Question

If $y=x^3\log \left(\frac{1}{x}\right)$, then prove that $x\frac{d^{2}y}{d{x}^2}-\frac{2dy}{dx}+3x^2=0$

Answer 1

Given,

$y=x^3\log \frac{1}{x}$

Differentiate on both sides w.r.t $x$

$\frac{dy}{dx}=3x^2\log \frac{1}{x}+x^3\frac{1}{\frac{1}{x}}\frac{-1}{x^2}$

$\frac{dy}{dx}=x^2(3\log \frac{1}{x}-1)$

Again differentiating on both sides w.r.t $x$

$\frac{d^{2}y}{d{x}^2}=2x(3\log \frac{1}{x}-1)+3x^2\left(\frac{-1}{x^2}\right)\left(\frac{1}{\frac{1}{x}}\right)$

$=2x(3\log \frac{1}{x}-1)-3x$

$x\frac{d^{2}y}{d{x}^2}-2\frac{dy}{dx}=6x^2\log \frac{1}{x}-5x^2-6x^2\log \frac{1}{x}+2x^2$

$x\frac{d^{2}y}{d{x}^2}-2\frac{dy}{dx}+3x^2=0$