Question

We would like to use the power series method to find the general solution to the differential equation d 2y dx2 − 4x dy dx + 12y = 0, 56 so we assume the solution is of the form y = X∞ n=0 and n , a power series centered at 0, and determine the coefficients an. a. As a first step, find the recursion formula for an+2 in terms of an. b. The coefficients a0 and a1 will be determined by the initial conditions. Use the recursion formula to determine an in terms of a0 and a1, for 2 ≤ n ≤ 9. c. Find a nonzero polynomial solution to this differential equation. d. Find a basis for the space of solutions to the equation. e. Find the solution to the initial value problem d 2y dx2 − 4x dy dx + 12y = 0, y(0) = 0, dy dx(0) = 1. f. To solve the differential equation d 2y dx2 − 4(x − 3) dy dx + 12y = 0, it would be most natural to assume that the solution has the form y = X∞ n=0 an(x − 3)n . Use this idea to find a polynomial solution to the differential equation d 2y dx2 − 4(x − 3) dy dx + 12y = 0

Answer 1

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

$\dfrac{\mathrm dy}{\mathrm dx}=\displaystyle\sum_{n\ge1}na_nx^{n-1}\implies4x\dfrac{\mathrm dy}{\mathrm dx}=4\sum_{n\ge1}na_nx^n=4\sum_{n\ge0}na_nx^n$

$\dfrac{\mathrm d^2y}{\mathrm dx^2}=\displaystyle\sum_{n\ge2}n(n-1)a_nx^{n-2}=\sum_{n\ge0}(n+2)(n+1)a_{n+2}x^n$

Substituting into the ODE

$\dfrac{\mathrm d^2y}{\mathrm dx^2}-4x\dfrac{\mathrm dy}{\mathrm dx}+12y=0$

gives

$\displaystyle\sum_{n\ge0}\bigg((n+2)(n+1)a_{n+2}-4na_n+12a_n\bigg)x^n=0$

so that the coefficients of the series are given according to

$\begin{cases}a_0=y(0)\\a_1=y'(0)\\a_{n+2}=\dfrac{4(n-3)a_n}{(n+2)(n+1)}&\text{for }n\ge0\end{cases}$

We can shift the index in the recursive part of this definition to get

$a_n=\dfrac{4(n-5)a_{n-2}}{n(n-1)}$

for $n\ge2$. There's dependency between coefficients that are 2 indices apart, so we can consider 2 cases:

• If $n=2k$, where $k\ge0$ is an integer, then

$k=0\implies n=0\implies a_0=a_0$

but since $y(0)=0$, we have $a_0=0$ and $a_{2k}=0$ for all $k\ge0$.

• If $n=2k+1$, then

$k=0\implies n=1\implies a_1=a_1$

$k=1\implies n=3\implies a_3=\dfrac{4(-2)a_1}{3\cdot2}=-\dfrac43a_1$

$k=2\implies n=5\implies a_5=0$

and so $a_{2k+1}=0$ for all $k\ge2$. If $y'(0)=1$, we then have $a_1=1$ and $a_3=-\dfrac43$.

So the ODE has solution

$y(x)=\displaystyle\sum_{k\ge0}(a_{2k}x^{2k}+a_{2k+1}x^{2k+1})\implies\boxed{y(x)=x-\dfrac43x^3}$