# Find the general solution of the differential equation:( (xlogx) dy/dx ) + y = (2/x) logx

Arun Kumar IIT Delhi
$\frac{\mathrm{d} y}{\mathrm{d} x}+y/(xlog(x))=2/x^2 \\=>\int 1/xlog(x)dx=log(log(x)) \\=>e^{log(log(x))}(\frac{\mathrm{d} y}{\mathrm{d} x}+y/(xlog(x))=2/x^2) \\=>e^{log(log(x))}*y=\int e^{log(log(x))}/x^2dx=\int log(x)/x^2dx=-e^{-log(x)}+c$