Differential Equations - Series Solutions: Examples

Differential Equations - Series Solutions: Examples - iv

Let’s work one final problem.

Example 4 Find the first four terms in each portion of the series solution around

x_{0} = 0

for the following differential equation.

(x^{2} + 1) y^{″} - 4 x y^{'} + 6 y = 0

We finally have a differential equation that doesn’t have a constant coefficient for the second derivative.

p (x) = x^{2} + 1 p (0) = 1 \neq 0

x_{0} = 0

is an ordinary point for this differential equation. We first need the solution and 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}

Plug these into the differential equation.

(x^{2} + 1) \sum_{n = 2}^{\infty} n (n - 1) a_{n} x^{n - 2} - 4 x \sum_{n = 1}^{\infty} n a_{n} x^{n - 1} + 6 \sum_{n = 0}^{\infty} a_{n} x^{n} = 0

Now, break up the first term into two so we can multiply the coefficient into the series and multiply the coefficients of the second and third series in as well.

\sum_{n = 2}^{\infty} n (n - 1) a_{n} x^{n} + \sum_{n = 2}^{\infty} n (n - 1) a_{n} x^{n - 2} - \sum_{n = 1}^{\infty} 4 n a_{n} x^{n} + \sum_{n = 0}^{\infty} 6 a_{n} x^{n} = 0

We will only need to shift the second series down by two to get all the exponents the same in all the series.

\sum_{n = 2}^{\infty} n (n - 1) a_{n} x^{n} + \sum_{n = 0}^{\infty} (n + 2) (n + 1) a_{n + 2} x^{n} - \sum_{n = 1}^{\infty} 4 n a_{n} x^{n} + \sum_{n = 0}^{\infty} 6 a_{n} x^{n} = 0

At this point we could strip out some terms to get all the series starting at

n = 2

, but that’s actually more work than is needed. Let’s instead note that we could start the third series at

n = 0

if we wanted to because that term is just zero. Likewise, the terms in the first series are zero for both

n = 1

and

n = 0

and so we could start that series at

n = 0

. If we do this all the series will now start at

n = 0

and we can add them up without stripping terms out of any series.

\begin{aligned} \sum_{n = 0}^{\infty} [n (n - 1) a_{n} + (n + 2) (n + 1) a_{n + 2} - 4 n a_{n} + 6 a_{n}] x^{n} & = 0 \\ \sum_{n = 0}^{\infty} [(n^{2} - 5 n + 6) a_{n} + (n + 2) (n + 1) a_{n + 2}] x^{n} & = 0 \\ \sum_{n = 0}^{\infty} [(n - 2) (n - 3) a_{n} + (n + 2) (n + 1) a_{n + 2}] x^{n} & = 0 \end{aligned}

Now set coefficients equal to zero.

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

Solving this gives,

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

Now, we plug in values of

n

n = 0 : a_{2} = - 3 a_{0}

n = 1 : a_{3} = - \frac{1}{3} a_{1}

n = 2 : a_{4} = - \frac{0}{12} a_{2} = 0

n = 3 : a_{5} = - \frac{0}{20} a_{3} = 0

Now, from this point on all the coefficients are zero. In this case both of the series in the solution will terminate. This won’t always happen, and often only one of them will terminate.
The solution in this case is,

y (x) = a_{0} {1 - 3 x^{2}} + a_{1} {x - \frac{1}{3} x^{3}}

Comments

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