Miscellaneous Exercise Class 12 Maths Differential Equations

NCERT Solutions Class 12 Maths Chapter 9 Miscellaneous | LearnCBSEHub

💡 Chapter 9 Summary

This exercise combines all concepts:

Order & Degree: Highest derivative and its power.
Verification: Differentiating a solution to match the D.E.
Methods: Variable Separable, Homogeneous Equations, Linear Equations.

Question 01

Find Order and Degree.

(i) $\frac{d^2y}{dx^2} + 5x(\frac{dy}{dx})^2 – 6y = \log x$

Highest derivative is $\frac{d^2y}{dx^2}$ (Order 2). Power is 1 (Degree 1).

Order: 2, Degree: 1

(ii) $(\frac{dy}{dx})^3 – 4(\frac{dy}{dx})^2 + 7y = \sin x$

Highest derivative is $\frac{dy}{dx}$ (Order 1). Highest power is 3 (Degree 3).

Order: 1, Degree: 3

(iii) $\frac{d^4y}{dx^4} – \sin(\frac{d^3y}{dx^3}) = 0$

Order is 4. Since the equation involves $\sin(y”’)$, it is not a polynomial in derivatives.

Order: 4, Degree: Not Defined

Question 02

Verify that the function is a solution.

(i) $xy = ae^x + be^{-x} + x^2$

Differentiate w.r.t x:
$xy’ + y = ae^x – be^{-x} + 2x \dots (1)$
Differentiate again:
$xy” + y’ + y’ = ae^x + be^{-x} + 2$
$xy” + 2y’ = (xy – x^2) + 2$ (using given equation for $ae^x+be^{-x}$)
$xy” + 2y’ – xy + x^2 – 2 = 0$.

Verified

(iii) $y = x \sin 3x$

$y’ = \sin 3x + 3x \cos 3x$.
$y” = 3\cos 3x + 3(\cos 3x – 3x \sin 3x) = 6\cos 3x – 9x \sin 3x$.
LHS $= y” + 9y – 6\cos 3x = (6\cos 3x – 9x \sin 3x) + 9(x \sin 3x) – 6\cos 3x = 0$.

Verified

Questions 03 — 05

Solving Differential Equations.

3. $(x^3 – 3xy^2)dx = (y^3 – 3x^2y)dy$

$\frac{dy}{dx} = \frac{x^3 – 3xy^2}{y^3 – 3x^2y}$. Homogeneous (degree 3).
Put $y=vx \implies v + x\frac{dv}{dx} = \frac{1-3v^2}{v^3-3v}$.
$x\frac{dv}{dx} = \frac{1-3v^2 – v(v^3-3v)}{v^3-3v} = \frac{1-v^4}{v^3-3v}$.
Integrate $\int \frac{v^3-3v}{1-v^4} dv = \int \frac{dx}{x}$.
Result leads to $x^2 – y^2 = c(x^2+y^2)^2$.

4. $\frac{dy}{dx} + \frac{\sqrt{1-y^2}}{\sqrt{1-x^2}} = 0$

Variable Separable: $\frac{dy}{\sqrt{1-y^2}} = -\frac{dx}{\sqrt{1-x^2}}$.
$\sin^{-1}y = -\sin^{-1}x + C \implies \sin^{-1}x + \sin^{-1}y = C$.

5. $\frac{dy}{dx} + \frac{y^2+y+1}{x^2+x+1} = 0$

$\int \frac{dy}{y^2+y+1} = -\int \frac{dx}{x^2+x+1}$. Complete squares.
$\frac{2}{\sqrt{3}}\tan^{-1}\frac{2y+1}{\sqrt{3}} = -\frac{2}{\sqrt{3}}\tan^{-1}\frac{2x+1}{\sqrt{3}} + C$.
Using $\tan^{-1}A + \tan^{-1}B$ formula leads to:
$x+y+1 = A(1-x-y-2xy)$.

Questions 06 — 10

Geometric and Advanced Problems.

6. Curve through $(0, \pi/4)$

$\sin x \cos y dx + \cos x \sin y dy = 0 \implies \tan x dx + \tan y dy = 0$.
$\log|\sec x| + \log|\sec y| = \log C \implies \sec x \sec y = C \implies \cos y = C’ \sec x$.
At $(0, \pi/4): \cos(\pi/4) = C'(1) \implies C’ = 1/\sqrt{2}$.
Ans: $\cos y = \frac{1}{\sqrt{2}}\sec x \implies \sqrt{2}\cos y = \sec x$.

8. Solve $y e^{x/y} dx = (x e^{x/y} + y^2) dy$

Rearrange: $\frac{dx}{dy} = \frac{x e^{x/y} + y^2}{y e^{x/y}} = \frac{x}{y} + \frac{y}{e^{x/y}}$.
Put $x = vy \implies \frac{dx}{dy} = v + y\frac{dv}{dy}$.
$v + y\frac{dv}{dy} = v + y e^{-v} \implies \frac{dv}{dy} = e^{-v}$.
$e^v dv = dy \implies e^v = y + C \implies e^{x/y} = y + C$.

10. Solve $[\frac{e^{-2\sqrt{x}}}{\sqrt{x}} – \frac{y}{\sqrt{x}}] \frac{dx}{dy} = 1$

Flip to $\frac{dy}{dx}$: $\frac{dy}{dx} + \frac{1}{\sqrt{x}}y = \frac{e^{-2\sqrt{x}}}{\sqrt{x}}$. (Linear)
I.F. $= e^{\int x^{-1/2} dx} = e^{2\sqrt{x}}$.
Solution: $y(e^{2\sqrt{x}}) = \int \frac{e^{-2\sqrt{x}}}{\sqrt{x}} \cdot e^{2\sqrt{x}} dx = \int \frac{1}{\sqrt{x}} dx = 2\sqrt{x} + C$.
Ans: $y = (2\sqrt{x} + C)e^{-2\sqrt{x}}$.

Questions 11 — 12

Particular Solutions.

11. $(x-y)(dx+dy) = dx-dy; \quad y=-1, x=0$

Let $x-y=t$. Solving gives $\log(x-y) = x+y+C$.
At $(0,-1): \log(1) = -1+C \implies C=1$.
Ans: $\log(x-y) = x+y+1$.

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

Separable: $\frac{dy}{2e^{-y}-1} = \frac{dx}{x+1}$. Multiply LHS num/den by $e^y$.
$\int \frac{e^y}{2-e^y} dy = \int \frac{dx}{x+1}$. Let $2-e^y=u$.
$-\log|2-e^y| = \log|x+1| + \log C$.
Using initial conditions gives specific C.
Ans: $y = \log|\frac{x+1}{2-e^y}| \dots$ (Simplified form).

Questions 13 — 15

Multiple Choice Questions.

13. General solution of $\frac{y dx – x dy}{y} = 0$

$y dx – x dy = 0 \implies \frac{dx}{x} = \frac{dy}{y}$.
$\log x = \log y + \log C \implies x = Cy$. (Or $y=Cx$).

Correct Option: (C) $y = Cx$

14. General solution of $\frac{dx}{dy} + P_1 x = Q_1$

For a linear equation in $x$, the integrating factor is $e^{\int P_1 dy}$ and solution is $x(I.F.) = \int (Q_1 \cdot I.F.) dy + C$.

Correct Option: (C)

15. General solution of $e^x dy + (y e^x + 2x) dx = 0$

Group terms: $(e^x dy + y e^x dx) + 2x dx = 0$.
$d(y e^x) + 2x dx = 0$. Integrate.
$y e^x + x^2 = C$.

Correct Option: (C) $y e^x + x^2 = C$