Here is what I'm looking at:

Use the method of reduction of order to find a second solution to

$\displaystyle x^2 y'' + 3xy' + y = 0$

$\displaystyle x > 0$

$\displaystyle y_1 = x^{-1}$

Here is what I did:

$\displaystyle y'' + \frac{3}{x} y' + x^{-2}y = 0$

$\displaystyle p(x) = 3x^{-1}$

$\displaystyle y_2 = \int {\frac{e^{-3ln(x)}}{(x^{-1})^2}} dx$

$\displaystyle y_2 = \int{\frac{x^{-3}}{x^{-2}}} dx$

$\displaystyle y_2 = \int{x^{-1}} dx$

$\displaystyle y_2 = ln|x| + c$

Now, when I plug the original ODE into a solver, the general solution comes up with this:

$\displaystyle c_1x^{-1} + c_2 \frac{ln(x)}{x}$

Does that mean that my solution to $\displaystyle y_2$ is wrong? And if so, where did I make a mistake?