Question

The curve satisfying $\frac{dy}{dx}=\frac{\sin y+x}{\sin 2y-x\cos y}$ passes through $(1,\pi )$ and $(h,\frac{\pi }{2})$ then the maximum value of $|h|$ is

Answer 1

$\frac{dy}{dx}=\frac{\sin y+x}{\sin 2y-x\cos y}\Rightarrow \cos y\frac{dy}{dx}=\frac{\sin y+x}{2\sin y-x}$

Let $\sin y=t\Rightarrow dt=\cos ydy$
$\Rightarrow \frac{dt}{dx}=\frac{t+x}{2t-x}$
Assuming $t=vx\Rightarrow \frac{dt}{dx}=v+x\frac{dv}{dx}$
$\Rightarrow v+x\frac{dv}{dx}=\frac{v+1}{2v-1}\Rightarrow x\frac{dv}{dx}=\frac{-2v^2+2v+1}{2v-1}\Rightarrow \int \frac{2v-1}{-2v^2+2v+1}dv=\int \frac{dx}{x}$

Putting $-2v^2+2v+1=z\Rightarrow dz=(-4v+2)dv$
$\Rightarrow -\frac{1}{2}\ln |z|=\ln |x|+\ln |C|\Rightarrow -\frac{1}{2}\ln |-2v^2+2v+1|=\ln |x|+\ln |C|\Rightarrow -\frac{1}{2}\ln |-\frac{2{\sin }^{2}y}{x^2}+\frac{2\sin y}{x}+1|=\ln |x|+\ln |C|\Rightarrow \ln |-\frac{2{\sin }^{2}y}{x^2}+\frac{2\sin y}{x}+1|=-2\ln |Cx|\Rightarrow -\frac{2{\sin }^{2}y}{x^2}+\frac{2\sin y}{x}+1=\frac{1}{(Cx{)}^2}\Rightarrow {\sin }^{2}y=x\sin y+\frac{x^2}{2}-\frac{2}{C^2}$

The curve passes through $(1,\pi )$
$0=0+\frac{1}{2}-\frac{2}{C^2}\Rightarrow C^2=4$
Therefore the required curve is,
${\sin }^{2}y=x\sin y+\frac{x^2}{2}-\frac{1}{2}$
Putting $x=h,y=\frac{\pi }{2}$
$1=h+\frac{h^2}{2}-\frac{1}{2}\Rightarrow h^2+2h=3\Rightarrow (h+1{)}^2=2\Rightarrow h=1,-3$

Hence, the maximum value of $|h|=3$