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`

`(\mathrm dy)/(\mathrm dx)=\displaystyle\sum_(n\ge1)na_nx^(n-1)\implies4x(\mathrm dy)/(\mathrm dx)=4\sum_(n\ge1)na_nx^n=4\sum_(n\ge0)na_nx^n`

`(\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

`(\mathrm d^2y)/(\mathrm dx^2)-4x(\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)=(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=(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=(4(-2)a_1)/(3\cdot2)=-\frac43a_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=-\frac43`.

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-\frac43x^3}`