I'm trying to practice the Frobenius method of solving ODEs, and I keep getting the answer wrong. It seems to be down to the shifting of limits of the sums, although it is not clear in the solutions I am using.

For example, consider the differential equation $$y''+\frac{y'}{x}+\left(-1-\frac{1}{4x^2}\right)y$$ By assuming the solution is of the form $y=\sum\limits_{n=0}^\infty c_nx^{n+p}$, this can be expressed in series: $$\sum\limits_{n=2}^\infty (n+p)(n+p-1)c_nx^{n+p-2}+\sum\limits_{n=1}^\infty (n+p)c_nx^{n+p-2}-\sum\limits_{n=0}^\infty c_nx^{n+p}-\frac{1}{4}\sum\limits_{n=0}^\infty c_nx^{n+p-2}$$

Then by shifting limits to make the sums correspond: $$\sum\limits_{n=0}^\infty (n+p+2)(n+p+1)c_{n+2}x^{n+p}+\sum\limits_{n=0}^\infty (n+p+1)c_{n+1}x^{n+p-1}-\sum\limits_{n=0}^\infty c_nx^{n+p}-\frac{1}{4}\sum\limits_{n=0}^\infty c_nx^{n+p-2}$$

But I am not sure what happens next. Clearly the indices of $x$ don't match up, but I'm pretty sure setting $n=-1$ in the summation is not the right thing to do. How can I get these terms all into the same summation so I can find a recurrence relation?

Using the indical equation, I can calculate that $p=\pm \frac{1}{2}$, but I don't think this helps at this stage.

If I shift the limits so that all the sums begin at $n=2$ I still have the problem of then the indices of $x$ not corresponding.

EDIT: Considering $n+p=z$ and multiplying the equation up by $x^2$ to get $x^2y''+xy'-x^2y-\frac{1}{4}y$, I arrive at $$\sum\limits_{n=2}^\infty (z)(z-1)c_nx^{z}+\sum\limits_{n=1}^\infty (z)c_nx^{z}-\sum\limits_{n=0}^\infty c_nx^{z+2}-\frac{1}{4}\sum\limits_{n=0}^\infty c_nx^{z}$$ where I can extend the limits of the first two sums to $0$ as their values at $n=0,1$ are $0$. This gives me $$\sum\limits_{n=0}^\infty \left((z)(z-1)+(z)-\frac{1}{4}\right)c_nx^{z}-\sum\limits_{n=2}^\infty c_{n-2}x^{z}$$

Thus giving a recurrence relation $$c_{n}=\frac{c_{n-2}}{z^2-\frac{1}{4}}$$

This is the correct solution. However, I only arrived there because multiplying up made the solution simpler on this occasion. Is this a general rule? Is there another way to see this solution?

For instance (sorry this question is getting so long, just trying to fully illustrate my point) if I am given $y''+\frac{1}{x}y'-\frac{4}{x^2}y$, multiplying up by $x^2$ does not make the answer clearer. Using the same method as above I arrive at $\sum\limits_{n=0}^\infty \left((z)(z-1)+(z)-4\right)c_nx^z=0 \implies (z^2-4)c_n=0$ which tells me nothing about the coefficients.

  • 2
    $\begingroup$ Why are you taking a series with $x^{n + p}$ and not just $x^{n}$? If you are arbitrarily choosing the power of $x$, why not just take $x^{n + p + 2}$ so that your 'lowest' $x$ power would be $x^{n + p}$. Just multiplying your equation by $x^{2}$ would make life easier.. $\endgroup$ – Mattos Apr 21 '15 at 9:48
  • $\begingroup$ In response to your edit, always multiply through so you have no fractions if possible, it makes calculations easier to deal with. $\endgroup$ – Mattos Apr 21 '15 at 10:55

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