Question

For each the differential equations given, find the general solution :
$(1+x^2)dy+2xydx=\cot xdx(x\ne 0)$

Answer 1

$(1+x^2)dy+2xydx=\cot xdx$

$\frac{dy}{dx}+\left(\frac{2x}{1+x^2}\right)y=\frac{\cot x}{1+x^2}$

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

$P=\frac{2x}{1+x^2},Q=\frac{\cot x}{1+x^2}$

$I.F.=e^{\int Pdx}$

$=e^{\int \frac{2x}{1+x^2}dx}$

$=1+x^2$

$y\times I.F=\int Q\times I.Fdx+c$

$y\times (1+x^2)=\int \left(\frac{\cot x}{1+x^2}\right)(1+x^2)dx+c$

$y\times (1+x^2)=\int \cot xdx+c$

$y(1+x^2)=\log \sin x+c$

$y=(1+x^2{)}^{-1}\log \sin x+C$