# Q13.    Find a particular solution of the differential equation $\frac{dy}{dx} + y \cot x = 4x \textup{cosec} x\ (x\neq 0)$, given that $y = 0 \ \textup{when}\ x = \frac{\pi}{2}$.

G Gautam harsolia

Given equation is
$\frac{dy}{dx} + y \cot x = 4x \textup{cosec} x\ (x\neq 0)$
This is  $\frac{dy}{dx} + py = Q$  type where $p =\cot x$ and $Q = 4xcosec x$$Q = 4x \ cosec x$
Now,
$I.F. = e^{\int pdx}= e^{\int \cot xdx}= e^{\log |\sin x|}= \sin x$
Now, the solution of given differential equation is given by relation
$y(I.F.) =\int (Q\times I.F.)dx +C$
$y(\sin x ) =\int (\sin x\times 4x \ cosec x)dx +C$
$y(\sin x) =\int(\sin x\times \frac{4x}{\sin x}) +C\\ \\ y(\sin x) = \int 4x + C\\ y\sin x= 2x^2+C$
Now, by using boundary conditions we will find the value of C
It is given that  y = 0 when $x= \frac{\pi}{2}$
at   $x= \frac{\pi}{2}$
$0.\sin \frac{\pi}{2 } = 2.\left ( \frac{\pi}{2} \right )^2+C\\ \\ C = - \frac{\pi^2}{2}$
Now, put the value of C
$y\sin x= 2x^2-\frac{\pi^2}{2}$
Therefore, the particular solution is  $y\sin x= 2x^2-\frac{\pi^2}{2}, (sinx\neq0)$

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