dy}{dx}=\sin^{-1}x

Find the general solution: e^x\tan y\,dx+(1-e^x)\sec^2y\,dy=0

Know exactly which chapters are costing you marks

Find a particular solution: (x^3+x^2+x+1)\dfrac{dy}{dx}=2x^2+x; y=1 when x=0.

Find a particular solution: x(x^2-1)\dfrac{dy}{dx}=1; y=0 when x=2.

Find a particular solution: \cos\left(\dfrac{dy}{dx}\right)=a (a\in\mathbb{R}); y=1 when x=0.

Find a particular solution: \dfrac{dy}{dx}=y\tan x; y=1 when x=0.

Find the equation of a curve passing through the point (0,0) and whose differential equation is y'=e^x\sin x.

For the differential equation xy\dfrac{dy}{dx}=(x+2)(y+2), find the solution curve passing through the point (1,-1).

Find the equation of a curve passing through the point (0,-2) given that at any point (x,y) on the curve, the product of the slope of its tangent and y coordinate of the point is equal to the x coordinate of the point.

At any point (x,y) of a curve, the slope of the tangent is twice the slope of the line segment joining the point of contact to the point (-4,-3). Find the equation of the curve given that it passes through (-2,1).

The volume of a spherical balloon being inflated changes at a constant rate. If initially its radius is 3 units and after 3 seconds it is 6 units, find the radius of the balloon after t seconds.

In a bank, principal increases continuously at the rate of r\% per year. Find the value of r if Rs 100 double itself in 10 years (\log_e 2=0.6931).

In a bank, principal increases continuously at the rate of 5\% per year. An amount of Rs 1000 is deposited with this bank, how much will it be worth after 10 years (e^{0.5}=1.648)?

In a culture, the bacteria count is 1{,}00{,}000. The number is increased by 10\% in 2 hours. In how many hours will the count reach 2{,}00{,}000, if the rate of growth of bacteria is proportional to the number present?

The general solution of the differential equation \dfrac{dy}{dx}=e^{x+y} is

FAQs — Class 12 Maths NCERT Solutions Chapter 9 Ex 9.3

How many questions are there in Exercise 9.3?

What does "variables separable" mean?

Where can I find the official NCERT textbook for this exercise?

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