Class 12 Maths NCERT Solutions Chapter 9 Miscellaneous Exercise – Differential Equations

The highest order derivative is \dfrac{d^2y}{dx^2}, so order = 2.

The equation is a polynomial in \dfrac{d^2y}{dx^2} and \dfrac{dy}{dx}, with the highest power of \dfrac{d^2y}{dx^2} being 1.

The highest order derivative is \dfrac{dy}{dx}, so order = 1. The highest power raised to \dfrac{dy}{dx} is 3.

The highest order derivative is \dfrac{d^4y}{dx^4}, so order = 4. But \dfrac{d^3y}{dx^3} sits inside a sine function, so the equation isn't a polynomial in its derivatives.

Differentiating xy=ae^x+be^{-x}+x^2: y+xy'=ae^x-be^{-x}+2x. Differentiating again: 2y'+xy''=ae^x+be^{-x}+2.

Since ae^x+be^{-x}=xy-x^2 from the original equation, substituting gives 2y'+xy''=xy-x^2+2, i.e. xy''+2y'-xy+x^2-2=0. ✓

Differentiating y=e^x(a\cos x+b\sin x) twice using the product rule gives y''=e^x(2b\cos x-2a\sin x), while y'=e^x[(a+b)\cos x+(b-a)\sin x].

Substituting into y''-2y'+2y, the \cos x and \sin x coefficients each cancel to zero, confirming the equation holds. ✓

Differentiating y=x\sin 3x: y'=\sin 3x+3x\cos 3x, and y''=6\cos 3x-9x\sin 3x.

y''+9y-6\cos3x=(6\cos3x-9x\sin3x)+9x\sin3x-6\cos3x=0. ✓

Differentiating x^2=2y^2\log y implicitly: 2x=4yy'\log y+2yy'=2yy'(2\log y+1), so x=yy'(2\log y+1).

From the original equation, 2\log y=x^2/y^2, so 2\log y+1=(x^2+y^2)/y^2. Substituting: x=y'(x^2+y^2)/y, i.e. (x^2+y^2)y'-xy=0. ✓

Differentiating x^2-y^2=c(x^2+y^2)^2 implicitly with respect to x: 2x-2yy'=4c(x^2+y^2)(x+yy').

From the original equation, c=\dfrac{x^2-y^2}{(x^2+y^2)^2}. Substituting this in and multiplying both sides by (x^2+y^2):

(2x-2yy')(x^2+y^2)=4(x^2-y^2)(x+yy'). Expanding both sides and collecting terms:

-2x^3+6xy^2-6x^2yy'+2y^3y'=0, which simplifies to x^3-3xy^2=y'(y^3-3x^2y).

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

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

Separating variables: \dfrac{dy}{y^2+y+1}=-\dfrac{dx}{x^2+x+1}. Both denominators complete the square the same way: t^2+t+1=\left(t+\frac{1}{2}\right)^2+\frac{3}{4}.

Integrating both sides gives \tan^{-1}\left(\dfrac{2y+1}{\sqrt3}\right)+\tan^{-1}\left(\dfrac{2x+1}{\sqrt3}\right)=C.

Applying the tangent addition formula \tan^{-1}A+\tan^{-1}B=\tan^{-1}\left(\dfrac{A+B}{1-AB}\right) with A=\frac{2y+1}{\sqrt3}, B=\frac{2x+1}{\sqrt3}, and simplifying the resulting fraction:

Separating variables: \tan x\,dx=-\tan y\,dy.

Integrating: -\log|\cos x|=\log|\cos y|+C_1, so \cos x\cos y=C.

At \left(0,\frac{\pi}{4}\right): \cos0\cdot\cos\frac{\pi}{4}=1\cdot\frac{1}{\sqrt2}=C.

Separating variables: \dfrac{dy}{1+y^2}=-\dfrac{e^x}{1+e^{2x}}dx. Substituting u=e^x on the right:

Integrating: \tan^{-1}y=-\tan^{-1}(e^x)+C.

At x=0,y=1: \dfrac{\pi}{4}=-\dfrac{\pi}{4}+C\Rightarrow C=\dfrac{\pi}{2}.

This is homogeneous in x/y, so put x=vy, giving \dfrac{dx}{dy}=v+y\dfrac{dv}{dy}.

Substituting: v+y\dfrac{dv}{dy}=\dfrac{vye^v+y^2}{ye^v}=v+ye^{-v}, so e^v\,dv=dy.

Integrating: e^v=y+C. Replacing v=x/y:

Expanding and grouping: dx[(x-y)-1]+dy[(x-y)+1]=0. With t=x-y, this gives \dfrac{dy}{dx}=\dfrac{1-t}{1+t}.

Since \dfrac{dy}{dx}=1-\dfrac{dt}{dx}, substituting: 1-\dfrac{dt}{dx}=\dfrac{1-t}{1+t}\Rightarrow\dfrac{dt}{dx}=\dfrac{2t}{1+t}.

Separating: \left(\dfrac{1}{t}+1\right)dt=2\,dx. Integrating: \log|t|+t=2x+C.

Replacing t=x-y: \log|x-y|=x+y+C. At x=0,y=-1 (t=1): 0=0-1+C\Rightarrow C=1.

Rearranging: \dfrac{dy}{dx}=\dfrac{e^{-2\sqrt{x}}}{\sqrt{x}}-\dfrac{y}{\sqrt{x}}, i.e. \dfrac{dy}{dx}+\dfrac{y}{\sqrt{x}}=\dfrac{e^{-2\sqrt{x}}}{\sqrt{x}} — linear, with P=\dfrac{1}{\sqrt{x}}.

I.F. =e^{\int dx/\sqrt{x}}=e^{2\sqrt{x}}.

y\,e^{2\sqrt{x}}=\displaystyle\int\dfrac{e^{-2\sqrt{x}}}{\sqrt{x}}\cdot e^{2\sqrt{x}}dx+C=\displaystyle\int\dfrac{dx}{\sqrt{x}}+C=2\sqrt{x}+C.

I.F. =e^{\int\cot x\,dx}=e^{\log\sin x}=\sin x.

y\sin x=\displaystyle\int4x\csc x\sin x\,dx+C=\displaystyle\int4x\,dx+C=2x^2+C.

At x=\pi/2,y=0: 0=2\left(\dfrac{\pi}{2}\right)^2+C=\dfrac{\pi^2}{2}+C\Rightarrow C=-\dfrac{\pi^2}{2}.

Separating variables: \dfrac{dy}{2e^{-y}-1}=\dfrac{dx}{x+1}. Multiplying numerator and denominator on the left by e^y and substituting u=e^y:

\dfrac{du}{2-u}=\dfrac{dx}{x+1}. Integrating: -\log|2-u|=\log|x+1|+C_1, so 2-e^y=\dfrac{C}{x+1}.

At x=0,y=0: 2-1=C\Rightarrow C=1. So 2-e^y=\dfrac{1}{x+1}, i.e. e^y=\dfrac{2x+1}{x+1}.

The equation simplifies to y\,dx=x\,dy, i.e. \dfrac{dx}{x}=\dfrac{dy}{y}. Integrating: \log x=\log y+\log C, so x=Cy, equivalently y=\dfrac{1}{C}x.

When the equation is linear in x with P_1,Q_1 functions of y, the integrating factor is e^{\int P_1\,dy}, and the solution follows the same pattern as the standard y-linear case, but with x and y swapped throughout.

Dividing by e^x: \dfrac{dy}{dx}+y=-2xe^{-x}, so I.F. =e^x.

y\,e^x=\displaystyle\int-2xe^{-x}\cdot e^x\,dx+C=\displaystyle\int-2x\,dx+C=-x^2+C.