Question

List 1 contains different functions and List 2 contains differential equations. Match options in List 1 with corresponding differential equation from List 2

Answer 1

(A) $e^{\sqrt{x}}+e^{-\sqrt{x}}=y$
$\Rightarrow \frac{dy}{dx}=\frac{1}{2\sqrt{x}}{e}^{\sqrt{x}}-\frac{1}{2\sqrt{x}}{e}^{-\sqrt{x}}$
$\Rightarrow 2\sqrt{x}\frac{dy}{dx}=e^{\sqrt{x}}-e^{-\sqrt{x}}$
$\Rightarrow 2(\sqrt{x}\frac{d^{2}y}{d{x}^2}+\frac{1}{2\sqrt{x}}\frac{dy}{dx})=\frac{1}{2\sqrt{x}}.y$
$\Rightarrow 4x\frac{d^{2}y}{d{x}^2}+2\frac{dy}{dx}-y=0$
(B) $4x^2+9y^2=36\Rightarrow 8x+18y\frac{dy}{dx}=0$
$\Rightarrow \frac{dy}{dx}+\frac{4x}{9y}=0$
(C) $y^2=x^2-cx\Rightarrow 2y\frac{dy}{dx}=2x-c$
$\therefore$ substituting the value of c in the given equation
$y^2-(2x-2y\frac{dy}{dx})x=x^2$
i.e., $2xy\frac{dy}{dx}=3x^2-y^2$
(D) $y=Acos2x+Bsin2x$
$\therefore \frac{dy}{dx}-2Asin2x+2Bcos2x$
$\Rightarrow \frac{d^{2}y}{d{x}^2}=-4Acos2x-4Bsin2x=-4y$
$\Rightarrow \frac{d^{2}y}{d{x}^2}+4y=0$