Question

Find particular solution of the differential equation sin 2x $\frac{dy}{dx}$ - y = tan x given that x = $\frac{\pi }{4}$ when y = 0.

Answer 1

By rearranging the given differential eqn we get linear differential eqn in $y$

i.e. $\frac{dy}{dx}-\csc 2x.y=\frac{\tan x}{\sin 2x}[\because \frac{dy}{dx}+P(x)y=Q(x\left)\right]$

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

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

I.E $=e^{-\int \csc 2xdx}=e^{-\frac{\ln (\csc 2x-\cot 2x)}{2}}$

$=\frac{1}{\sqrt{\csc 2x-\cot 2x}}$

$\therefore$ Sol is (form eqn $\left(1\right)$)

$y\left(\frac{1}{\sqrt{\csc 2x-\cot 2x}}\right)=\int \frac{1}{2}{\sec }^{2}x\frac{1}{\sqrt{\frac{1}{\sin 2x}-\frac{\cos 2x}{\sin 2x}}}dx+c$

$=\frac{-1}{2}\int {\sec }^2\frac{1}{\sqrt{\frac{2{\sin }^{2}x}{2\sin x\cos x}}}dx+C$

$=\frac{1}{2}\int \frac{{\sec }^{2}x}{\sqrt{\tan x}}dx+C$

$=\frac{1}{2}2\sqrt{\tan x}+C(\because \int \frac{f\left(x\right)}{\sqrt{f\left(x\right)}}=2\sqrt{f\left(x\right)})$

$y.\left(\frac{1}{\sqrt{\csc 2x-\cot 2x}}\right)=\sqrt{\tan x}+C$

$y.\frac{1}{\sqrt{\tan x}}=\sqrt{\tan x}+C$

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