CUET Preparation Today
For $y=xe^x$, which of the following is correct ? |
$x\frac{d^2y}{dx^2}-x\frac{dy}{dx}-y=0$ $x\frac{d^2y}{dx^2}-y\frac{dy}{dx}-x=0$ $x\frac{d^2y}{dx^2}+x\frac{dy}{dx}+y=0$ $y\frac{d^2y}{dx^2}-y\frac{dy}{dx}-x=0$ |
$x\frac{d^2y}{dx^2}-x\frac{dy}{dx}-y=0$ |
The correct answer is Option (1) → $x\frac{d^2y}{dx^2}-x\frac{dy}{dx}-y=0$ $y=xe^x$ so $\frac{dy}{dx}=xe^x+e^x$ or $\frac{dy}{dx}=y+e^x$ so $\frac{d^2y}{dx^2}=\frac{dy}{dx}+e^x$ [we know that $\frac{y}{x}=e^x$] so $\frac{d^2y}{dx^2}=\frac{dy}{dx}+\frac{y}{x}$ or $x\frac{d^2y}{dx^2}-x\frac{dy}{dx}-y=0$ |
