Initial Value Problem to find series solution

Question

Consider the following initial value problem: $$e^{-x}y'' + \ln(1 + x)y' - x^2y = 0$$ $$y(0) = 1$$ $$y'(0) = 2$$

1. (done) Show that $x = 0$ is an ordinary point in the differential equation.
2. Find the first $4$ non-zero terms of the series solution centered in $x = 0$;
3. Which is the radius of convergence of the series solution found in $(2)$?

I have solved number 1, but number 2 seems to be giving me way too much trouble. First, I can't find the series solutions. I feel like I'm missing some quick corners, as I may not need to spend one full hour trying to work out the math in this. Could anyone help me?

Answer 1

This answer specifically addresses one way to work through part (2).

Assuming a solution of the form

$$y=\sum_{n\ge0}a_nx^n=a_0+a_1x+a_2x^2+\cdots$$

notice that $y(0)=a_0=1$, and similarly you have $y'(0)=a_1=2$.

Differentiate twice to find

$$y'=\sum_{n\ge0}na_nx^{n-1}=\sum_{n\ge1}na_nx^{n-1}=\sum_{n\ge0}(n+1)a_{n+1}x^n\\ y''=\sum_{n\ge1}n(n-1)a_nx^{n-2}=\sum_{n\ge2}n(n-1)a_nx^{n-2}=\sum_{n\ge0}(n+2)(n+1)a_{n+2}x^n$$

Recall that

$$e^{-x}=\sum_{n\ge0}\frac{(-x)^n}{n!}=\sum_{n\ge0}\frac{(-1)^n}{n!}x^n\\ \ln(1+x)=\sum_{n\ge1}\frac{(-x)^n}n=\sum_{n\ge1}\frac{(-1)^n}nx^n$$

Now in the ODE replace everything you can with these series expansions:

$$e^{-x}y''+\ln(1+x)y'-x^2y=0$$

becomes

$$\left(\sum_{n\ge0}\frac{(-1)^n}{n!}x^n\right)\left(\sum_{n\ge0}(n+2)(n+1)a_{n+2}x^n\right)+\left(\sum_{n\ge1}\frac{(-1)^n}nx^n\right)\left(\sum_{n\ge0}(n+1)a_{n+1}x^n\right)-\sum_{n\ge0}a_nx^{n+2}=0$$

Simplifying this looks like a sordid business, but fortunately the exercises asks you for just first several non-zero terms of the series expansion of $y$. To that end, we only care about extracting what information we can about the first several $a_n$, so consider truncating the expansions up to, say, $4$ terms each:

$$\color{red}{\left(1-x+\frac{x^2}2-\frac{x^3}6\right)\left(2\cdot1a_2+3\cdot2a_3x+4\cdot3a_4x^2+5\cdot4a_5x^3\right)}\\ \quad\quad+\color{green}{\left(x-\frac{x^2}2+\frac{x^3}3-\frac{x^4}4\right)\left(a_1+2a_2x+3a_3x^2+4a_4x^3\right)}\\ \quad\quad-\color{blue}{\left(a_0x^2+a_1x^3+a_2x^4+a_3x^5\right)}=0$$

Expand the left side and collect terms with the same power of $x$:

$$\begin{array}{c|rl} x^0&\color{red}{2a_2}+\color{green}{0}-\color{blue}{0}&=0\\ x^1&\color{red}{(-2a_2+6a_3)}+\color{green}{a_1}-\color{blue}{0}&=0\\ x^2&\color{red}{(a_2-6a_3+12a_4)}+\color{green}{\left(-\frac{a_1}2+2a_2\right)}-\color{blue}{a_0}&=0\\ x^3&\color{red}{\left(-\frac{a_2}3+3a_3-12a_4+20a_5\right)}+\color{green}{\left(\frac{a_1}3-a_2+3a_3\right)}-\color{blue}{a_1}&=0 \end{array}$$

Now, with $a_0=1$ and $a_1=2$, compute the next few coefficients:

$$\begin{align} 2a_2=0\implies a_2&=0\\ a_1-2a_2-6a_3=0\implies a_3&=-\frac13\\ -a_0-\frac{a_1}2+3a_2-6a_3+12a_4=0\implies a_4&=0\\ -\frac{2a_1}3-\frac{4a_2}3+6a_3-12a_4+20a_5=0\implies a_5&=-\frac16 \end{align}$$

So we find

$$y(x)\approx1+2x+\color{lightgray}{0x^2}-\frac{x^3}3+\color{lightgray}{0x^4}-\frac{x^5}6$$