Repeated root case for ay''+by'+cy=0 To solve the ODE $a y''(t)+b y'(t) +c y(t) = 0$, where $a,b,c$ are constant, we solve the characteristic equation $ar^{2}+br+c=0$. In the case when the roots are two repeated roots, i.e,. $r=r_{1}=r_{2}$, we get two linearly independent solutions $y_{1}(t)=e^{r_{1}t}$ and $y_{2}(t)= t e^{r_{1}}$.
I understand that the solution $y_{2}(t)= t e^{r_{1}}$ works and there are several approaches to get this solution, as mentioned in a previous post, and in the textbook by Boyce and DiPrima also shows three approaches, see: . 
I do not have questions about these approaches. My question is, all these approaches require some computation. Is there any way to show $t e^{r_{1}}$ is a solution, without any computation? Thanks in advance for any insights!
 A: Well I'm taking your question as asking "Why does there have to be 2 solutions?"
If you have taken a Linear Algebra course then you could see the differential equation as asking for a basis for the eigenspace. Your solutions to the characteristic equation are your eigenvalues and your solutions to the DE are your eigenfunctions. 
Since the DE is second order you know that you have a two dimensional space, and you need two linearly independent solutions to form a basis for the space.
The characteristic equation only gives one solution so you need to find the other.
We then assume that the other solution, $y_2$, is of the form $f(t)y_1(t)$ and then compute $f(t)$
A: Well, if you're just looking for something to convince you that this is true, just think back to when you first did taylor series. You wanted the terms to be linearly independent so for each coefficient, $c_n$, you multiplied it by $x^n$
So your Taylor Series would be
$$c_0+c_1x+c_2x^2+...+c_nx^n$$
You will see a similar pattern occur with your repeated roots
If the root $r_1$ is repeated n times your solutions would be:
$$c_1e^{r_1t}+c_2te^{r_1t}+c_3t^2e^{r_1t}+...+c_nt^{n-1}e^{r_1t}$$
