solve
xy'+2y=x2lnx
xy′+2y=x2lnxxy'+2y=x^2\ln xxy′+2y=x2lnx
x2y′+2xy=x3lnxx^2y'+2xy=x^3\ln xx2y′+2xy=x3lnx
(x2y)′=x3lnx(x^2y)'=x^3\ln x(x2y)′=x3lnx
x2y=∫x3lnxdx=14∫lnxdx4=x44lnx−14∫x4dlnx=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=x2y=∫x3lnxdx=41∫lnxdx4=4x4lnx−41∫x4dlnx=
=x44lnx−14∫x3dx=x44lnx−x416+C=\frac{x^4}{4}\ln x-\frac{1}{4}\int x^3dx=\frac{x^4}{4}\ln x-\frac{x^4}{16}+C=4x4lnx−41∫x3dx=4x4lnx−16x4+C
y=x216(4lnx−1)+Cx−2y=\frac{x^2}{16}(4\ln x-1)+Cx^{-2}y=16x2(4lnx−1)+Cx−2
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