Question

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

• A. True
• B. False

Answer 1

The correct option is B False

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

$\Rightarrow \frac{dy}{dx}=x^2\frac{d}{dx}\left(\log \left(\frac{1}{x}\right)\right)+\log \left(\frac{1}{x}\right)\frac{d}{dx}\left(x^2\right)$

$\Rightarrow \frac{dy}{dx}=x^2\times \frac{1}{\frac{1}{x}}\frac{d}{dx}\left(\frac{1}{x}\right)+\log \left(\frac{1}{x}\right)2x$

$\Rightarrow \frac{dy}{dx}=-x+2x\log \left(\frac{1}{x}\right)$

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

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

$\Rightarrow x\frac{d^{2}y}{d{x}^2}=-x+2y$

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

$\Rightarrow x\frac{d^{2}y}{d{x}^2}=-x+2y$

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

$=2y+x-2x\log \left(\frac{1}{x}\right)+3x^2\ne 0$

Hence the given statement is false.