← 2025 Paper 1

UPSC Maths 2025 Paper 1 Q7c-ii — Solution

10 marks · Section B

Question

Find the complete solution of $x^3\dfrac{d^3 y}{dx^3} + 3x^2\dfrac{d^2 y}{dx^2} + x\dfrac{dy}{dx} + y = x\log x$.

Technique

This is a Cauchy–Euler equation. Substitute $x = e^t$ (so $t=\log x$) to convert it to a constant-coefficient ODE in $t$, using $x^k D^k = \theta(\theta-1)\cdots(\theta-k+1)$ with $\theta = \dfrac{d}{dt}$.

Solution

With $x=e^t$, $\theta=\dfrac{d}{dt}$, the standard operator identities are $xD = \theta,\quad x^2D^2 = \theta(\theta-1),\quad x^3D^3 = \theta(\theta-1)(\theta-2).$

The left side becomes $\theta(\theta-1)(\theta-2) + 3\theta(\theta-1) + \theta + 1.$ Expand: $\theta(\theta-1)(\theta-2) = \theta^3 - 3\theta^2 + 2\theta,\qquad 3\theta(\theta-1) = 3\theta^2 - 3\theta,$ $\text{sum} = \theta^3 - 3\theta^2 + 2\theta + 3\theta^2 - 3\theta + \theta + 1 = \theta^3 + 1.$

So the equation transforms to (RHS: $x\log x = e^t\,t$) $(\theta^3 + 1)\,y = t\,e^t.$

Complementary function. Auxiliary equation $m^3 + 1 = 0 \Rightarrow (m+1)(m^2 - m + 1)=0$, roots $m = -1,\qquad m = \frac{1 \pm i\sqrt3}{2}.$ Thus $y_c = C_1 e^{-t} + e^{t/2}\!\left[C_2\cos\!\frac{\sqrt3}{2}t + C_3\sin\!\frac{\sqrt3}{2}t\right].$ Back to $x$ ($e^t=x$, $t=\log x$): $y_c = \frac{C_1}{x} + \sqrt{x}\left[C_2\cos\!\Bigl(\tfrac{\sqrt3}{2}\log x\Bigr) + C_3\sin\!\Bigl(\tfrac{\sqrt3}{2}\log x\Bigr)\right].$

Particular integral. $y_p = \dfrac{1}{\theta^3+1}\,t\,e^t$. Shift: $\dfrac{1}{\theta^3+1}e^t\,t = e^t\,\dfrac{1}{(\theta+1)^3 + 1}\,t$. Now $(\theta+1)^3 + 1 = \theta^3 + 3\theta^2 + 3\theta + 2$; applied to $t$ keep terms up to first order ($\theta^2 t=\theta^3 t=0$): $\frac{1}{2 + 3\theta + \cdots}\,t = \frac{1}{2}\,\frac{1}{1 + \tfrac{3}{2}\theta + \cdots}\,t = \frac{1}{2}\Bigl(1 - \tfrac{3}{2}\theta + \cdots\Bigr)t = \frac{1}{2}\Bigl(t - \tfrac{3}{2}\Bigr).$ So $y_p = e^t\cdot\frac{1}{2}\Bigl(t - \frac{3}{2}\Bigr) = \frac{x}{2}\Bigl(\log x - \frac{3}{2}\Bigr) = \frac{x\log x}{2} - \frac{3x}{4}.$

Complete solution. $y = \frac{C_1}{x} + \sqrt{x}\left[C_2\cos\!\Bigl(\tfrac{\sqrt3}{2}\log x\Bigr) + C_3\sin\!\Bigl(\tfrac{\sqrt3}{2}\log x\Bigr)\right] + \frac{x\log x}{2} - \frac{3x}{4}.$

Answer

$\boxed{\;y = \frac{C_1}{x} + \sqrt{x}\Bigl[C_2\cos\bigl(\tfrac{\sqrt3}{2}\log x\bigr) + C_3\sin\bigl(\tfrac{\sqrt3}{2}\log x\bigr)\Bigr] + \frac{x\log x}{2} - \frac{3x}{4}.\;}$