# General solution for degree 2 differential equation given 3 solutions

Given that $e^{t}$, $e^{2t}$ and $e^{3t}$ solve the linear differential equation of the form $y''+a_1(t)y'+a_2(t)y=f(t)$, find the general solution.

I have tried to look for the solution to the associated homogeneous equation to no avail. I can write $e^{t}+a_1(t)e^{t}+a_2(t)e^{t}=f(t)$ if I assume that $e^{t}$ solves the homogeneous equation. However, I cannot seem to prove that this implies that $f(t)=0$. The same follows for the other two given solutions.

Without knowing $a_1(t)$ or $a_2(t)$ or $f(t)$, this seems impossible. Can anyone provide a hint or some insight?

The basic fact is:

For every homogeneous linear equation, the space of solutions is a vector space. For every non-homogeneous linear equation, the space of solutions is an affine space. In both cases, the dimension of the space of solutions is equal to the degree of the equation.

In the case at hand, the equation is non-homogeneous of degree 2. Hence, the space of solutions is a 2-dimensional affine space. Such a space is determined uniquely by three points which are not colinear. Since you are given three such points, you're done.

Edit: Another way to express the above idea, more or less, is as follows. If both $f_1$ and $f_2$ are solutions to a linear equation, then $f_1-f_2$ solves the associated homogeneous equation.

• So, could write something like $c_1e^{t}+c_2e^{2t}+e^{3t}$? May 9, 2017 at 1:30
• @zz20s Not exactly, but something similar to that... May 9, 2017 at 1:32
• Can you elaborate further then? I don't seem to understand. May 9, 2017 at 1:32
• @zz20s First, you need to make sure you know what a $d$-dimensional affine space is. Then, you need to recall how such a space is determined by $d+1$ points on it. May 9, 2017 at 1:35
• @zz20s I edited my answer. Perhaps it will be more clear to you now. May 9, 2017 at 1:44