Question

Solve the following differential equation: $(1+x^2)dy+2xydx=\cot xdx;x\ne 0$

Answer 1

$(1+x^2)dy+2xydx=cotxdx$

This can be written as

$\frac{dy}{dx}+P\left(x\right)y=Q\left(x\right)$ and here, $P\left(x\right)=\frac{2x}{1+x^2},Q\left(x\right)=\frac{cotx}{1+x^2}$

The integrating factor $(I.F.)=e^{\int Pdx}=e^{\int \left(\frac{2xdx}{1+x^2}\right)}$

Let $x^2=t,2xdx=dt$

$\therefore I.F.=e^{\int \frac{dt}{1+t}}=e^{\ln (1+t)}=1+t=1+x^2$

Now, $I.F.\times y=\int (I.F.)Q\left(x\right)dx$

$\Rightarrow y(1+x^2)=\int (1+x^2)\times \frac{cotx}{1+x^2}dx$

$\Rightarrow y(1+x^2)=\int cotxdx$

$\Rightarrow y(1+x^2)=\ln sinx+c$