Question

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

$\frac{dy}{dx}+2ytanx=sinx,y=0whenx=\frac{\pi }{3}.$

Answer 1

Given, $\frac{dy}{dx}+2ytanx=sinx$
On comparing with the form $\frac{dy}{dx}+Py+Q,$ we get
$P=2tanx,Q=sinx\therefore IF=e^{2\int tanxdx}=e^{2log|secx|\Rightarrow IF=se{c}^{2}x}$
The general solution is given by
$y.IF=\int Q\times IFdx+C$ ...(i)
$\Rightarrow yse{c}^{2}x=\int (sinx.se{c}^{2}x)dx)+C\Rightarrow yse{c}^{2}x=\int sinx.\frac{1}{co{s}^{2}x}dx+C\Rightarrow yse{c}^{2}x=\int tanxsecxdx+C\Rightarrow yse{c}^{2}x=secx+C$ ...(ii)
Put, y=0 and $x=\frac{\pi }{3}$ in Eq. (ii), we get
$0\times se{c}^2\frac{\pi }{3}=sec\frac{pi}{3}+C\Rightarrow 0=2+C\Rightarrow C=-2$
On putting the value of C in Eq. (ii), we get
$yse{c}^{2}x=secx-2\Rightarrow y=cosx-2co{s}^{2}x$
Hence, the required solution of the given differential equation is $y=cosx-2co{s}^{2}x.$