Differential Equations - Series Solutions: Examples

Differential Equations - Series Solutions: Examples - ii

Example 2 Find a series solution around

x_{0} = 0

for the following differential equation.

y^{″} - x y = 0

As with the first example

p (x) = 1

and so again for this differential equation every point is an ordinary point. Now we’ll start this one out just as we did the first example. Let’s write down the form of the solution and get its derivatives.

y (x) = \sum_{n = 0}^{\infty} a_{n} x^{n} y^{'} (x) = \sum_{n = 1}^{\infty} n a_{n} x^{n - 1} y^{″} (x) = \sum_{n = 2}^{\infty} n (n - 1) a_{n} x^{n - 2}

Plugging into the differential equation gives,

\sum_{n = 2}^{\infty} n (n - 1) a_{n} x^{n - 2} - x \sum_{n = 0}^{\infty} a_{n} x^{n} = 0

Unlike the first example we first need to get all the coefficients moved into the series.

\sum_{n = 2}^{\infty} n (n - 1) a_{n} x^{n - 2} - \sum_{n = 0}^{\infty} a_{n} x^{n + 1} = 0

Now we will need to shift the first series down by 2 and the second series up by 1 to get both of the series in terms of

x^{n} n

\sum_{n = 0}^{\infty} (n + 2) (n + 1) a_{n + 2} x^{n} - \sum_{n = 1}^{\infty} a_{n - 1} x^{n} = 0

Next, we need to get the two series starting at the same value of

n

. The only way to do that for this problem is to strip out the

n = 0

term.

\begin{aligned} (2) (1) a_{2} x^{0} + \sum_{n = 1}^{\infty} (n + 2) (n + 1) a_{n + 2} x^{n} - \sum_{n = 1}^{\infty} a_{n - 1} x^{n} & = 0 \\ 2 a_{2} + \sum_{n = 1}^{\infty} [(n + 2) (n + 1) a_{n + 2} - a_{n - 1}] x^{n} & = 0 \end{aligned}

We now need to set all the coefficients equal to zero. We will need to be careful with this however. The

n = 0

coefficient is in front of the series and the

n = 1, 2, 3, \dots

are all in the series. So, setting coefficient equal to zero gives,

\begin{aligned} n = 0 & 2 a_{2} & = 0 \\ n = 1, 2, 3, \dots & (n + 2) (n + 1) a_{n + 2} - a_{n - 1} & = 0 \end{aligned}

Solving the first as well as the recurrence relation gives,

\begin{aligned} n = 0 & a_{2} & = 0 \\ n = 1, 2, 3, \dots & a_{n + 2} & = \frac{a_{n - 1}}{(n + 2) (n + 1)} \end{aligned}

Now we need to start plugging in values of

n

$a_{3} = \frac{a_{0}}{(3) (2)}$	$a_{4} = \frac{a_{1}}{(4) (3)}$	$a_{5} = \frac{a_{2}}{(5) (4)} = 0$

$\begin{aligned} a_{6} & = \frac{a_{3}}{(6) (5)} \\ = \frac{a_{0}}{(6) (5) (3) (2)} \end{aligned}$	$\begin{aligned} a_{7} & = \frac{a_{4}}{(7) (6)} \\ = \frac{a_{1}}{(7) (6) (4) (3)} \end{aligned}$	$a_{8} = \frac{a_{5}}{(8) (7)} = 0$

$⋮$	$⋮$	$⋮$

$\begin{aligned} a_{3 k} & = \frac{a_{0}}{(2) (3) (5) (6) \dots (3 k - 1) (3 k)} \\ k & = 1, 2, 3, \dots \end{aligned}$	$\begin{aligned} a_{3 k + 1} & = \frac{a_{1}}{(3) (4) (6) (7) \dots (3 k) (3 k + 1)} \\ k & = 1, 2, 3, \dots \end{aligned}$	$\begin{aligned} a_{3 k + 2} & = 0 \\ k & = 0, 1, 2, \dots \end{aligned}$

There are a couple of things to note about these coefficients. First, every third coefficient is zero. Next, the formulas here are somewhat unpleasant and not all that easy to see the first time around. Finally, these formulas will not work for

k = 0

unlike the first example.
Now, get the solution,

\begin{aligned} y (x) & = a_{0} + a_{1} x + a_{2} x^{2} + a_{3} x^{3} + a_{4} x^{4} + \dots + a_{3 k} x^{3 k} + a_{3 k + 1} x^{3 k + 1} + \dots \\ = a_{0} + a_{1} x + \frac{a_{0}}{6} x^{3} + \frac{a_{1}}{12} x^{4} \dots + \frac{a_{0} x^{3 k}}{(2) (3) (5) (6) \dots (3 k - 1) (3 k)} + \\ \frac{a_{1} x^{3 k + 1}}{(3) (4) (6) (7) \dots (3 k) (3 k + 1)} + \dots \end{aligned}

Again, collect up the terms that contain the same coefficient, factor the coefficient out and write the results as a new series,

y (x) = a_{0} {1 + \sum_{k = 1}^{\infty} \frac{x^{3 k}}{(2) (3) (5) (6) \dots (3 k - 1) (3 k)}} + a_{1} {x + \sum_{k = 1}^{\infty} \frac{x^{3 k + 1}}{(3) (4) (6) (7) \dots (3 k) (3 k + 1)}}

We couldn’t start our series at

k = 0

this time since the general term doesn’t hold for

k = 0

Now, we need to work an example in which we use a point other that

x = 0

. In fact, let’s just take the previous example and rework it for a different value of

x_{0}

. We’re also going to need to change up the instructions a little for this example.

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