Question

Solve `x.dy/dx-y=2x^2y`?

Answer 1

`y=xe^[[x^2+c]]`

Explanation:

`xdy/dx=2x^2y+y`, separating the variables.....

`1/ydy/dx=[2x^2+1/x]` so `dy/y=2x^2dx+dx/x`, integrating...
`int1/ydy=int2x^2dx+int1/xdx`, we have `lny=x^2+lnx+C`

Which gives `ln[y/x]=x^2+C`.

`e^[ln[y/x]]= e^[[x^2+c]]` [ theory of logs]

i.e, `y/x= e^[[x^2+c]]` and `so, y=xe^[[x^2+c]]]`