Question

Find the equation of the curve passing through the point $(0,\frac{\pi }{4})$ whose differential equation is $\sin x\cos ydx+\cos x\sin ydy=0$.

Answer 1

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

$\Rightarrow \cos x\sin ydy=-\sin x\cos ydx$

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

Integrate both sides,

$\int \frac{\sin y}{\cos y}dy=\int \frac{-\sin x}{\cos x}dx$

Let $\cos y=u$ and $\cos x=v$. Then, $-\sin ydy=du$ and $-\sin xdx=dv$

$\Rightarrow \int \frac{-du}{u}=\int \frac{dv}{v}$

$\Rightarrow -\ln u=\ln v+C$

$\Rightarrow \ln u+\ln v=c$, where $c$ is the constant of integration.

$\Rightarrow \ln \left(uv\right)=c$

$\Rightarrow uv=k$, where $k=e^c=$ constant.

Resubstitute the values for $u$ and $v$,

$\cos x\cos y=k$

This is the general solution of the given differential equation.

This curve passes through $(0,\frac{\pi }{4})$

$\Rightarrow \cos 0\cos \frac{\pi }{4}=k$

$\Rightarrow k=\frac{1}{\sqrt{2}}$

Hence, the equation of the curve is

$\cos x\cos y=\frac{1}{\sqrt{2}}$