Question

Find the equation of the curve passing through the point $(0,\frac{\pi }{4})$ whose differential equation is $sinxcosydx+cosxsinydy=0$

Answer 1

The differential equation of the given curve is
$sinxcosydx+cosxsinydy=0\Rightarrow \frac{sinx}{cosx}dx+\frac{siny}{cosy}dy=0\Rightarrow tanxdx+tanydy=0$
On integrating both sides, we get
$\int tanxdx+\int tanydy=logC\Rightarrow log\left(secx\right)+log\left(secy\right)=logC\Rightarrow secx.secy=C$
The curve passes through the point $(0,\frac{\pi }{4})$, therefore put x=0, $y=\frac{\pi }{4}$, we get $sec0sec\frac{\pi }{4}=C\Rightarrow C=\sqrt{2}$
On putting the value of C in Eq. (i), we get $secx.secy=\sqrt{2}$
$\Rightarrow secx.\frac{1}{cosy}=\sqrt{2}\Rightarrow cosy=\frac{secx}{\sqrt{2}}$
Hence, the required equation of the curve is $cosy=\frac{secx}{\sqrt{2}}$