Question

For the differential equation given in the question find a particular solution satisfying the given condition.

$\frac{dy}{dx}-3ycotx=sin2x,wherey=2andx=\frac{\pi }{2}$

Answer 1

Given, $\frac{dy}{dx}-3ycotx=sin2x$
On comparing with the form $\frac{dy}{dx}+Py=Q,$ we get
P=-3cotx, Q=sin 2x
$IF=e^{-3\int cotxdx}=e^{-3log|sinx|}=e^{log|sinx{|}^{-3}}\Rightarrow IF=\frac{1}{si{n}^{3}x}$ ...(i)
The general solution of the given differential equation is given by
$y\times IF=\int Q\times IFdx+C\Rightarrow y\times \frac{1}{si{n}^{3}x}=\int \frac{1}{si{n}^{3}x}sin2xdx\Rightarrow y\times \frac{1}{si{n}^{3}x}=2\int \frac{sinxcosx}{si{n}^{3}x}dx+C\Rightarrow \frac{1}{si{n}^{3}x}\times y=2\int \frac{cosx}{si{n}^{2}x}dx+C\Rightarrow \frac{y}{si{n}^{3}x}=2\int cotxcosecxdx+C\Rightarrow \frac{y}{si{n}^{3}x}=-2cosecx+C\Rightarrow y=-2(\frac{1}{sinx}\times si{n}^{3}x)+Csi{n}^{3}x\Rightarrow y=-2si{n}^{2}x+Csi{n}^{3}x$ ...(iii)
Given, y=2 and $x=\frac{\pi }{2},$ then from Eq. (ii), we get
On $2=-2si{n}^2\left(\frac{\pi }{2}\right)+Csi{n}^3\left(\frac{\pi }{2}\right)\Rightarrow 2=-2+C\Rightarrow C=4$
On putting the value of C in Eq. (ii), we get
$y=-2si{n}^{2}x+4si{n}^{3}x\Rightarrow y=4si{n}^{3}x-2si{n}^{2}x$