Find the solution of differential equation given below: d y d x + y · cosec 2 ( x ) = cosec 2 ( x )…

Given the differential equation is,

$\frac{dy}{dx}+\cos e{c}^2\left(x\right)·y=\cos e{c}^2\left(x\right)·cotx..........\left(1\right)$

Comparing with (1), we get

$\frac{dy}{dx}+py=Q$

Which is a linear differential equation,

$$\begin{gathered} P=\cos e{c}^{2}x \\ Q=\cos e{c}^{2}x·cotx \\ I.F.=e^{\int Pdx} \\ =e^{\int \cos e{c}^{2}x dx} \\ =e^{-cotx} \end{gathered}$$

The solution of (1) is given by,

$$\begin{gathered} y·I.F.=\int Q·IF+C \\ y·e^{-cotx}=\int \cos e{c}^{2}x·cotx·e^{-cotx} dx+C. \end{gathered}$$

Now, put $cotx=t$ in RHS

$$\begin{gathered} -\cos e{c}^{2}x dx=dt \\ \cos e{c}^{2}x=-dt \\ y·e^{-cotx}=\int e^{-t}·t (-dt) \\ =-\int t·e^{-t} dt \\ =-[t\int e^{-t} dt-\int \frac{d}{dt}t(\int e^{-t} dt\left)\right] \\ =-[-t{e}^{-t}-\int -e^{-t} dt] \\ =-[-t{e}^{-t}-e^{-t}]+C \\ =t{e}^{-t}+e^{-t}+C \\ =e^{-t}(1+t)+C \\ =e^{-cotx}(1+cotx)+C. \end{gathered}$$