# Frobenius method differential equations

I'm trying to solve this equation using Frobenius method. $$xy''-y'-4x^3y=0$$

$$y=\sum a_n x^{n+r}$$ However doing the indicial equations I get an incompatible system. $$(r(r-1)-r)a_0x^{r-2}=0 \\ (r+1)(r-1))a_1x^{r-1}=0$$

• Why do you think that you get two indicial equations? For $a_0\ne 0$ (why else would you start the power series at index $0$) you get $r(r-2)=0$, so that you have a solution for $r=0$ and possibly another independent one for $r=2$. It might also be that the second solution requires order reduction and contains some logarithmic term. – Lutz Lehmann Oct 19 '19 at 17:56
• But what happens with the other equation? If these terms aren't 0 I cant equal the series one to 0 – Marco Villalobos Oct 19 '19 at 18:28
• Of course you can satisfy the second equation, just set $a_1=0$. – Lutz Lehmann Oct 19 '19 at 18:52

For $$r=0$$ you get the coefficient iteration $$n(n+1)a_{n+1}-(n+1)a_{n+1}-4a_{n-3}=0$$ where it is understood that $$a_k=0$$ for $$k<0$$. This iteration formula tells us that the power series coefficients split into 4 independent sub-sequences $$(a_{4k+i})_k$$, $$i=0,1,2,3$$. Only the ones with $$i=0$$ and $$i=2$$ are non-zero. The second series for $$i=2$$ also accounts for the case $$r=2$$. As these sub-sequences give independent solutions, this is a full generating system, a solution basis.
If you set, inspired by the coefficient structure, $$y(x)=f(x^2/2)$$, then $$0=x[x^2f''(x^2/2)+f'(x^2/2)]-xf'(x^2/2)-4x^3f(x^2/2)=x^3(f''(x^2/2)-4f(x^2/2)]$$ which implies that $$f(t)=c_1e^{2t}+c_2e^{-2t}=d_1\cosh(2t)+d_2\sinh(2t)$$, so that $$y(x)=a_0\cosh(x^2)+a_1\sinh(x^2).$$
• The formula collects the coefficients of $x^n$. There are 4 subsequences because $n(n-2)a_{n}=4a_{n-4}$ connects only the elements of $(a_{4k+i})_k$. And what do you mean with "solve"? What you can do is to set $y(x)=u(x^4)$ or $y(x)=x^2v(x^4)$ and try to get something close to a Bessel equation. – Lutz Lehmann Oct 19 '19 at 18:56
• You get $a_{4k+4}=\frac{4}{(4k+4)(4k+2)}a_{4k}=\frac1{(2k+2)(2k+1)}a_{4k}$, so that $(2(k+1))!a_{4(k+1)}=(2k)!a_{4k}$ which gives $a_{4k}=\frac{a_0}{(2k)!}$ which are the coefficients of the hyperbolic cosine. – Lutz Lehmann Oct 19 '19 at 19:06