Find the equation of the curve passing through the point
Question:

Find the equation of the curve passing through the point $\left(0, \frac{\pi}{4}\right)$ whose differential equation is, $\sin x \cos y d x+\cos x \sin y d y=0$

Solution:

The differential equation of the given curve is:

$\sin x \cos y d x+\cos x \sin y d y=0$

$\Rightarrow \frac{\sin x \cos y d x+\cos x \sin y d y}{\cos x \cos y}=0$

$\Rightarrow \tan x d x+\tan y d y=0$

Integrating both sides, we get:

$\log (\sec x)+\log (\sec y)=\log \mathrm{C}$

$\log (\sec x \cdot \sec y)=\log \mathrm{C}$

$\Rightarrow \sec x \cdot \sec y=\mathrm{C}$            $\ldots(1)$

The curve passes through point $\left(0, \frac{\pi}{4}\right)$.

$\therefore 1 \times \sqrt{2}=\mathrm{C}$

$\Rightarrow \mathrm{C}=\sqrt{2}$

On substituting $\mathrm{C}=\sqrt{2}$ in equation (1), we get:

$\sec x \cdot \sec y=\sqrt{2}$

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

$\Rightarrow \cos y=\frac{\sec x}{\sqrt{2}}$

Hence, the required equation of the curve is $\cos y=\frac{\sec x}{\sqrt{2}}$.