We have a second order linear homogeneous ODE with variable coefficients of the form:

$$y''(x)+ \frac{b(x)}{a(x)}y'(x)+\frac{c(x)}{a(x)} y(x)=0.$$

  • For $y=c_1\eta(x)+c_2\eta'(x)$, find $a(x)$, $b(x)$ and $c(x)$.

  • Use the result when $y = \log x$.

  • Check the result obtained.

I'm having trouble with the first part, mainly it seems like such a general form of an ODE that not much can be done. I was thinking of getting a system of three equations and solving that, two of them being immediate:

$$a(x)\eta''(x)+b(x)\eta'(x)+c(x)\eta(x)=0$$ $$a(x)\eta'''(x)+b(x)\eta''(x)+c(x)\eta'(x)=0$$

I don't know where a third linearly independent equation could come from. Also, are $c_1$ and $c_2$ important enough that I should work with them?


  • 1
    $\begingroup$ You don't say what $c_1$ and $c_2$ are. Are they arbitrary? $\endgroup$ – John B Jan 19 '16 at 23:17
  • $\begingroup$ That's all the information I'm given, but I gather that they are arbitrary constants that depend on the initial values. $\endgroup$ – Gonate Jan 19 '16 at 23:23
  • $\begingroup$ A hint then: write $\frac{b(x)}{a(x)}y'(x)+\frac{c(x)}{a(x)} y(x)=-y''(x)$, replace $y(x)$ and vary $c_1$ and $c_2$. $\endgroup$ – John B Jan 19 '16 at 23:27

The phrasing of the question is singularly vague; if this is indeed the exact way the question was put to you, I cannot blame you the least for not knowing how to proceed.

I interpret the first bullet as follows:

Assume that there exists a function $\eta(x)$ such that every solution to the above ODE can be written as a linear combination of $\eta(x)$ and $\eta'(x)$, i.e. $y(x) = c_1 \eta(x) + c_2 \eta'(x)$ is a solution to the ODE for every choice of $c_{1,2}$.

Call $q(x) = \frac{b(x)}{a(x)}$ and $r(x) = \frac{c(x)}{a(x)}$. Then, by substituting $\eta(x)$ resp. $\eta'(x)$ in the original ODE, you obtain expressions for $q(x)$ and $r(x)$ in terms of $\eta$ and its derivatives. To obtain the third equation you're looking for, an idea would be to take the derivative of the entire ODE, and use the fact that $\eta'(x)$ solves the original ODE. After some cancellations, this yields \begin{equation} r'(x) \eta(x) + q'(x) \eta'(x) = 0. \tag{1} \end{equation} The first two expressions obtained enable you to write \begin{align} q(x) &= \frac{\eta(x) \eta'''(x) - \eta'(x) \eta''(x)}{\eta'(x)^2-\eta(x) \eta''(x)},\tag{2}\\ r(x) &= \frac{\eta''(x)^2 - \eta'(x) \eta'''(x)}{\eta'(x)^2-\eta(x) \eta''(x)}. \tag{3} \end{align} Note that $q(x) \eta'(x) + r(x) \eta(x) = -\eta''(x)$. Taking the derivative of this equation, and using the fact that $\eta'(x)$ solves the original ODE, you end up with $(1)$. In other words: there are only two independent equations, so you can only solve for $q(x)$ and $r(x)$, and not for $a,b,c$.

Note that an obvious choice for $a,b,c$ based on $(2)$ and $(3)$ would be \begin{align} a(x) &= \eta'(x)^2-\eta(x) \eta''(x),\\ b(x) &= \eta(x) \eta'''(x) - \eta'(x) \eta''(x),\\ c(x) &= \eta''(x)^2 - \eta'(x) \eta'''(x), \end{align} but as you can see, this choice is not uniquely determined: you could multiply these choices of $a,b,c$ with an arbitrary function of $x$ and still satisfy $(2)$ and $(3)$.

For the rest: I sincerely hope you inadvertently made a typo, and that it should read

For $\eta(x) = \log(x)$, apply the above results.

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