Question on Reduction of Order for Second Linear DEs? Can someone explain how the method of reduction of order can be used to explain the solution of second linear ODEs with repeated roots in the form of:
$$
ax^2 y'' + bxy' +cy = 0 ?
$$
I know that if there is a repeated root, then one solution is $y(x)=x^{r_1}$. Also, the method of reduction of order essentially states that there is a function $u$ by which the first solution can be multiplied in order to arrive at a second solution, however, I find myself struggling with a purely proof-based example using only a, b, and c.
With a numerically based example, I have no problem, but symbolically that appears to be were I am struggling.
Could anyone clue me in?
Many thanks.
 A: I believe that this is what you are looking for, but if not please comment.
We can rewrite the DE as:
$$x^2y''+axy'+by=0$$
If $y_1$ is a solution to the DE, then we will look for a second solution of the form $y_2=y_1v$.
$$y_2'=y_1'v+y_1v'  $$
$$y_2''=y_1''v+2y_1'v'+y_1v''$$
Plugging these equations into the original DE we find:
$$x^2y_1''v+2x^2y_1'v'+x^2y_1v''+axy_1'v+axy_1v'+by_1v=0  $$
Now, because $y_1$ is itself a solution, the first, fourth, and last terms are summed to $0$.  So we are left with:
$$x^2y_1v''+2x^2y_1'v'+axy_1v'=0  $$
Let's now make the following substitutions $u=v'$, and $y_1=x^r$.
$$x^{r+2}u'+2rx^{r+1}u+ax^{r+1}u=0   $$
$$u'+\frac{2r+a}{x}u=0  $$
Do you see how to proceed from here?
EDIT
Our characteristic equation is $r^2+(a-1)r+b=0$.  Since we have repeated roots, the derivative of the characteristic equation must be zero as the roots occur at the same place on the $r$ axis.  Thus,
$$2r+a-1=0$$
$$2r+a=1$$
Now, just substitute this into that last differential equation, and solve for $u$.  Then integrate again to find $v$. Also, you should arrive at $u=1/x$ and $v=\ln x$.
