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UPSC Maths 2013 Paper 1 Q6c — Solution

15 marks · Section B

Question

Find the general solution of the equation

Technique

Cauchy-Euler $x=e^{t}$ substitution; constant-coefficient ODE with resonant RHS; complex-method particular solution ($y=e^{it}u$ followed by integrating-factor solution of $u''+2iu'=t$).

Solution

Strategy. Cauchy-Euler equation: substitute $x=e^{t}$ to convert to a constant-coefficient ODE. The RHS becomes $t\sin t$ — resonant with the homogeneous solution, requiring a higher-order ansatz.

Step 1 — Substitute $x=e^{t}$, $t=\ln x$

With $D=d/dt$:

The ODE becomes

Step 2 — Homogeneous (complementary) solution

$D^{2}+1=0$ gives $D=\pm i$, so

Step 3 — Particular solution via complex method

Write $t\sin t=\operatorname{Im}(t\,e^{it})$. Solve $(D^{2}+1)y=t\,e^{it}$; take the imaginary part for the real solution.

The RHS contains the resonant factor $e^{it}$. Substitute $y=e^{it}u$. Then

(computed via $y'=e^{it}(u'+iu),\;y''=e^{it}(u''+2iu'-u)$, then $y''+y=e^{it}(u''+2iu')$.)

Setting $=t\,e^{it}$:

Let $v=u'$: $v'+2iv=t$. Integrating factor $e^{2it}$ gives $(e^{2it}v)'=te^{2it}$.

Integrating by parts:

So $v=\dfrac{t}{2i}+\dfrac{1}{4}+(\text{transient})=-\dfrac{it}{2}+\dfrac{1}{4}$ (drop homogeneous part).

Integrate: $u=-\dfrac{it^{2}}{4}+\dfrac{t}{4}$.

Hence

Multiply out and take imaginary part:

Step 4 — Verify

$y_p=\dfrac{t\sin t}{4}-\dfrac{t^{2}\cos t}{4}$.

$y_p'=\dfrac{\sin t+t\cos t}{4}-\dfrac{2t\cos t-t^{2}\sin t}{4}=\dfrac{\sin t-t\cos t+t^{2}\sin t}{4}$.

$y_p''=\dfrac{\cos t-(\cos t-t\sin t)+2t\sin t+t^{2}\cos t}{4}=\dfrac{3t\sin t+t^{2}\cos t}{4}$.

$y_p''+y_p=\dfrac{3t\sin t+t^{2}\cos t}{4}+\dfrac{t\sin t-t^{2}\cos t}{4}=\dfrac{4t\sin t}{4}=t\sin t$ ✓.

Step 5 — Back-substitute and write general solution

Replace $t=\ln x$:

Answer