UPSC 2017 Maths Optional Paper 1 Q7b-i — Step-by-Step Solution

9 marks · Section B

Particular integral via operator method · ODEs · asked 7× in 13 yrs · Read the full method →

Question

Solve the differential equation $x\dfrac{d^2y}{dx^2}-\dfrac{dy}{dx}-4x^3y=8x^3\sin(x^2)$ .

Substitution $t=x^2$ turns the variable-coefficient ODE into $y_{tt}-y=2\sin t$ ; solve by CF + operator PI.

Let $t=x^2$ , so $\dfrac{dt}{dx}=2x$ . Then

\frac{dy}{dx}=\frac{dy}{dt}\cdot2x,\qquad \frac{d^2y}{dx^2}=\frac{d}{dx}\!\Big(2x\frac{dy}{dt}\Big)=2\frac{dy}{dt}+2x\cdot\frac{d^2y}{dt^2}\cdot2x=2\frac{dy}{dt}+4x^2\frac{d^2y}{dt^2}.

x\Big(2\frac{dy}{dt}+4x^2\frac{d^2y}{dt^2}\Big)-2x\frac{dy}{dt}-4x^3y=8x^3\sin(x^2).

The $2x\,\dfrac{dy}{dt}$ terms cancel:

4x^3\frac{d^2y}{dt^2}-4x^3y=8x^3\sin(x^2).

Divide by $4x^3$ (and $x^2=t$ ):

\frac{d^2y}{dt^2}-y=2\sin t.

Complementary function: $m^2-1=0\Rightarrow m=\pm1$ , so $y_c=C_1e^{t}+C_2e^{-t}$ .

Particular integral for $2\sin t$ (with $\phi(D)=D^2-1$ , $D^2\to-1$ ):

y_p=\frac{1}{D^2-1}\,2\sin t=\frac{2\sin t}{-1-1}=-\sin t.

Thus $y=C_1e^{t}+C_2e^{-t}-\sin t.$

\boxed{\;y=C_1e^{x^2}+C_2e^{-x^2}-\sin(x^2).\;}