Method of Frobenius, constants replaced by power series

The source I am using is Differential Equations with Applications and Historical Notes (Simmons). I appologize in advance if I did not include enough information from the text about my question.

The method of Frobenius is motivated with the $$z = log(x)$$ substitution in the Euler equation

$$x^2y'' + pxy' + qy = 0,$$

which has solutions $$x^{m_1}$$ and $$x^{m_2}$$ (or $$x^{m_1}\log{x}$$) corresponding to $$m^2 + (p-1)m + q = 0$$.

It then says that if $$p$$ and $$q$$ are replaced by power series,

$$y'' + \frac{(p_0 + p_1x + p_2x^2 + \dots)}{x}y' + \frac{(q_0 + q_1x + q_2x^2 + \dots)}{x^2}y=0$$

then it is natural to guess that the solutions (to the equations with power series) might be found by replacing the original solutions with solutions of the form

$$y_1 = x^m(a_0 + a_1x + a_2x^2 + \dots),\\ y_2 = x^m\log{x}(a_0 + a_1x + a_2x^2 + \dots)$$

Why would this be a natural guess? I assume that there is more to this than simply multiplying our solutions by a power series simply because power series now appear in the differential equation. But I also feel like there is something really obvious that I am missing.

Thanks in advance.

• Is your question about why it is natural for $y_2$ to have $\log x$ added to it? May 21, 2017 at 18:37
• Why is it natural to multiply by power series May 21, 2017 at 18:38
• Well, for small $x$ the solution is approximately that for an Euler equation. What is natural is up to the reader. To me a power series solution is a natural ansatz when the functions $p$ and $q$ can be represented as power series. If this wasn't the case, it may still be a way to find a solution, but more difficult to solve. May 21, 2017 at 18:41

1 Answer

Locally, in the neighborhood of $x=0$, your equation can be approximated as $y''+\frac{p_0}{x}y'+\frac{q_0}{x^2}y=0$ which is Cauchy-Euler equation.

So the presence of power series factor in $y(x)=x^r(a_0+a_1x+....)$ is to account for the deviation of $y(x)$, away from $x=0$, from its asymptotic behavior $y(x)\rightarrow a_0.x^r$ as $x\rightarrow 0$.