Solver the following differential equation :
(1) `x(dy)/(dx) + 2y =x^(2) .log x`

Explanation

` x (dy)/(dx) +2y=x ^(2) . Log x `
`therefore (dy)/(dx) +(2/x) .y= x.log x `
This is the linear differential equation of the form
` (dy)/(dx) + P.y = Q , " where " P = 2/x and Q = x.log x `
`therefore I.F. e^(int p dx) = e^(int 2/x dx) = e^(2 int 1/x dx)`
` e^(2 log x) =e^(log x^(2)) =x^(2)`
The solution of (1) is given by
`y.(I.F) = int Q. (I.F.) dx + c`
`y.x^(2) = int x.log x.x^(2) dx+c`
` x^(2).y = int x^(3) . log x dx +c`
` = (log x ) int x^(3) dx - int [ d/dx (log x) int x^(3) dx] dx +c`
` = (log x).x^(4)/4 - int 1/x . x^(4)/4 (dx) +c`
` 1/4 x^(4) logx -1/4 int x^(3) dx +c`
` x^(2).y = 1/4 x^(4) log x -1/4 .x^(4)/4 +c`
` y.x^(2) = (x^(4) log x)/4 - x^(4)/16 +c`
This is the general solution .