I'm trying to solve a 2nd order ODE with frobenius method

Question

I was trying to solve the ODE $$2x(1+x)y'' + (3+x)y' - xy = 0$$ but I couldn't seem to get the recurrence expression. I already got the roots of the indicial equation which are $r=0$ and $r=-1/2$ but after that I'm stuck. Could someone please help me with that, thank you!

Answer 1

Since there are two indicial solution, I show you how to prodeed for the $r=0$ one, and let you do the one with $r=-\frac 12$.

$\displaystyle-xy=-x\sum\limits_{i=0}^{\infty}a_nx^n=\sum\limits_{i=0}^{\infty}(-a_n)x^{n+1}=\sum\limits_{i=1}^{\infty}(-a_{n-1})x^n$

• First step is to make $(-x)$ enter the summation.
• Then we shift the index to get a proper $x^n$

$\begin{align}(3+x)y'&=(3+x)\sum\limits_{i=1}^{\infty} na_nx^{n-1}=\sum\limits_{i=1}^{\infty} (3na_n)x^{n-1}+\sum\limits_{i=1}^{\infty} (na_n)x^{n}\\ &=\sum\limits_{i=0}^{\infty} (3(n+1)a_{n+1})x^{n}+\sum\limits_{i=1}^{\infty} (na_n)x^{n} \end{align}$

• Here in the first step we split the sum according to powers of $x$ so that the index shifting is facilitated
• Notice the derivation makes the index start at $1$ instead of $0$

$\begin{align}2x(1+x)y'' &=(2x+2x^2)\sum\limits_{i=2}^{\infty} n(n-1)a_nx^{n-2}\\ &=\sum\limits_{i=2}^{\infty} (2n(n-1)a_n)x^{n-1}+\sum\limits_{i=2}^{\infty} (2n(n-1)a_n)x^{n}\\ &=\sum\limits_{i=1}^{\infty} (2n(n+1)a_{n+1})x^{n}+\sum\limits_{i=2}^{\infty} (2n(n-1)a_n)x^{n} \end{align}$

• Here the double derivation makes the index start at $2$ instead of $0$
  

$\begin{align}\text{ODE} &=\Big(\overbrace{-a_0x}^{y\text{ term}}+\overbrace{3a_1+6a_2x+a_1x}^{y'\text{ term}}+\overbrace{4a_2x}^{y''\text{ term}}\Big)+\\ &\quad\sum\limits_{i=2}^{\infty}\Big(\underbrace{-a_{n-1}}_{y\text{ term}}+\underbrace{3(n+1)a_{n+1}+na_n}_{y'\text{ term}}+\underbrace{2n(n+1)a_{n+1}+2n(n-1)a_n}_{y''\text{ term}}\Big)x^n\end{align}$

• Notice that we have various starting index series $i=0$, $i=1$ and $i=2$, so the common part of all these terms is from $i=2$.
• We isolate all the terms with index $i<2$, i.e the constant terms in $x^0$ and the linear terms in $x^1$.
• We regroup the rest into the same big summation

Now, identification with the zero series gives:

$\begin{cases} 3a_1=0\\ -a_0+a_1+10a_2=0\\ (2n+3)(n+1)a_{n+1}+n(2n-1)a_n-a_{n-1}=0 \end{cases}$

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For the other one, let set $y(x)=\dfrac{u(x)}{\sqrt{x}}$, the ode is transformed to

$$2x(1+x)u''+(1-x)u'+(1-x)u=0$$

Can you take it from here and apply the same method for the new $(1-x)u'$ and $(1-x)u$ terms ?