Question

If $x\cos (a+y)=\cos y,$ then prove that $\frac{dy}{dx}=\frac{{cos}^2(a+y)}{\sin a}$. Show that $\sin a\frac{d^{2}y}{d{x}^2}+\sin 2(a+y)\frac{dy}{dx}=0$.

Answer 1

$x\cos (a+y)=\cos y$,

Which implies $x=\frac{\cos y}{\cos (a+y)}$,

Now differentiate on both sides , we get

$dx=\frac{\cos (a+y)(-\sin y)-\cos y(-\sin (a+y\left)\right)}{{\cos }^2(a+y)}dy=\frac{\sin (a+y-y)}{{\cos }^2(a+y)}dy=\frac{\sin a}{{\cos }^2(a+y)}dy$

$\Rightarrow \frac{dy}{dx}=\frac{{\cos }^2(a+y)}{\sin a}$

$\Rightarrow \frac{d^{2}y}{d{x}^2}=\frac{2\cos (a+y)(-\sin (a+y)}{\sin a}\times \frac{dy}{dx}=\frac{-\sin 2(a+y)}{\sin a}\times \frac{dy}{dx}$

Therefore, $\sin a\frac{d^{2}y}{d{x}^2}+\sin 2(a+y)\frac{dy}{dx}=0$