Section 7.7 : 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 2nd order differential equations so we’re going to do a quick example here involving a 3rd order differential equation so we can make sure and say that we’ve done at least one example with an order larger than 2.
Recall that we can only find a series solution about x0=0 if this point is an ordinary point, or in other words, if the coefficient of the highest derivative term is not zero at x0=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(x)=∞∑n=0anxny′(x)=∞∑n=1nanxn−1y‴(x)=∞∑n=3n(n−1)(n−2)anxn−3Plugging into the differential equation gives,
∞∑n=3n(n−1)(n−2)anxn−3+x2∞∑n=1nanxn−1+x∞∑n=0anxn=0Now, move all the coefficients into the series and do appropriate shifts so that all the series are in terms of xn.
∞∑n=3n(n−1)(n−2)anxn−3+∞∑n=1nanxn+1+∞∑n=0anxn+1=0∞∑n=0(n+3)(n+2)(n+1)an+3xn+∞∑n=2(n−1)an−1xn+∞∑n=1an−1xn=0Next, 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,
∞∑n=0(n+3)(n+2)(n+1)an+3xn+∞∑n=1[(n−1)+1]an−1xn=0∞∑n=0(n+3)(n+2)(n+1)an+3xn+∞∑n=1nan−1xn=0So, 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,
∞∑n=0[(n+3)(n+2)(n+1)an+3+nan−1]xn=0We can now set the coefficients equal to get a fairly simply recurrence relation.
(n+3)(n+2)(n+1)an+3+nan−1=0n=0,1,2,…Solving the recurrence relation gives,
an+3=−nan−1(n+1)(n+2)(n+3)n=0,1,2,…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:a3=0n=1:a4=−a0(2)(3)(4)
n=2:a5=−2a1(3)(4)(5)
n=3:a6=−3a2(4)(5)(6)
n=4:a7=−4a3(5)(6)(7)=0
n=5:a8=−5a4(6)(7)(8)=5a0(2)(3)(4)(6)(7)(8)
n=6:a9=−6a5(7)(8)(9)=(2)(6)a1(3)(4)(5)(7)(8)(9)
n=7:a10=−7a6(8)(9)(10)=(3)(7)a2(4)(5)(6)(8)(9)(10)
n=8:a11=−8a7(9)(10)(11)=0
n=9:a12=−9a8(10)(11)(12)=−(5)(9)a0(2)(3)(4)(6)(7)(8)(10)(11)(12)
n=10:a13=−10a9(11)(12)(13)=−(2)(6)(10)a1(3)(4)(5)(7)(8)(9)(11)(12)(13)
n=11:a14=−11a10(12)(13)(14)=−(3)(7)(11)a2(4)(5)(6)(8)(9)(10)(12)(13)(14)
Okay, we can now break the coefficients down into 4 sub cases given by a4k, a4k+1, a4k+2 and a4k+3 for k=0,1,2,3,… We’ll give a semi-detailed derivation for a4k and then leave the rest to you with only couple of comments as they are nearly identical derivations.
First notice that all the a4k terms have a0 in them and they will alternate in sign. Next notice that we can turn the denominator into a factorial, (4k)! 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,
a4k=(−1)k(1)2(5)2⋯(4k−3)2a0(4k)!k=1,2,3,…and notice that this will only work starting with k=1 as we won’t get a0 out of this formula as we should by plugging in k=0.
Now, for a4k+1 the derivation is almost identical and so the formula is,
a4k+1=(−1)k(2)2(6)2⋯(4k−2)2a1(4k+1)!k=1,2,3,…and again notice that this won’t work for k=0
The formula for a4k+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,
a4k+2=2(−1)k(3)2(7)2⋯(4k−1)2a2(4k+2)!k=1,2,3,…noticing yet one more time that this won’t work for k=0.
Finally, we have a4k+3=0 for k=0,1,2,3,…
Now that we have all the coefficients let’s get the solution,
y(x)=a0+a1x+a2x2+a3x3+⋯+a4kx4k+a4k+1x4k+1+a4k+3x4k+3+a4k+3x4k+3+⋯=a0+a1x+a2x2⋯+(−1)k(1)2(5)2⋯(4k−3)2a0(4k)!x4k+(−1)k(2)2(6)2⋯(4k−2)2a1(4k+1)!x4k+1+2(−1)k(3)2(7)2⋯(4k−1)2a2(4k+2)!x4k+2+⋯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,
y(x)=a0{1+∞∑k=1(−1)k(1)2(5)2⋯(4k−3)2x4k(4k)!}+a1{x+∞∑k=1(−1)k(2)2(6)2⋯(4k−2)2x4k+1(4k+1)!}+a2{x2+∞∑k=12(−1)k(3)2(7)2⋯(4k−1)2x4k+2(4k+2)!}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 2nd 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.