Question

For the given differential equation find the general solution.

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

Answer 1

Given, $(1+x^2)dy+2xydx=cotxdx$
$\Rightarrow (1+x^2)dy=dx(cotx-2xy)\Rightarrow \frac{dy}{dx}=\frac{cotx-2xy}{1+x^2}\Rightarrow \frac{dy}{dx}+\frac{2xy}{1+x^2}=\frac{cotx}{1+x^2}$
On comparing with the form $\frac{dy}{dx}+Py=Q,$ we get
$\therefore P=\frac{2x}{1+x^2}andQ=\frac{cotx}{1+x^2}\therefore IF=e^{\int Pdx}=e^{\int \frac{2x}{1+x^2}dx}Let1+x^2=t\Rightarrow 2x=\frac{dt}{dx}\Rightarrow dx=\frac{dt}{2x}\therefore IF=e^{\int \frac{2x}{t}\times \frac{dt}{2x}}\Rightarrow IF=e^{log|t|}=t=1+x^2$ ...(i)
The general solution of the given diffrential equation is given by
$y.IF=\int Q\times IFdx+C\Rightarrow (1+x^2)y=\int \left(\right(1+x^2\left)\frac{cotx}{(1+x^2)}\right)dx+C\Rightarrow (1+x^2)y=log|sinx|+C\Rightarrow y=log|sinx|(1+x^2{)}^{-1}+C(1+x^2{)}^{-1}$