Question

Find the particular solution of the following differential equation:
${\sec }^y(1+x^2)dy+2x\tan ydx=0$, given that $y=\frac{\pi }{4}$, when $x=1$.

Answer 1

$\sec y(1+x^2)dy=-2x\tan ydx$

$\Rightarrow -\csc ydy=\frac{2x}{1+x^2}dx$

$\Rightarrow \log \mid \csc y+\cot y\mid =\log (1+x^2)+C$

Now, put $x=1$ and $y=\frac{z}{4}$, we get

$\log \mid \csc \left(\frac{z}{4}\right)+\cot \left(\frac{z}{4}\right)\mid =\log (1+1)+C$

$\Rightarrow \log \mid \sqrt{2}+1\mid =\log 2+C$

$\Rightarrow C=\log \mid \frac{\sqrt{2}+1}{2}\mid$

Equation is

$\frac{\csc y+\cot y}{1+x^2}=\frac{\sqrt{2}+1}{2}$