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?


1 Answer 1


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.

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

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