Question

If $y=y\left(x\right),y\in [0,\frac{\pi }{2})$ is the solution of the differential equation $\sec y\frac{dy}{dx}-\sin (x+y)-\sin (x-y)=0$ with $y\left(0\right)=0$, then $5y^{\prime }\left(\frac{\pi }{2}\right)$ is equal to

Answer 1

Given: $\sec y\frac{dy}{dx}-\sin (x+y)-\sin (x-y)=0$
$\Rightarrow \sec y\frac{dy}{dx}-2\sin \left(x\right)\cdot \cos \left(y\right)=0$
$\Rightarrow {\sec }^{2}ydy=2\sin xdx$
$\Rightarrow \tan y=-2\cos x+C$
Since $y\left(0\right)=0,$ therefore $C=2$
$\therefore \tan y=-2\cos x+2$
Put $x=\frac{\pi }{2}$
$\Rightarrow \tan y=2$
$\Rightarrow \cos y=\frac{1}{\sqrt{5}}$
Now, $\frac{dy}{dx}=2\sin x\cdot {\cos }^{2}y$
Put $x=\frac{\pi }{2},$ we have
$y^{\prime }\left(\frac{\pi }{2}\right)=2\cdot \left(\frac{1}{5}\right)$
$\therefore 5y^{\prime }\left(\frac{\pi }{2}\right)=2$