# Two functions cannot be solutions to a general second order homogenous differential equation on some interval

I recently asked a question about when $x^2$ cannot be a solution to a homogenous second order differential equation under a specific condition.

My book on differential equations has now gone onto say that given the equation $y''+p(x)y'+q(x)y=0$ and given that $p(x)$ and $q(x)$ are both continuous on the open interval $x\in(\frac{1}{2},2)$ then $y_{1}=e^{2x}$ and $y_{2}=x^{2}$ cannot both be solutions in this interval.

I can't see how this works. I was thinking of Abel's Theorem, or something along those lines using the Wronskian but it doesn't appear to be working. Could someone explain this to me please?

This one can also be solved just by plugging in the different values.

Suppose $y = e^{2x}$ is a solution to the differential equation. Then you'd have

\begin{align} 4e^{2x} + 2e^{2x} p + e^{2x}q &= 0 & \implies \\ 4 + 2p + q &=0 & \implies \\ q &= -4 -2p \end{align}

If we require that $y = x^2$ be a solution, we get the following relation for $p$ and $q$ $$2 + 2xp + x^2q = 0$$ Inserting the relation between $p$ and $q$ that we found previously, we have \begin{align} 2 + 2xp + x^2(-4 -2p) &= 0 & \implies \\ p(x -x^2) &= 2x^2 -1 & \implies \\ p &= \frac{2x^2 -1}{x(1-x)}.\end{align}

This is the only function $p$ which would be compatible with $y=x^2$ and $y=e^{2x}$ both being solutions to the differential equation. However, it's not continuous in the domain $(\frac12, 2)$, and hence is not allowed. Therefore, the two functions can't both be solutions.

• I seem to be overthinking all of these. Your help is much appreciated!
– user221122
Apr 29, 2015 at 13:01