Question

Find the particular solution of the differential equation

${\displaystyle \tan x\cdot \frac{dy}{dx}=2x\tan x+x^2-y}$; ${\displaystyle (\tan x\ne 0)}$ given that y = 0 when ${\displaystyle x=\frac{\pi }{2}}$

Answer 1

The given differential equation is

${\displaystyle \tan x\cdot \frac{dy}{dx}=2x\tan x+x^2-y}$; ${\displaystyle (\tan x\ne 0)}$

${\displaystyle \Rightarrow \frac{dy}{dx}=2x+x^2\cot x-y\cot x}$

${\displaystyle \Rightarrow \frac{dy}{dx}+\left(\cot x\right)y=2x+x^2\cot x}$

This is a linear differential equation.

Here, P = cot x, Q = 2x + x² cot x

:. I.F. = ${\displaystyle e^{\int Pdx}=e^{\int \cot sdx}=e^{\log \left|\sin x\right|}=\sin x}$

The general solution of this linear differential equation is given by

y(I.F.) = ∫Q(I.F.)dx + C

${\displaystyle \Rightarrow y\cdot \sin x=\int (2x+x^2\cot x)\sin xdx+C}$

${\displaystyle \Rightarrow y\cdot \sin x=\int 2x\sin xdx+\int x^2\cos xdx+C}$     

${\displaystyle y\cdot \sin x=\int 2x\sin xdx+x^2\sin x-\int 2x\sin x+C}$     (Applying integration by parts in the 2nd integral)

${\displaystyle \Rightarrow y\cdot \sin x=x^2\sin x+C}$......1

When y = 0,  ${\displaystyle x=\frac{\pi }{2}}$  (Given)

${\displaystyle \therefore 0\times \sin \frac{\pi }{2}=\frac{{\pi }^2}{4}\sin \frac{\pi }{4}+C}$

${\displaystyle \Rightarrow C=-\frac{{\pi }^2}{4}}$

Substituting the value of C in (1), we get

${\displaystyle y\sin x=x^2\sin x-\frac{{\pi }^2}{4}}$

${\displaystyle \Rightarrow (x^2-y)\sin x=\frac{{\pi }^2}{4}}$

This is the particular solution of the given differential equation