Exercise 9.3 Class 12 Maths Differential Equations

NCERT Solutions Class 12 Maths Chapter 9 Ex 9.3 | LearnCBSEHub

💡 Method: Separating Variables

If a differential equation can be expressed in the form $f(x) dx = g(y) dy$, we say the variables are separable.

Solution Steps:

Move all $x$ terms to one side with $dx$.
Move all $y$ terms to the other side with $dy$.
Integrate both sides: $\int f(x) dx = \int g(y) dy + C$.

Questions 01 — 05

Find the General Solution.

1. $\frac{dy}{dx} = \frac{1-\cos x}{1+\cos x}$

Using identities: $\frac{1-\cos x}{1+\cos x} = \frac{2\sin^2(x/2)}{2\cos^2(x/2)} = \tan^2(x/2)$.
$\int dy = \int (\sec^2(x/2) – 1) dx$.
$y = \frac{\tan(x/2)}{1/2} – x + C = 2\tan(x/2) – x + C$.

2. $\frac{dy}{dx} = \sqrt{4-y^2}$

Separate variables: $\int \frac{dy}{\sqrt{2^2-y^2}} = \int dx$.
$\sin^{-1}(\frac{y}{2}) = x + C \implies \frac{y}{2} = \sin(x+C) \implies y = 2\sin(x+C)$.

3. $\frac{dy}{dx} + y = 1$

$\frac{dy}{dx} = 1-y \implies \int \frac{dy}{1-y} = \int dx$.
$-\log|1-y| = x + C \implies \log|1-y| = -x – C$.
$1-y = e^{-x-C} = A e^{-x} \implies y = 1 – A e^{-x}$ (where $A = \pm e^{-C}$).

4. $\sec^2 x \tan y dx + \sec^2 y \tan x dy = 0$

Divide by $\tan x \tan y$: $\frac{\sec^2 x}{\tan x} dx + \frac{\sec^2 y}{\tan y} dy = 0$.
Integrate: $\log|\tan x| + \log|\tan y| = \log C$.
$\log|\tan x \tan y| = \log C \implies \tan x \tan y = C$.

5. $(e^x+e^{-x})dy – (e^x-e^{-x})dx = 0$

$\frac{dy}{dx} = \frac{e^x-e^{-x}}{e^x+e^{-x}} \implies dy = \frac{e^x-e^{-x}}{e^x+e^{-x}} dx$.
Put $e^x+e^{-x}=t \implies (e^x-e^{-x})dx = dt$.
$y = \int \frac{dt}{t} = \log|t| + C = \log(e^x+e^{-x}) + C$.

Questions 06 — 10

Logarithmic and Exponential Forms.

6. $\frac{dy}{dx} = (1+x^2)(1+y^2)$

$\int \frac{dy}{1+y^2} = \int (1+x^2) dx$.
$\tan^{-1} y = x + \frac{x^3}{3} + C$.

7. $y \log y dx – x dy = 0$

$y \log y dx = x dy \implies \frac{dx}{x} = \frac{dy}{y \log y}$.
Integrate RHS: Let $\log y = t \implies \frac{1}{y}dy = dt$.
$\log x = \log(\log y) + \log C \implies x = C \log y \implies y = e^{x/C}$.

9. $\frac{dy}{dx} = \sin^{-1} x$

$y = \int \sin^{-1} x \cdot 1 dx$. (Integration by Parts).
$y = x \sin^{-1} x – \int \frac{x}{\sqrt{1-x^2}} dx$. Let $1-x^2=t$.
$y = x \sin^{-1} x + \sqrt{1-x^2} + C$.

10. $e^x \tan y dx + (1-e^x) \sec^2 y dy = 0$

$\frac{e^x}{1-e^x} dx + \frac{\sec^2 y}{\tan y} dy = 0$.
Integrate: $-\log|1-e^x| + \log|\tan y| = \log C$.
$\frac{\tan y}{1-e^x} = C \implies \tan y = C(1-e^x)$.

Questions 11 — 14

Find Particular Solution given $y(x_0)=y_0$.

11. $(x^3+x^2+x+1)\frac{dy}{dx} = 2x^2+x; \quad y=1, x=0$

$\frac{dy}{dx} = \frac{x(2x+1)}{(x^2+1)(x+1)}$. Using Partial Fractions:
$\int dy = \int (\frac{1}{2}\frac{2x}{x^2+1} + \frac{1}{2}\frac{1}{x^2+1} – \frac{1}{2(x+1)}) dx$.
$y = \frac{1}{2}\log(x^2+1) + \frac{1}{2}\tan^{-1}x – \frac{1}{2}\log|x+1| + C$.
Put $x=0, y=1: 1 = 0 + 0 – 0 + C \implies C=1$.
Ans: $y = \frac{1}{4}\log(\frac{(x^2+1)^2}{(x+1)^2}) + \frac{1}{2}\tan^{-1}x + 1$.

12. $x(x^2-1)\frac{dy}{dx} = 1; \quad y=0, x=2$

$dy = \frac{dx}{x(x-1)(x+1)}$. Partial Fractions: $\frac{-1}{x} + \frac{1/2}{x-1} + \frac{1/2}{x+1}$.
$y = -\log x + \frac{1}{2}\log(x^2-1) + C = \frac{1}{2}\log(\frac{x^2-1}{x^2}) + C$.
At $x=2, y=0: 0 = \frac{1}{2}\log(3/4) + C \implies C = \frac{1}{2}\log(4/3)$.
Ans: $y = \frac{1}{2}\log[\frac{4(x^2-1)}{3x^2}]$.

13. $\cos(\frac{dy}{dx}) = a; \quad y=1, x=0$

$\frac{dy}{dx} = \cos^{-1} a \implies \int dy = \int \cos^{-1} a dx$.
$y = x \cos^{-1} a + C$.
At $x=0, y=1 \implies C=1$.
$y = x \cos^{-1} a + 1 \implies \frac{y-1}{x} = \cos^{-1} a \implies \cos(\frac{y-1}{x}) = a$.

Questions 15 — 18

Curves and Slopes.

15. Curve through (0,0) with slope $y’ = e^x \sin x$

$y = \int e^x \sin x dx$. Using formula from Ex 7.6 Q21:
$y = \frac{e^x}{2}(\sin x – \cos x) + C$.
Passes through (0,0): $0 = \frac{1}{2}(0-1) + C \implies C = 1/2$.
Ans: $2y = e^x(\sin x – \cos x) + 1$.

17. Curve through (0,-2) where (slope $\times$ y) = x

$y \frac{dy}{dx} = x \implies \int y dy = \int x dx$.
$\frac{y^2}{2} = \frac{x^2}{2} + C \implies y^2 – x^2 = 2C$.
At (0,-2): $4 – 0 = 2C \implies 2C = 4$.
Ans: $y^2 – x^2 = 4$ (Hyperbola).

Questions 19 — 22

Growth, Decay, and Rate of Change.

19. Spherical Balloon Volume

$\frac{dV}{dt} = k$. Volume $V = \frac{4}{3}\pi r^3$.
$\frac{d}{dt}(\frac{4}{3}\pi r^3) = k \implies 4\pi r^2 \frac{dr}{dt} = k$.
$\int 4\pi r^2 dr = \int k dt \implies \frac{4}{3}\pi r^3 = kt + C$.
$t=0, r=3: \frac{4}{3}\pi(27) = C \implies C=36\pi$.
$t=3, r=6: \frac{4}{3}\pi(216) = 3k + 36\pi \implies 288\pi = 3k + 36\pi \implies k = 84\pi$.
Equation: $\frac{4}{3}\pi r^3 = 84\pi t + 36\pi$. Divide by $4\pi/3$: $r^3 = 63t + 27$.
Ans: $r = (63t + 27)^{1/3}$.

20. Principal increasing at $r\%$

$\frac{dP}{dt} = \frac{r}{100}P \implies \int \frac{dP}{P} = \int \frac{r}{100} dt$.
$\log P = \frac{rt}{100} + C$.
Given $P$ doubles in 10 years. Let initial $P=P_0$. at $t=10, P=2P_0$.
$\log(2P_0) – \log P_0 = \frac{r(10)}{100} \implies \log 2 = \frac{r}{10}$.
$r = 10 \log 2 = 10(0.6931) = 6.931\%$.

22. Bacteria Growth

$\frac{dN}{dt} = kN \implies N = C e^{kt}$.
$t=0, N=100000 \implies C=100000$. So $N = 100000 e^{kt}$.
$t=2, N=110000 \implies 1.1 = e^{2k} \implies 2k = \log(1.1) \implies k = \frac{1}{2}\log(1.1)$.
Find $t$ when $N=200000$: $2 = e^{kt} \implies kt = \log 2$.
$t = \frac{\log 2}{k} = \frac{2 \log 2}{\log 1.1}$.

Question 23

$\frac{dy}{dx} = e^{x+y}$

$\frac{dy}{dx} = e^x \cdot e^y \implies e^{-y} dy = e^x dx$.
$-e^{-y} = e^x + C \implies e^x + e^{-y} = -C = K$.

Correct Option: (A)