Question

Find the particular solution of the following differential equation:
$\cos ydy+\cos x\sin ydx=0$, given that $y=\frac{\pi }{2}$, when $x=\frac{\pi }{2}$.

Answer 1

$\cos ydy+\cos x\sin ydx=0$

$\Rightarrow +\cot ydy=-\cos xdx$

Now integrating, we get

$\ln \sin y=-\sin x+C$

$\Rightarrow \ln \sin y+\sin x=C$

Put $x=\frac{z}{2}$ and $y=\frac{z}{2}$, we get

$\ln \sin \frac{z}{2}+\sin \frac{z}{2}=C$

$\ln \left(1\right)+1=C$

$C=1$

Equation is

$\ln \sin y+\sin x=1$