Question

For differential equation $(x+\tan y)dy=\sin 2ydx;y\left(2\right)=\frac{\pi }{4}$, then find the value of the constant.

Answer 1

Given, $(x+\tan y)dy=\sin 2ydx$

$\Rightarrow \sin 2y\frac{dx}{dy}-x=\tan y$

$\Rightarrow \frac{dy}{dx}-\frac{x}{\sin 2y}=\frac{1}{2{\cos }^{2}y}$ ...(1)
Here $P=\frac{-1}{{\sin }^{2}y}\Rightarrow \int Pdy=-\int \frac{1}{{\sin }^{2}y}dy$

$=\frac{1}{2}\log \left(\cot y\right)=\log \left(\sqrt{\cot y}\right)$
$\therefore I.F=e^{\log \left(\sqrt{\cot y}\right)}=\sqrt{\cot y}$
Multiplying (1) by $I.F,$ we get
$\sqrt{\cot y}\frac{dx}{dy}-\frac{x\sqrt{\cot y}}{{\sin }^{2}x}=\frac{\sqrt{\cot y}}{2{\cos }^{2}y}$
Integrating both sides
$x\sqrt{\cot y}=\int \frac{\sqrt{\cot y}}{2{\cos }^{2}y}dy+c\Rightarrow x\sqrt{\cot y}=\sqrt{\cot y}.\tan y+c$

$\Rightarrow x=\tan y+c\sqrt{\tan y}$
As $y\left(2\right)=\frac{\pi }{4}$

$\therefore 2=1+c$

$\Rightarrow c=1$