Question

Solve: $(x^2+1)\frac{dy}{dx}+4xy=\frac{1}{x^2+1}$

Answer 1

Given,

$(x^2+1)\frac{dy}{dx}+4xy=\frac{1}{x^2+1}$

$\frac{dy}{dx}+\frac{4x}{(x^2+1)}y=\frac{1}{(x^2+1{)}^2}$

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

$P=\frac{4x}{(x^2+1)},Q=\frac{1}{(x^2+1{)}^2}$

$I.F=e^{\int Pdx}=e^{\int \frac{4x}{(x^2+1)}dx}=(x^2+1{)}^2$

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

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

$y(1+x^2{)}^2=\int dx$

$y(1+x^2{)}^2=x+c$