Question

For the differential equation $\sin 2x\frac{dy}{dx}-y=\tan x;$ and $y\left(\frac{\pi }{4}\right)=2$ Find the value of the constant.

Answer 1

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

$\Rightarrow \frac{dy}{dx}-y\csc 2x=\frac{\sin x}{\cos x}.\frac{1}{2\sin x\cos x}$

$\Rightarrow \frac{dy}{dx}-y\csc 2x=\frac{{\sec }^{2}x}{2}$ ...(1)

Here $P=\csc 2x\Rightarrow \int Pdx=\int -\csc 2xdx=-\frac{1}{2}\log \left(\tan x\right)=\log {\left(\tan x\right)}^{\frac{-1}{2}}$

$\therefore I.F.=e^{\log {\left(\tan x\right)}^{\frac{-1}{2}}}={\left(\tan x\right)}^{\frac{-1}{2}}$

Multiplying (1) by $I.F.$ we get

${\left(\tan x\right)}^{\frac{-1}{2}}\frac{dy}{dx}-y{\left(\tan x\right)}^{\frac{-1}{2}}\csc 2x={\left(\tan x\right)}^{\frac{-1}{2}}\frac{{\sec }^{2}x}{2}$

Integrating both sides, we get

$y=\tan x+c\sqrt{\tan x}$

As $y\left(\frac{\pi }{4}\right)=2$

$\Rightarrow 2=1+c$

$\Rightarrow c=1$