Solving xy' + 3y = x² with integrating factor

[tomeatworld]

Homework Statement

x y' + 3y = x²

Homework Equations

The Attempt at a Solution

So I tried using an integrating factor (as I couldn't separate variables).
So I've said that the function p(x)= 1/x and q(x)=x. So the integrating factor is e^log(x) or x. Putting this in:
x y = $$\int x^{2}$$ so
x y = $$\frac{1}{3} x^{3}$$ + c
and finally: y(x) = $$\frac{1}{3} x^{2} + \frac{c}{x}$$ but this isn't the right answer. Where have I gone wrong?

 

[Mark44]

You don't have the right integrating factor. Starting from y' + (3/x)y = x, your integrating factor should be x³. Multiplying by the integrating factor gives x³y' + 3x²y = x⁴.

This equation can be written as d/dx(x³y) = x⁴. Can you take it from there?

 

[tomeatworld]

ahh I see! yeah, that's great! thanks a load!