Solveeit Logo
Login

Question

Question: \(y = \frac{x}{x + 1}\)is a solution of the differential equation...

y=xx+1y = \frac{x}{x + 1}is a solution of the differential equation

A

y2dydx=x2y^{2}\frac{dy}{dx} = x^{2}

B

x2dydx=y2x^{2}\frac{dy}{dx} = y^{2}

C

y2dydx=x2y^{2}\frac{dy}{dx} = x^{2}

D

xdydx=yx\frac{dy}{dx} = y

Answer

x2dydx=y2x^{2}\frac{dy}{dx} = y^{2}

Explanation

Solution

We have y=xx+1y = \frac{x}{x + 1} ⇒ 1y=x+1x=1+1x\frac{1}{y} = \frac{x + 1}{x} = 1 + \frac{1}{x}

Differentiating w.r.t. x,

−1y2dydx=0−1x2- \frac{1}{y^{2}}\frac{dy}{dx} = 0 - \frac{1}{x^{2}}

∴ x2dydx=y2x^{2}\frac{dy}{dx} = y^{2}