Differential Equations - Higher Order: Series Solutions

The purpose of this section is not to do anything new with a series solution problem. Instead it is here to illustrate that moving into a higher order differential equation does not really change the process outside of making it a little longer.

Back in the Series Solution chapter we only looked at 2^nd order differential equations so we’re going to do a quick example here involving a 3^rd order differential equation so we can make sure and say that we’ve done at least one example with an order larger than 2.

Example 1 Find the series solution around $x_0=0$ for the following differential equation.$$y^{\prime \prime \prime }+x^2{y}^{\prime }+xy=0$$

Recall that we can only find a series solution about $x_0=0$ if this point is an ordinary point, or in other words, if the coefficient of the highest derivative term is not zero at $x_0=0$. That is clearly the case here so let’s start with the form of the solutions as well as the derivatives that we’ll need for this solution.
$$y\left(x\right)=\sum _{n=0}^{\infty }{a}_n{x}^n{y}^{\prime }\left(x\right)=\sum _{n=1}^{\infty }n{a}_n{x}^{n-1}{y}^{\prime \prime \prime }\left(x\right)=\sum _{n=3}^{\infty }n\left(n-1\right)\left(n-2\right)a_n{x}^{n-3}$$Plugging into the differential equation gives,
$$\sum _{n=3}^{\infty }n\left(n-1\right)\left(n-2\right)a_n{x}^{n-3}+x^2\sum _{n=1}^{\infty }n{a}_n{x}^{n-1}+x\sum _{n=0}^{\infty }{a}_n{x}^n=0$$Now, move all the coefficients into the series and do appropriate shifts so that all the series are in terms of $x^n$.
$$\begin{array}{cc}\sum _{n=3}^{\infty }n\left(n-1\right)\left(n-2\right)a_n{x}^{n-3}+\sum _{n=1}^{\infty }n{a}_n{x}^{n+1}+\sum _{n=0}^{\infty }{a}_n{x}^{n+1}& =0\\ \sum _{n=0}^{\infty }\left(n+3\right)\left(n+2\right)\left(n+1\right)a_{n+3}{x}^n+\sum _{n=2}^{\infty }\left(n-1\right)a_{n-1}{x}^n+\sum _{n=1}^{\infty }{a}_{n-1}{x}^n& =0\end{array}$$Next, let’s notice that we can start the second series at $n=1$ since that term will be zero. So let’s do that and then we can combine the second and third terms to get,
$$\begin{array}{cc}\sum _{n=0}^{\infty }\left(n+3\right)\left(n+2\right)\left(n+1\right)a_{n+3}{x}^n+\sum _{n=1}^{\infty }\left[\left(n-1\right)+1\right]a_{n-1}{x}^n& =0\\ \sum _{n=0}^{\infty }\left(n+3\right)\left(n+2\right)\left(n+1\right)a_{n+3}{x}^n+\sum _{n=1}^{\infty }n{a}_{n-1}{x}^n& =0\end{array}$$So, we got a nice simplification in the new series that will help with some further simplification. The new second series can now be started at $n=0$ and then combined with the first series to get,
$$\sum _{n=0}^{\infty }\left[\left(n+3\right)\left(n+2\right)\left(n+1\right)a_{n+3}+n{a}_{n-1}\right]x^n=0$$We can now set the coefficients equal to get a fairly simply recurrence relation.
$$\left(n+3\right)\left(n+2\right)\left(n+1\right)a_{n+3}+n{a}_{n-1}=0n=0,1,2,\dots$$Solving the recurrence relation gives,
$$a_{n+3}=\frac{-n{a}_{n-1}}{\left(n+1\right)\left(n+2\right)\left(n+3\right)}n=0,1,2,\dots$$Now we need to start plugging in values of $n$ and this will be one of the main areas where we can see a somewhat significant increase in the amount of work required when moving into a higher order differential equation.
$n=0:a_3=0$

$n=1:a_4=\frac{-a_0}{\left(2\right)\left(3\right)\left(4\right)}$

$n=2:a_5=\frac{-2a_1}{\left(3\right)\left(4\right)\left(5\right)}$

$n=3:a_6=\frac{-3a_2}{\left(4\right)\left(5\right)\left(6\right)}$

$n=4:a_7=\frac{-4a_3}{\left(5\right)\left(6\right)\left(7\right)}=0$

$n=5:a_8=\frac{-5a_4}{\left(6\right)\left(7\right)\left(8\right)}=\frac{5a_0}{\left(2\right)\left(3\right)\left(4\right)\left(6\right)\left(7\right)\left(8\right)}$

$n=6:a_9=\frac{-6a_5}{\left(7\right)\left(8\right)\left(9\right)}=\frac{\left(2\right)\left(6\right)a_1}{\left(3\right)\left(4\right)\left(5\right)\left(7\right)\left(8\right)\left(9\right)}$

$n=7:a_{10}=\frac{-7a_6}{\left(8\right)\left(9\right)\left(10\right)}=\frac{\left(3\right)\left(7\right)a_2}{\left(4\right)\left(5\right)\left(6\right)\left(8\right)\left(9\right)\left(10\right)}$

$n=8:a_{11}=\frac{-8a_7}{\left(9\right)\left(10\right)\left(11\right)}=0$

$n=9:a_{12}=\frac{-9a_8}{\left(10\right)\left(11\right)\left(12\right)}=\frac{-\left(5\right)\left(9\right)a_0}{\left(2\right)\left(3\right)\left(4\right)\left(6\right)\left(7\right)\left(8\right)\left(10\right)\left(11\right)\left(12\right)}$

$n=10:a_{13}=\frac{-10a_9}{\left(11\right)\left(12\right)\left(13\right)}=\frac{-\left(2\right)\left(6\right)\left(10\right)a_1}{\left(3\right)\left(4\right)\left(5\right)\left(7\right)\left(8\right)\left(9\right)\left(11\right)\left(12\right)\left(13\right)}$

$n=11:a_{14}=\frac{-11a_{10}}{\left(12\right)\left(13\right)\left(14\right)}=\frac{-\left(3\right)\left(7\right)\left(11\right)a_2}{\left(4\right)\left(5\right)\left(6\right)\left(8\right)\left(9\right)\left(10\right)\left(12\right)\left(13\right)\left(14\right)}$


Okay, we can now break the coefficients down into 4 sub cases given by $a_{4k}$, $a_{4k+1}$, $a_{4k+2}$ and $a_{4k+3}$ for $k=0,1,2,3,\dots$ We’ll give a semi-detailed derivation for $a_{4k}$ and then leave the rest to you with only couple of comments as they are nearly identical derivations.

First notice that all the $a_{4k}$ terms have $a_0$ in them and they will alternate in sign. Next notice that we can turn the denominator into a factorial, $\left(4k\right)!$ to be exact, if we multiply top and bottom by the numbers that are already in the numerator and so this will turn these numbers into squares. Next notice that the product in the top will start at 1 and increase by 4 until we reach $4k-3$. So, taking all of this into account we get,
$$a_{4k}=\frac{{\left(-1\right)}^k{\left(1\right)}^2{\left(5\right)}^2\cdots {\left(4k-3\right)}^2{a}_0}{\left(4k\right)!}k=1,2,3,\dots$$and notice that this will only work starting with $k=1$ as we won’t get $a_0$ out of this formula as we should by plugging in $k=0$.
Now, for $a_{4k+1}$ the derivation is almost identical and so the formula is,
$$a_{4k+1}=\frac{{\left(-1\right)}^k{\left(2\right)}^2{\left(6\right)}^2\cdots {\left(4k-2\right)}^2{a}_1}{\left(4k+1\right)!}k=1,2,3,\dots$$and again notice that this won’t work for $k=0$
The formula for $a_{4k+2}$ is again nearly identical except for this one note that we also need to multiply top and bottom by a 2 in order to get the factorial to appear in the denominator and so the formula here is,
$$a_{4k+2}=\frac{2{\left(-1\right)}^k{\left(3\right)}^2{\left(7\right)}^2\cdots {\left(4k-1\right)}^2{a}_2}{\left(4k+2\right)!}k=1,2,3,\dots$$noticing yet one more time that this won’t work for $k=0$.
Finally, we have $a_{4k+3}=0$ for $k=0,1,2,3,\dots$
Now that we have all the coefficients let’s get the solution,
$$\begin{array}{cc}y\left(x\right)& =a_0+a_{1}x+a_2{x}^2+a_3{x}^3+\cdots +a_{4k}{x}^{4k}+a_{4k+1}{x}^{4k+1}+a_{4k+3}{x}^{4k+3}+a_{4k+3}{x}^{4k+3}+\cdots \\ & =a_0+a_{1}x+a_2{x}^2\cdots +\frac{{\left(-1\right)}^k{\left(1\right)}^2{\left(5\right)}^2\cdots {\left(4k-3\right)}^2{a}_0}{\left(4k\right)!}{x}^{4k}+\\ & \frac{{\left(-1\right)}^k{\left(2\right)}^2{\left(6\right)}^2\cdots {\left(4k-2\right)}^2{a}_1}{\left(4k+1\right)!}{x}^{4k+1}+\\ & \frac{2{\left(-1\right)}^k{\left(3\right)}^2{\left(7\right)}^2\cdots {\left(4k-1\right)}^2{a}_2}{\left(4k+2\right)!}{x}^{4k+2}+\cdots \end{array}$$Collecting up the terms that contain the same coefficient (except for the first one in each case since the formula won’t work for those) and writing everything as a set of series gives us our solution,
$$\begin{array}{cc}y\left(x\right)& =a_0\left\{1+\sum _{k=1}^{\infty }\frac{{\left(-1\right)}^k{\left(1\right)}^2{\left(5\right)}^2\cdots {\left(4k-3\right)}^2{x}^{4k}}{\left(4k\right)!}\right\}+\\ & a_1\left\{x+\sum _{k=1}^{\infty }\frac{{\left(-1\right)}^k{\left(2\right)}^2{\left(6\right)}^2\cdots {\left(4k-2\right)}^2{x}^{4k+1}}{\left(4k+1\right)!}\right\}+\\ & a_2\left\{x^2+\sum _{k=1}^{\infty }\frac{2{\left(-1\right)}^k{\left(3\right)}^2{\left(7\right)}^2\cdots {\left(4k-1\right)}^2{x}^{4k+2}}{\left(4k+2\right)!}\right\}\end{array}$$

So, there we have it. As we can see the work in getting formulas for the coefficients was a little messy because we had three formulas to get, but individually they were not as bad as even some of them could be when dealing with 2^nd order differential equations. Also note that while we got lucky with this problem and we were able to get general formulas for the terms the higher the order the less likely this will become.