Question

A. Seek power series solutions of the given differential equation about the given point x0; find the recurrence relation that the coefficients must satisfy.
b. Find the first four nonzero terms in each of two solutions y1 and y2 (unless the series terminates sooner).


c. By evaluating the Wronskian W[y1, y2](x0), show that y1 and y2 form a fundamental set of solutions.


d. If possible, find the general term in each solution.


y" + 3y' = 0, x0 = 0

Answer 1

We want to find a solution such that

$y=\displaystyle\sum_{n\ge0}a_nx^n$

$y'=\displaystyle\sum_{n\ge0}(n+1)a_{n+1}x^n$

$y''=\displaystyle\sum_{n\ge0}(n+1)(n+2)a_{n+2}x^n$

With these conditions, $y(0)=a_0$ and $y'(0)=a_1$.

Substituting the series into the ODE gives

$\displaystyle\sum_{n\ge0}(n+1)(n+2)a_{n+2}x^n+3\sum_{n\ge0}(n+1)a_{n+1}x^n=0$

$\displaystyle\sum_{n\ge0}\bigg((n+1)(n+2)a_{n+2}+3(n+1)a_{n+1}\bigg)x^n=0$

so that the coefficients of $y$ satisfy

$(n+1)(n+2)a_{n+2}+3(n+1)a_{n+1}=0$

for $n\ge0$, or

$(n-1)na_n+3(n-1)a_{n-1}=0\implies a_n=-\dfrac3na_{n-1}$

for $n\ge2$. Notice that this implies a dependency of all $a_n$ beyond $n=1$ on $a_1$, while $a_0$ takes on whatever initial value $y(0)$ is given. In particular,

$a_2=-\dfrac32a_1$

$a_3=-\dfrac33a_2=\dfrac{3^2}{3\cdot2}a_1=\dfrac{3^2}{3!}a_1$

$a_4=-\dfrac34a_3=-\dfrac{3^3}{4\cdot3\cdot2}a_1=-\dfrac{3^3}{4!}a_1$

and so on up to

$a_n=\dfrac{(-3)^{n-1}}{n!}a_1$

So we can extract two fundamental solutions $y_1,y_2$ such that $y=y_1+y_2$, where

$y_1=a_0$

$y_2=\displaystyle-3a_1\sum_{n\ge1}\frac{(-3x)^n}{n!}$

Recall that

$e^x=\displaystyle\sum_{n\ge0}\frac{x^n}{n!}$

which tells us

$y_2=-3a_1(e^{-3x}-1)=-3a_1e^{-3x}+3a_1$

but $y_1$ is a constant solution and already accounts for the constant term in $y_2$, and $-3a_1$ can be reduced to a simpler constant $a_1$, leaving us with

$y_2=a_1e^{-3x}$

The Wronskian is

$W(y_1,y_2)=\begin{vmatrix}y_1&y_2\\{y_1}'&{y_2}'\end{vmatrix}=\begin{vmatrix}a_0&a_1e^{-3x}\\0&-3a_1e^{-3x}\end{vmatrix}=-3a_0a_1e^{-3x}$

$\implies W(y_1,y_2)(0)=-3a_0a_1$

so the two solutions are indeed independent as long as neither initial value is 0.