Answer to Question #222725 in Differential Equations for maxie

Question

solve

xy'+2y=x²lnx

Answer 1

$xy'+2y=x^2\ln x$

$x^2y'+2xy=x^3\ln x$

$(x^2y)'=x^3\ln x$

$x^2y=\int x^3\ln x dx=\frac{1}{4}\int \ln x dx^4=\frac{x^4}{4}\ln x-\frac{1}{4}\int x^4d\ln x=$

$=\frac{x^4}{4}\ln x-\frac{1}{4}\int x^3dx=\frac{x^4}{4}\ln x-\frac{x^4}{16}+C$

$y=\frac{x^2}{16}(4\ln x-1)+Cx^{-2}$