MathJax TeX Test Page
This is just a special form of differential equation that slightly differs from constant coefficients. It looks like this:
$$a_1x^ny^{(n)} + a_{n-1}x^{n-1}y^{(n-1)} + ... a_2x^2y"+ a_1xy' + a_0y = 0$$
Note that this is **homogeneous**. If that equation equals $\cos(x)$ or $e^x$, there'll be no (easily solvable) answer.

MathJax TeX Test Page
Just like with constant coefficients, there are three cases. The first is the normal case: two distinct roots. The second is repeat roots, and last, we have imaginary roots.

Let's begin with an example:
$$2x^2y" + xy'-15y = 0$$
We begin by guessing $x^r$
$$2x^2(r)(r-1)x^{r-2} + x(r)x^{r-1} - 15x^r = 0 \to 2r(r-1)x^r + rx^r - 15x^r = 0$$
We divide through by $x^r$ to get our roots.
$$2r(r-1) + r - 15 = 0 \to 2r^2 - r + 15 = 0 \to (r-3)(2r + 5) = 0$$
We get our answer:
$$y = c_1x^3 + c_2x^{-\frac{5}{2}}$$

MathJax TeX Test Page
This is the case where you get two of the same r.
Here is an example:
$$x^2y'' + xy' = 0$$
So, we guess $x^r$. Skipping a step (dividing out $x^r$), we get:
$$r(r-1) + r = 0 \to r^2 = 0$$
We know one root, and using reduction of order, which I'll post about next, we can get the second root. The first root is $c_1x^0 = c_1$, and the second root is $c_2x^0\ln(x) = c_2\ln(x)$.
Our answer becomes: $$y = c_1 + c_2\ln(x)$$

MathJax TeX Test Page
This method is exactly the same as with constant coefficients. Let's look at an example.
$$x^2y'' + 7xy' + 13y = 0$$
Once again, we plug in $x^r$, and as before, we get an equation for r.
$$r^2 + 6r + 13 = 0 \to r = -3 \pm 2i$$
$$y = x^{-3 \pm 2i} = e^{(-3 \pm 2i)\ln(x)} = e^{-3\ln(x)}e^{\pm 2i\ln(x)} = $$
$$c_1x^{-3}(\cos(3\ln(x)) + i\sin(2\ln(x)) + c_2x^{-3}(\cos(-3\ln(x)) + i\sin(-2\ln(x))$$
We can combine the terms into one and take out the i, so our final answer is
$$y = c_1\dfrac{\cos(2\ln(x)}{x^3} + c_2\dfrac{\sin(2\ln(x)}{x^3}$$