Question

Let $0<x<\pi$ and $y(x)$ be given by $(1+\sin x)y^3-(\cos x)y^2+2(1+\sin x)y-2cosx=0$ The derivative of y with respect to $\frac{x}{2}$ at $x=\frac{\pi }{2}$ is

• A. $-\frac{1}{2}$
• B. $\frac{1}{2}$
• C. $-2$
• D. $2$

Answer 1

Answer: (C)

Given equation is $(1+\sin x)y^3-(\cos x)y^2+2(1+\sin x)y-2\cos x=0$ Put $x=\frac{\pi }{2},$ we get $\left(1+\sin \frac{\pi }{2}\right)y^2-\left(\cos \frac{\pi }{2}\right)y^2+2\left(1+\sin \frac{\pi }{2}\right)y-2\cos \frac{\pi }{2}=0$
$\Rightarrow$ $2y^3+4y=0\Rightarrow y=0,\pm \sqrt{2}i.$ Let $\frac{x}{2}=t$
$\therefore$ $(1+\sin 2t)y^3-(\cos 2t)y^2$ $+2(1+\sin 2t)y-2\cos 2t=0$
On differentiating w.r.t.t, we get $3y^2\frac{dy}{dt}(1+\sin 2t)+2y^3(\cos 2t)$ $-2y\frac{dy}{dt}(\cos 2t)+2y^2\sin 2t+2\frac{dy}{dt}(1+\sin 2t)$ $+2y(2\cos 2t)+4\sin 2t=0$ when $2t=\frac{\pi }{2},y=0,$ $2\frac{dy}{dt}(1+1)+4=0$
$\Rightarrow$ $\frac{dy}{dt}=-1$
$\Rightarrow$ $\frac{dy}{\frac{dx}{2}}=-1$
$\Rightarrow$ $\frac{dy}{dx}=-2$