Question

asked Jan 2, 2018 166k views

1 Answer

Dan Hixon · Answer 1 · 2018-01-08T12:15:22+0000

Assuming we're looking for a power series solution centered around
x=0

, take

$y=\displaystyle\sum_(n\ge0)a_nx^n$

$y'=\displaystyle\sum_(n\ge1)na_nx^(n-1)$

$y''=\displaystyle\sum_(n\ge2)n(n-1)a_nx^(n-2)$

Substituting into the ODE yields

$\displaystyle\sum_(n\ge2)n(n-1)a_nx^(n-2)+\sum_(n\ge1)na_nx^(n+1)-\sum_(n\ge0)a_nx^(n+1)=0$

The first series starts with a constant term; the second series starts at
x^2

; the last starts at
x^1

. So, extract the first two terms from the first series, and the first term from the last series so that each new series starts with a
x^2

term. We have

$\displaystyle\sum_(n\ge2)n(n-1)a_nx^(n-2)=2a_2+6a_3x+\sum_(n\ge4)n(n-1)a_nx^(n-2)$

$\displaystyle\sum_(n\ge0)a_nx^(n+1)=a_0x+\sum_(n\ge1)a_nx^(n+1)$

Re-index the first sum to have it start at
n=1

(to match the the other two sums):

$\displaystyle\sum_(n\ge4)n(n-1)a_nx^(n-2)=\sum_(n\ge1)(n+3)(n+2)a_(n+3)x^(n+1)$

So now the ODE is

$\displaystyle\left(2a_2+6a_3x+\sum_(n\ge1)(n+3)(n+2)a_(n+3)x^(n+1)\right)+\sum_(n\ge1)na_nx^(n+1)-\left(a_0x+\sum_(n\ge1)a_nx^(n+1)\right)=0$

Consolidate into one series starting
n=1

:

$\displaystyle2a_2+(6a_3-a_0)x+\sum_(n\ge1)\bigg[(n+3)(n+2)a_(n+3)+(n-1)a_n\bigg]x^(n+1)=0$

Suppose we're given initial conditions
y(0)=a_0

and

(which follow from setting
x=0

in the power series representations for

and

, respectively). From the above equation it follows that

$\begin{cases}2a_2=0\\6a_3-a_0=0\\(n+3)(n+2)a_(n+3)+(n-1)a_n=0&\text{for }n\ge2\end{cases}$

Let's first consider what happens when
n=3k-2

, i.e.
$n\in\{1,4,7,10,\ldots\}$ . The recurrence relation tells us that

$a_4=-(1-1)/((1+3)(1+2))a_1=0\implies a_7=0\implies a_(10)=0$

and so on, so that
a_(3k-2)=0

except for when
k=1

.

Now let's consider
n=3k-1

, or
$n\in\{2,5,8,11,\ldots\}$ . We know that
a_2=0

, and from the recurrence it follows that
a_(3k-1)=0

for all

.

Finally, take
n=3k

, or
$n\in\{0,3,6,9,\ldots\}$ . We have a solution for
a_3

in terms of
a_0

, so the next few terms (
k=2,3,4

) according to the recurrence would be

$a_6=-\frac2{6\cdot5}a_3=-\frac2{6\cdot5\cdot3\cdot2}a_0=-(a_0)/(6\cdot3\cdot5)$

$a_9=-\frac5{9\cdot8}a_6=(a_0)/(9\cdot6\cdot3\cdot8)$

$a_(12)=-\frac8{12\cdot11}a_9=-(a_0)/(12\cdot9\cdot6\cdot3\cdot11)$

and so on. The reordering of the product in the denominator is intentionally done to make the pattern clearer. We can surmise the general pattern for
n=3k

as

$a_(3k)=((-1)^(k+1)a_0)/((3k\cdot(3k-3)\cdot(3k-2)\cdot\cdots\cdot6\cdot3\cdot(3k-1))$

$a_(3k)=((-1)^(k+1)a_0)/(3^k(k\cdot(k-1)\cdot\cdots\cdot2\cdot1)\cdot(3k-1))$

a_(3k)=((-1)^(k+1)a_0)/(3^kk!(3k-1))

So the series solution to the ODE is given by

$y=\displaystyle\sum_(n\ge0)a_nx^n$

$y=a_1x+\displaystyle\sum_(k\ge0)((-1)^(k+1)a_0)/(3^kk!(3k-1))$

Attached is a plot of a numerical solution (blue) to the ODE with initial conditions sampled at
a_0=y(0)=1