Series Solutions Near an Ordinary Point I am attempting to solve this problem for practice: $y"-(x-3)y' - y = 0$ at $x_{0} = 3$.

But it appears as though I don't have an idea of the best approach to employ to go about solving it.

Can someone help me understand the steps in solving a DE such as this?  I know that a power series is required, but the garden variety method of solution doesn't appear to be working for me.

Any suggestions?
It looks much simpler than that. Try differentiating $\exp(-3x+\frac{1}{2}x^2)$.

This seems to be attracting down votes. So maybe I should add a little more. If you do that you find that it is a solution to $y''-(x-3)y'-y=0$. Note also that the equation is linear, so you can take any constant multiple and the solution still works. Of course, there is another solution which is harder to find. But you have not specified the initial conditions clearly, so you may not need it.
It looks much simpler than that. Try differentiating $\exp(-3x+\frac{1}{2}x^2)$.

This seems to be attracting down votes. So maybe I should add a little more. If you do that you find that it is a solution to $y''-(x-3)y'-y=0$. Note also that the equation is linear, so you can take any constant multiple and the solution still works. Of course, there is another solution which is harder to find. But you have not specified the initial conditions clearly, so you may not need it.