Question

Solve the following initial value problem:-
\[\tan x\left( \frac{dy}{dx} \right) = 2x\tan x + x^2 - y; \tan x \neq 0\] given that y = 0 when \[x = \frac{\pi}{2}\]

Answer 1

\[\tan x\left( \frac{dy}{dx} \right) = 2x\tan x + x^2 - y\]
\[ \Rightarrow \frac{dy}{dx} + \frac{1}{\tan x}y = \frac{2x\tan x + x^2}{\tan x}\]
\[ \Rightarrow \frac{dy}{dx} + \left( \cot x \right)y = 2x + x^2 \cot x\]
This is a linear differential equation of the form \[\frac{dy}{dx} + Py = Q\]
Integrating factor, I.F. = \[e^{\int Pdx} = e^{\int\cot xdx} = e^{log\sin x} = \sin x\]
The solution of the given differential equation is given by
\[y \times \left( I . F . \right) = \int Q \times \left( I . F . \right)dx + C\]
\[ \Rightarrow y \times \sin x = \int\left( 2x + x^2 \cot x \right)\sin xdx + C\]
\[ \Rightarrow y\sin x = \int2x\sin xd x + \int x^2 \cos xdx + C\]
\[ \Rightarrow y\sin x = \int2x\sin xdx + \left[ x^2 \int\cos xdx - \int\left( \frac{d}{dx} x^2 \times \int\cos xdx \right)dx \right] + C\]
\[\Rightarrow y\sin x = \int2x\sin xdx + x^2 \sin x - \int2x\sin xdx + C\]
\[ \Rightarrow y\sin x = x^2 \sin x + C\]
\[ \Rightarrow y = x^2 + \text{ cosec }x \times C . . . . . \left( 1 \right)\]
It is given that, y = 0 when \[x = \frac{\pi}{2}\]
\[\therefore 0 = \left( \frac{\pi}{2} \right)^2 +\text{ cosec }\frac{\pi}{2} \times C\]
\[ \Rightarrow C = - \frac{\pi^2}{4}\]
Putting \[C = - \frac{\pi^2}{4}\] in (1), we get

\[y = x^2 - \frac{\pi^2}{4}\text{ cosec }x\]

Hence, 

\[y = x^2 - \frac{\pi^2}{4}\text{ cosec }x\] is the required solution.