# Can I solve an Euler differential equation by using the Frobenius method?

I'm having some trouble by trying to solve Euler equations by using the Frobenius method. For example, I'm asked to solve the Euler differential equation

$$x^2y'' + xy' - y = 0$$

using a power series solution.

I start by assuming there is at least one solution with the form $y = x^\sigma\sum{a_nx^n}$.

First, I divide the whole differential equation by $x^2$. Then I substitute the expression above so I get

$$\sum{(n+\sigma)(n+\sigma-1)a_nx^{n+\sigma-2}} + \frac{1}{x}\sum{(n+\sigma)a_nx^{n+\sigma-1}}-\frac{1}{x^2}\sum{a_nx^{n+\sigma}}$$

And now, dividing by $x^{\sigma-2}$, I get

$$\sum{((n+\sigma)(n+\sigma-1)+(n+\sigma) - 1})a_nx^n$$

Now I don't know how to find the recurrence relation I'm looking for in order to find the form of $a_n$. In all the examples I've been able to find, in the last expression one always finds terms of $a_{n-1}$, for example, but here I don't know how to continue.

Did I do something wrong? I'm confused because I believe the equation given fits the requeriments needed in order to the Frobenius method to be applicable, but this happens to me every time I try to solve an Euler equation using it.

Thank you very much in advance.

• You have to set that $n \geq 0$, i.e., $a_{n<0}=0$ and $a_0 \neq 0$, otherwise $\sigma$ is not well defined. Take then a look at the equation for $n=0$. Commented Sep 1, 2014 at 11:57
• In short, the Euler equation is a Frobenius method problem where only the indicial equation matters. Or, from my perspective teaching, the method of Frobenius is a generalization of the Euler ODE solution to other singular ODEs of the same managable (regular) singularity. Commented Sep 1, 2014 at 13:24

First you need the indicial equation: we have $$((n+\sigma)(n+\sigma-1)+(n+\sigma)-1)a_n=0\tag{*}$$ for all $n$, and you can assume that $a_0\ne0$, otherwise take a different value of $\sigma$. So taking $n=0$ gives $$\sigma(\sigma-1)+\sigma-1=0$$ and hence $\sigma=\pm1$. Considering $\sigma=-1$, equation $(*)$ becomes $$(n^2-2n)a_n=0$$ for all $n$; therefore $a_0$ and $a_2$ are arbitrary, all others must be zero. So your solution is $$y=x^{-1}(a_0+a_2x^2)=a_0x^{-1}+a_2x\ .$$ It is easy to check that this is a solution of your DE, and it has two arbitrary constants, as expected.
If you take $\sigma=1$, a similar process gives only $a_0$ non-zero, so $y=a_0x$, which is already included in the above solution. When the roots of the indicial equation differ by an integer, sometimes (though not here) it is necessary to seek an alternative form of solution. Try searching for "frobenius method solutions differ by an integer".