Question

$\text{If}x\cos (a+y)=\cos y\text{then prove that}$

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

$\text{Hence, show that}\sin a\frac{d^{2}y}{d{x}^2}+\sin 2(a+y)\frac{dy}{dx}=0$.

Answer 1

Given:
$x\cos (a+y)=\cos y$

$\therefore x=\frac{\cos y}{\cos (a+y)}$ .............(1)

Differentiating equation (1) w.r.t.y

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

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

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

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

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

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

Now we have $\sin a\frac{dy}{dx}={\cos }^2(a+y)$ ...............(3)

Differentiating equation (3) w.r.t. x, we will get

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

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

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

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

Hence Proved.