Exercise 9.5 Class 12 Maths Differential Equations

NCERT Solutions Class 12 Maths Chapter 9 Ex 9.5 | LearnCBSEHub

💡 The Method of Integrating Factors

Form 1: $\frac{dy}{dx} + Py = Q$

Integrating Factor (I.F.) $= e^{\int P dx}$
Solution: $y(\text{I.F.}) = \int (Q \cdot \text{I.F.}) dx + C$

Form 2: $\frac{dx}{dy} + P_1 x = Q_1$

Integrating Factor (I.F.) $= e^{\int P_1 dy}$
Solution: $x(\text{I.F.}) = \int (Q_1 \cdot \text{I.F.}) dy + C$

Questions 01 — 05

Find the General Solution.

1. $\frac{dy}{dx} + 2y = \sin x$

Here $P=2, Q=\sin x$.
I.F. $= e^{\int 2 dx} = e^{2x}$.
Solution: $y e^{2x} = \int e^{2x} \sin x dx$.
Use formula $\int e^{ax}\sin bx dx = \frac{e^{ax}}{a^2+b^2}(a\sin bx – b\cos bx)$.
$y e^{2x} = \frac{e^{2x}}{5}(2\sin x – \cos x) + C \implies y = \frac{1}{5}(2\sin x – \cos x) + C e^{-2x}$.

2. $\frac{dy}{dx} + 3y = e^{-2x}$

Here $P=3, Q=e^{-2x}$.
I.F. $= e^{\int 3 dx} = e^{3x}$.
$y e^{3x} = \int e^{-2x} e^{3x} dx = \int e^x dx = e^x + C$.
$y = e^{-2x} + C e^{-3x}$.

3. $\frac{dy}{dx} + \frac{y}{x} = x^2$

Here $P = 1/x, Q = x^2$.
I.F. $= e^{\int \frac{1}{x} dx} = e^{\log x} = x$.
$y(x) = \int (x^2 \cdot x) dx = \int x^3 dx = \frac{x^4}{4} + C$.
$y = \frac{x^3}{4} + \frac{C}{x}$.

4. $\frac{dy}{dx} + (\sec x)y = \tan x$

$P=\sec x, Q=\tan x$.
I.F. $= e^{\int \sec x dx} = e^{\log(\sec x + \tan x)} = \sec x + \tan x$.
$y(\sec x+\tan x) = \int \tan x(\sec x+\tan x) dx = \int (\sec x \tan x + \tan^2 x) dx$.
$= \int (\sec x \tan x + \sec^2 x – 1) dx = \sec x + \tan x – x + C$.
$y(\sec x+\tan x) = \sec x + \tan x – x + C$.

5. $\cos^2 x \frac{dy}{dx} + y = \tan x$

Divide by $\cos^2 x$: $\frac{dy}{dx} + (\sec^2 x)y = \sec^2 x \tan x$.
I.F. $= e^{\int \sec^2 x dx} = e^{\tan x}$.
$y e^{\tan x} = \int e^{\tan x} \sec^2 x \tan x dx$. Put $\tan x = t$.
$= \int t e^t dt = t e^t – e^t + C = e^{\tan x}(\tan x – 1) + C$.
$y = \tan x – 1 + C e^{-\tan x}$.

Questions 06 — 10

Logarithmic and Inverse Forms.

6. $x \frac{dy}{dx} + 2y = x^2 \log x$

$\frac{dy}{dx} + \frac{2}{x}y = x \log x$. I.F. $= e^{2\log x} = x^2$.
$y(x^2) = \int x^3 \log x dx$. (Use Integration by Parts)
$y x^2 = \frac{x^4}{4}\log x – \frac{x^4}{16} + C$.

7. $x \log x \frac{dy}{dx} + y = \frac{2}{x} \log x$

$\frac{dy}{dx} + \frac{1}{x \log x}y = \frac{2}{x^2}$.
I.F. $= e^{\int \frac{dx}{x \log x}} = e^{\log(\log x)} = \log x$.
$y \log x = \int \frac{2}{x^2} \log x \cdot \log x$? No, RHS becomes $\frac{2}{x^2}\log x \cdot \text{Wait}$.
Check RHS: $\frac{2}{x}\log x \cdot \frac{1}{x \log x} \dots$ No, divide equation by $x \log x$.
RHS becomes $\frac{2}{x^2}$. Solution: $y \log x = \int \frac{2}{x^2} \log x dx$.
$= -2\frac{\log x}{x} – \frac{2}{x} + C$.

9. $x \frac{dy}{dx} + y – x + xy \cot x = 0$

Rearrange: $x \frac{dy}{dx} + y(1 + x \cot x) = x$.
$\frac{dy}{dx} + (\frac{1}{x} + \cot x)y = 1$.
I.F. $= e^{\log x + \log \sin x} = x \sin x$.
$y(x \sin x) = \int x \sin x dx = -x \cos x + \sin x + C$.
$y = \frac{1}{x} – \cot x + \frac{C}{x \sin x}$.

10. $(x+y) \frac{dy}{dx} = 1$

Flip: $\frac{dx}{dy} = x + y \implies \frac{dx}{dy} – x = y$. (Linear in x)
I.F. $= e^{\int -1 dy} = e^{-y}$.
$x e^{-y} = \int y e^{-y} dy = -y e^{-y} – e^{-y} + C$.
$x = -y – 1 + C e^y$.

Questions 11 — 12

Form $\frac{dx}{dy} + P_1 x = Q_1$.

11. $y dx + (x-y^2) dy = 0$

$y \frac{dx}{dy} + x = y^2 \implies \frac{dx}{dy} + \frac{1}{y}x = y$.
I.F. $= e^{\log y} = y$.
$x(y) = \int y \cdot y dy = \frac{y^3}{3} + C$.
$x = \frac{y^2}{3} + \frac{C}{y}$.

12. $(x+3y^2) \frac{dy}{dx} = y$

$\frac{dx}{dy} = \frac{x+3y^2}{y} = \frac{x}{y} + 3y \implies \frac{dx}{dy} – \frac{1}{y}x = 3y$.
I.F. $= e^{-\log y} = \frac{1}{y}$.
$x(\frac{1}{y}) = \int 3y (\frac{1}{y}) dy = \int 3 dy = 3y + C$.
$x = 3y^2 + Cy$.

Questions 13 — 15

Particular Solutions (Find C).

13. $\frac{dy}{dx} + 2y \tan x = \sin x; \quad y=0, x=\pi/3$

I.F. $= e^{2\log \sec x} = \sec^2 x$.
$y \sec^2 x = \int \sin x \sec^2 x dx = \int \sec x \tan x dx = \sec x + C$.
At $x=\pi/3, y=0: 0 = 2 + C \implies C = -2$.
Ans: $y = \cos x – 2\cos^2 x$.

14. $(1+x^2)\frac{dy}{dx} + 2xy = \frac{1}{1+x^2}; \quad y=0, x=1$

$\frac{dy}{dx} + \frac{2x}{1+x^2}y = \frac{1}{(1+x^2)^2}$.
I.F. $= 1+x^2$.
$y(1+x^2) = \int \frac{1}{1+x^2} dx = \tan^{-1}x + C$.
At $x=1, y=0: 0 = \pi/4 + C \implies C = -\pi/4$.
Ans: $y(1+x^2) = \tan^{-1}x – \frac{\pi}{4}$.

15. $\frac{dy}{dx} – 3y \cot x = \sin 2x; \quad y=2, x=\pi/2$

I.F. $= e^{-3\log \sin x} = \text{cosec}^3 x$.
$y \text{cosec}^3 x = \int \sin 2x \frac{1}{\sin^3 x} dx = \int \frac{2\sin x \cos x}{\sin^3 x} dx = 2\int \cot x \text{cosec} x dx$.
$y \text{cosec}^3 x = -2 \text{cosec} x + C$.
At $x=\pi/2, y=2: 2(1) = -2(1) + C \implies C=4$.
Ans: $y = 2\sin^2 x + 4\sin^3 x$.

Questions 16 — 17

Geometric Problems.

16. Slope = Sum of coordinates. Origin (0,0).

$\frac{dy}{dx} = x + y \implies \frac{dy}{dx} – y = x$.
I.F. $= e^{-x}$.
$y e^{-x} = \int x e^{-x} dx = -x e^{-x} – e^{-x} + C$.
$y + x + 1 = C e^x$. At (0,0): $1=C$.
Ans: $x + y + 1 = e^x$.

17. Sum of coords exceeds slope by 5. Point (0,2).

$x + y = \frac{dy}{dx} + 5 \implies \frac{dy}{dx} – y = x – 5$.
I.F. $= e^{-x}$.
$y e^{-x} = \int (x-5)e^{-x} dx = -(x-5)e^{-x} – e^{-x} + C$.
$y = -(x-5) – 1 + C e^x = 4 – x + C e^x$.
At (0,2): $2 = 4 + C \implies C = -2$.
Ans: $y = 4 – x – 2e^x$.

Questions 18 — 19

Multiple Choice Questions.

18. Integrating Factor of $x \frac{dy}{dx} – y = 2x^2$

$\frac{dy}{dx} – \frac{1}{x}y = 2x$.
I.F. $= e^{\int -1/x dx} = e^{-\log x} = \frac{1}{x}$.

Correct Option: (C)

19. Integrating Factor of $(1-y^2)\frac{dx}{dy} + yx = ay$

$\frac{dx}{dy} + \frac{y}{1-y^2}x = \dots$
I.F. $= e^{\int \frac{y}{1-y^2} dy} = e^{-\frac{1}{2}\log(1-y^2)} = (1-y^2)^{-1/2} = \frac{1}{\sqrt{1-y^2}}$.

Correct Option: (D)