# General solution of $\frac{dy}{dx} + \frac{y}{x^2}=\frac{1}{x^3}$

I'm trying to find the general solution of the following first order differential equation (using the integrating factor method): $$\frac{dy}{dx} + \frac{y}{x^2}=\frac{1}{x^3}$$

I found the integrating factor to be $e^{-1/x}$ meaning I would have to integrate $$\frac{e^{-1/x}}{x^3}$$ and then divide through by $e^{-1/x}$ to get the general solution, but I don't know how to integrate this. Can anyone help?

Note that if type expressions in the form e^abc, you achieve $e^abc$. If you wish to have "abc" appear all in the exponent, then you should enclose the exponent in brackets. That is, e^{abc} yields $e^{abc}$. I have fixed this in your post, so you can view the $\TeX$ source or yourself. –  JavaMan Feb 20 '13 at 16:00
Hint: try a substitution $x = 1/u$ in the integral. Then $dx = -du/u^2$ and you get
$$\int dx \: \frac{e^{-1/x}}{x^3} = -\int du \: u \, e^{-u}$$