Question

For each of the given differential equation find the general solution.
$\frac{dy}{dx}+ysecx=tanx(0\le x\le \frac{\pi }{2}).$

Answer 1

The given differential equation is $\frac{dy}{dx}+ysecx=tanx(0\le x\le \frac{\pi }{2}).$
On comparing with the form $\frac{dy}{dx}+Py=Q$ we get P=secx and $Q=tanx$
$\therefore IF=e^{\int Pdx}\Rightarrow e^{\int secxdx}$
$\Rightarrow IF=e^{log|secx+tanx|}=secx+tanx$ ..(i)
The solution of the given differential equation is given by
$y.IF=\int Q\times IFdx+C\Rightarrow y(secx+tanx)=\int tanx(secx+tanx)dx+C\Rightarrow y(secx+tanx)=\int tanx.secxdx+\int ta{n}^{2}xdx+C\Rightarrow y(secx+tanx)=secx+\int se{c}^{2}xdx-\int 1dx+c$ $\because ta{n}^{2}x=se{c}^x-1$
$\Rightarrow y(secx+tanx)=secx+tanx+C$