7.4 Series solutions of differential equations

Calculus volume 3 Page 1 / 2

Use power series to solve first-order and second-order differential equations.

In Introduction to Power Series , we studied how functions can be represented as power series, $y (x) = \sum_{n = 0}^{\infty} a_{n} x^{n} .$ We also saw that we can find series representations of the derivatives of such functions by differentiating the power series term by term. This gives $y^{'} (x) = \sum_{n = 1}^{\infty} n a_{n} x^{n - 1}$ and $y ″ (x) = \sum_{n = 2}^{\infty} n (n - 1) a_{n} x^{n - 2} .$ In some cases, these power series representations can be used to find solutions to differential equations.

Be aware that this subject is given only a very brief treatment in this text. Most introductory differential equations textbooks include an entire chapter on power series solutions. This text has only a single section on the topic, so several important issues are not addressed here, particularly issues related to existence of solutions. The examples and exercises in this section were chosen for which power solutions exist. However, it is not always the case that power solutions exist. Those of you interested in a more rigorous treatment of this topic should consult a differential equations text.

Problem-solving strategy: finding power series solutions to differential equations

Assume the differential equation has a solution of the form $y (x) = \sum_{n = 0}^{\infty} a_{n} x^{n} .$
Differentiate the power series term by term to get $y^{'} (x) = \sum_{n = 1}^{\infty} n a_{n} x^{n - 1}$ and $y ″ (x) = \sum_{n = 2}^{\infty} n (n - 1) a_{n} x^{n - 2} .$
Substitute the power series expressions into the differential equation.
Re-index sums as necessary to combine terms and simplify the expression.
Equate coefficients of like powers of $x$ to determine values for the coefficients $a_{n}$ in the power series.
Substitute the coefficients back into the power series and write the solution.

Series solutions to differential equations

Find a power series solution for the following differential equations.

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

Assume $y (x) = \sum_{n = 0}^{\infty} a_{n} x^{n}$ (step 1). Then, $y^{'} (x) = \sum_{n = 1}^{\infty} n a_{n} x^{n - 1}$ and $y ″ (x) = \sum_{n = 2}^{\infty} n (n - 1) a_{n} x^{n - 2}$ (step 2). We want to find values for the coefficients $a_{n}$ such that

$\begin{matrix} y ″ - y & = & 0 \\ \sum_{n = 2}^{\infty} n (n - 1) a_{n} x^{n - 2} - \sum_{n = 0}^{\infty} a_{n} x^{n} & = & 0 (step 3). \end{matrix}$

We want the indices on our sums to match so that we can express them using a single summation. That is, we want to rewrite the first summation so that it starts with $n = 0 .$
To re-index the first term, replace n with $n + 2$ inside the sum, and change the lower summation limit to $n = 0 .$ We get

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

This gives

$\begin{matrix} \sum_{n = 0}^{\infty} (n + 2) (n + 1) a_{n + 2} x^{n} - \sum_{n = 0}^{\infty} a_{n} x^{n} & = & 0 \\ \sum_{n = 0}^{\infty} [(n + 2) (n + 1) a_{n + 2} - a_{n}] x^{n} & = & 0 (step 4). \end{matrix}$

Because power series expansions of functions are unique, this equation can be true only if the coefficients of each power of x are zero. So we have

$(n + 2) (n + 1) a_{n + 2} - a_{n} = 0 for n = 0, 1, 2,….$

This recurrence relationship allows us to express each coefficient $a_{n}$ in terms of the coefficient two terms earlier. This yields one expression for even values of n and another expression for odd values of n . Looking first at the equations involving even values of n , we see that

$\begin{matrix} a_{2} & = & \frac{a_{0}}{2} \\ a_{4} & = & \frac{a_{2}}{4 \cdot 3} = \frac{a_{0}}{4!} \\ a_{6} & = & \frac{a_{4}}{6 \cdot 5} = \frac{a_{0}}{6!} \\ ⋮. \end{matrix}$

Thus, in general, when n is even, $a_{n} = \frac{a_{0}}{n!}$ (step 5).
For the equations involving odd values of n , we see that

$\begin{matrix} a_{3} & = & \frac{a_{1}}{3 \cdot 2} = \frac{a_{1}}{3!} \\ a_{5} & = & \frac{a_{3}}{5 \cdot 4} = \frac{a_{1}}{5!} \\ a_{7} & = & \frac{a_{5}}{7 \cdot 6} = \frac{a_{1}}{7!} \\ ⋮. \end{matrix}$

Therefore, in general, when n is odd, $a_{n} = \frac{a_{1}}{n!}$ (step 5 continued).
Putting this together, we have

$\begin{matrix} y (x) & = \sum_{n = 0}^{\infty} a_{n} x^{n} \\ = a_{0} + a_{1} x + \frac{a_{0}}{2} x^{2} + \frac{a_{1}}{3!} x^{3} + \frac{a_{0}}{4!} x^{4} + \frac{a_{1}}{5!} x^{5} + ⋯. \end{matrix}$

Re-indexing the sums to account for the even and odd values of n separately, we obtain

$y (x) = a_{0} \sum_{k = 0}^{\infty} \frac{1}{(2 k)!} x^{2 k} + a_{1} \sum_{k = 0}^{\infty} \frac{1}{(2 k + 1)!} x^{2 k + 1} (step 6).$

Analysis for part a.
As expected for a second-order differential equation, this solution depends on two arbitrary constants. However, note that our differential equation is a constant-coefficient differential equation, yet the power series solution does not appear to have the familiar form (containing exponential functions) that we are used to seeing. Furthermore, since $y (x) = c_{1} e^{x} + c_{2} e^{− x}$ is the general solution to this equation, we must be able to write any solution in this form, and it is not clear whether the power series solution we just found can, in fact, be written in that form.
Fortunately, after writing the power series representations of $e^{x}$ and $e^{− x},$ and doing some algebra, we find that if we choose

$c_{0} = \frac{(a_{0} + a_{1})}{2}, c_{1} = \frac{(a_{0} - a_{1})}{2},$

we then have $a_{0} = c_{0} + c_{1}$ and $a_{1} = c_{0} - c_{1},$ and

$\begin{matrix} y (x) & = a_{0} + a_{1} x + \frac{a_{0}}{2} x^{2} + \frac{a_{1}}{3!} x^{3} + \frac{a_{0}}{4!} x^{4} + \frac{a_{1}}{5!} x^{5} + ⋯ \\ = (c_{0} + c_{1}) + (c_{0} - c_{1}) x + \frac{(c_{0} + c_{1})}{2} x^{2} + \frac{(c_{0} - c_{1})}{3!} x^{3} + \frac{(c_{0} + c_{1})}{4!} x^{4} + \frac{(c_{0} - c_{1})}{5!} x^{5} + ⋯ \\ = c_{0} \sum_{n = 0}^{\infty} \frac{x^{n}}{n!} + c_{1} \sum_{n = 0}^{\infty} \frac{{(− x)}^{n}}{n!} \\ = c_{0} e^{x} + c_{1} e^{− x} . \end{matrix}$

So we have, in fact, found the same general solution. Note that this choice of $c_{1}$ and $c_{2}$ is not obvious. This is a case when we know what the answer should be, and have essentially “reverse-engineered” our choice of coefficients.
Assume $y (x) = \sum_{n = 0}^{\infty} a_{n} x^{n}$ (step 1). Then, $y^{'} (x) = \sum_{n = 1}^{\infty} n a_{n} x^{n - 1}$ and $y ″ (x) = \sum_{n = 2}^{\infty} n (n - 1) a_{n} x^{n - 2}$ (step 2). We want to find values for the coefficients $a_{n}$ such that

$\begin{matrix} (x^{2} - 1) y ″ + 6 x y^{'} + 4 y & = & −4 \\ (x^{2} - 1) \sum_{n = 2}^{\infty} n (n - 1) a_{n} x^{n - 2} + 6 x \sum_{n = 1}^{\infty} n a_{n} x^{n - 1} + 4 \sum_{n = 0}^{\infty} a_{n} x^{n} & = & −4 \\ x^{2} \sum_{n = 2}^{\infty} n (n - 1) a_{n} x^{n - 2} - \sum_{n = 2}^{\infty} n (n - 1) a_{n} x^{n - 2} + 6 x \sum_{n = 1}^{\infty} n a_{n} x^{n - 1} + 4 \sum_{n = 0}^{\infty} a_{n} x^{n} & = & −4 . \end{matrix}$

Taking the external factors inside the summations, we get

$\sum_{n = 2}^{\infty} n (n - 1) a_{n} x^{n} - \sum_{n = 2}^{\infty} n (n - 1) a_{n} x^{n - 2} + \sum_{n = 1}^{\infty} 6 n a_{n} x^{n} + \sum_{n = 0}^{\infty} 4 a_{n} x^{n} = −4 (step 3).$

Now, in the first summation, we see that when $n = 0$ or $n = 1,$ the term evaluates to zero, so we can add these terms back into our sum to get

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

Similarly, in the third term, we see that when $n = 0,$ the expression evaluates to zero, so we can add that term back in as well. We have

$\sum_{n = 1}^{\infty} 6 n a_{n} x^{n} = \sum_{n = 0}^{\infty} 6 n a_{n} x^{n} .$

Then, we need only shift the indices in our second term. We get

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

Thus, we have

$\begin{matrix} \sum_{n = 0}^{\infty} n (n - 1) a_{n} x^{n} - \sum_{n = 0}^{\infty} (n + 2) (n + 1) a_{n + 2} x^{n} + \sum_{n = 0}^{\infty} 6 n a_{n} x^{n} + \sum_{n = 0}^{\infty} 4 a_{n} x^{n} & = & −4 (step 4). \\ \sum_{n = 0}^{\infty} [n (n - 1) a_{n} - (n + 2) (n + 1) a_{n + 2} + 6 n a_{n} + 4 a_{n}] x^{n} & = & −4 \\ \sum_{n = 0}^{\infty} [(n^{2} - n) a_{n} + 6 n a_{n} + 4 a_{n} - (n + 2) (n + 1) a_{n + 2}] x^{n} & = & −4 \\ \sum_{n = 0}^{\infty} [n^{2} a_{n} + 5 n a_{n} + 4 a_{n} - (n + 2) (n + 1) a_{n + 2}] x^{n} & = & −4 \\ \sum_{n = 0}^{\infty} [(n^{2} + 5 n + 4) a_{n} - (n + 2) (n + 1) a_{n + 2}] x^{n} & = & −4 \\ \sum_{n = 0}^{\infty} [(n + 4) (n + 1) a_{n} - (n + 2) (n + 1) a_{n + 2}] x^{n} & = & −4 \end{matrix}$

Looking at the coefficients of each power of x , we see that the constant term must be equal to $−4,$ and the coefficients of all other powers of x must be zero. Then, looking first at the constant term,

$\begin{matrix} 4 a_{0} - 2 a_{2} & = & −4 \\ a_{2} & = & 2 a_{0} + 2 (step 3). \end{matrix}$

For $n \geq 1,$ we have

$\begin{matrix} (n + 4) (n + 1) a_{n} - (n + 2) (n + 1) a_{n + 2} & = & 0 \\ (n + 1) [(n + 4) a_{n} - (n + 2) a_{n + 2}] & = & 0 . \end{matrix}$

Since $n \geq 1,$ $n + 1 \neq 0,$ we see that

$(n + 4) a_{n} - (n + 2) a_{n + 2} = 0$

and thus

$a_{n + 2} = \frac{n + 4}{n + 2} a_{n} .$

For even values of n , we have

$\begin{matrix} a_{4} & = & \frac{6}{4} (2 a_{0} + 2) = 3 a_{0} + 3 \\ a_{6} & = & \frac{8}{6} (3 a_{0} + 3) = 4 a_{0} + 4 \\ ⋮. \end{matrix}$

In general, $a_{2 k} = (k + 1) (a_{0} + 1)$ (step 5).
For odd values of n , we have

$\begin{matrix} a_{3} & = & \frac{5}{3} a_{1} \\ a_{5} & = & \frac{7}{5} a_{3} = \frac{7}{3} a_{1} \\ a_{7} & = & \frac{9}{7} a_{5} = \frac{9}{3} a_{1} = 3 a_{1} \\ ⋮. \end{matrix}$

In general, $a_{2 k + 1} = \frac{2 k + 3}{3} a_{1}$ (step 5 continued).
Putting this together, we have

$y (x) = \sum_{k = 0}^{\infty} (k + 1) (a_{0} + 1) x^{2 k} + \sum_{k = 0}^{\infty} (\frac{2 k + 3}{3}) a_{1} x^{2 k + 1} (step 6).$

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Source: OpenStax, Calculus volume 3. OpenStax CNX. Feb 05, 2016 Download for free at http://legacy.cnx.org/content/col11966/1.2

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