UPSC 2024 Maths Optional Paper 1 Q6c-ii — Step-by-Step Solution

10 marks · Section B

Method of variation of parameters · ODEs · asked 12× in 14 yrs · Read the full method →

Question

Find the general solution of $x^2 y''-2xy'+2y=x^3\sin x$ by the method of variation of parameters.

Solve the Euler–Cauchy homogeneous equation; convert to standard form (divide by $x^2$ ); apply the variation-of-parameters formula.

Step 1 — Homogeneous solution.

Try $y=x^m$ : $m(m-1)-2m+2=m^2-3m+2=0\Rightarrow m=1,2$ .

Complementary function: $y_1=x$ , $y_2=x^2$ .

Step 2 — Standard form.

Divide the ODE by $x^2$ : $y''-\dfrac{2}{x}y'+\dfrac{2}{x^2}y=x\sin x$ .

The right-hand side in standard form is $g(x)=x\sin x$ .

Step 3 — Wronskian.

W=\begin{vmatrix}x & x^2\\1 & 2x\end{vmatrix}=2x^2-x^2=x^2.

Step 4 — Particular solution.

y_p=-y_1\int\frac{y_2 g}{W}\,dx+y_2\int\frac{y_1 g}{W}\,dx.

\frac{y_2 g}{W}=\frac{x^2\cdot x\sin x}{x^2}=x\sin x,\qquad \frac{y_1 g}{W}=\frac{x\cdot x\sin x}{x^2}=\sin x.

\int x\sin x\,dx=-x\cos x+\sin x,\qquad\int\sin x\,dx=-\cos x.

y_p=-x(-x\cos x+\sin x)+x^2(-\cos x)=x^2\cos x-x\sin x-x^2\cos x=-x\sin x.

\boxed{\;y(x)=c_1 x+c_2 x^2-x\sin x.\;}