Find 2nd order homogenous ODE from solutions $x^2, e^{-x}$ I'm trying to find the 2nd order ODE given those two solutions. I usually achieve this by finding the characteristic polynomial by multiplying the root factors, however the solution $x^2$ is giving me some trouble. 
As far as I understand, $x^2$ implies that $0$ is a triple root, but by finding the characteristic polynomial $x^3(x+1)$ I only get a 4th degree non quadratic equation, which I do not know how to "convert" to a 2nd degree one.
Am I doing it wrong or should another method be used, ie. differentiating the solutions up to the 2nd derivative and proceeding by trying different polinomials to multiply each derivative?
Thanks in advance.
 A: If you are looking for an equation of the form $y'' + p(x)y' + q(x)y = 0$, then you can plug in your two solutions $y = x^2$ and $y = e^{-x}$, giving two equations in two unknowns $p(x)$ and $q(x)$. You can then use standard linear algebra techniques to find $p(x)$ and $q(x)$.
A: The differential equation must be
$$uy''+vy'+wy=0$$
where $u,v,w$ are functions of the variable $x$.
The solution $y=e^{-x}$ leads to $u-v+w=0$, or $v=u+w$.
The solution $y=x^2$ leads to $2u+2xv+x^2w=0$, which must be true for all $x$.
Replace $v=u+w$, then
$$2u+2x(u+w)+x^2w=0$$
$$2(x+1)u+x(x+2)w=0$$
Now choose functions $u$ and $w$ so that this equality holds. An obvious choice would be
$$u(x)=\frac{1}{x+1}$$
$$w(x)=\frac{-2}{x(x+2)}$$
And then
$$v(x)=\frac{1}{x+1}-\frac{2}{x(x+2)}$$
A: The method is very simple. Start with the general form of a homogeneous, second-order ODE
$$ y'' + a(x)y' + b(x)y = 0 $$
You know the two solutions, so you can plug them into the equation to get
\begin{align} 2 + 2xa(x) + x^2b(x) &= 0 \\ e^{-x} - e^{-x}a(x) + e^{-x}b(x) &= 0 \end{align}
This is a system of equations in $a(x)$ and $b(x)$. You can easily solve this algebraically to get
$$ a(x) = \frac{x^2-2}{x^2+2x}, \quad b(x) = -\frac{2x+2}{x^2+2x} $$
Note that the coefficients are non-constant here, so the characteristic equation will not help.
