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Series solutions of differential equations

  • Thread starter FaNgS
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  • #1
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i haven't been able to solve this question in my textbook:

--> Find a power series solution for y''(x) - y(x)=0

First i attempted to find singular and ordinary points for the differential equation and i found that x=0 is an ordinary point. then i set y=[tex]\sum[/tex]A(n)x^n for n=0 to infinity
subsituted into the differential equation after finding the second derivative.
then after some manipulations i found that
y=A(0)[tex]\sum[/tex]x^(2n)/(2n)! + A(1)[tex]\sum[/tex]x^(2n+1)/(2n+1)!

the solution says it is equal to e^x+e^(-x)

I tried using the series of e^x=[tex]\sum[/tex]x^n/n! but i couldnt get the solution

help people

THANKS
 

Answers and Replies

  • #2
exk
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Are there any initial conditions for this problem (otherwise there is no way you can get that solution from the series).

I get the following: go through the first steps (derivatives, etc...)
get the recurrence relation of A(n+2)=A(n)/[(n+2)(n+1)], solving the recurrence relation I get A(2k)=A(0)/k! and A(2k+1)=A(1)/k!

Then:
[itex]y(x)=\sum{A(2k)x^{2k}+A(2k+1)x^{2k+1}} = \sum{\frac{A(0)x^{2k}}{k!}+\frac{A(1)x^{2k+1}}{k!}}[/itex]

Which is certainly going to be 2 functions of the e^(x) type. However, without initial conditions you can't get A(0) and A(1), although it's possible to see (given that you know the final solution) that A(0) probably = 1 and A(1) probably = -1 (since that -1 can be changed into (-x)^(2k+1) preserving the -1 since the power is always odd).

Note that the summation is still from 0 to infinity.
 
Last edited:
  • #3
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exk there are no initial conditions the question asks for a general solution, i'm sorry for not specifying that before.
the question is basically trying to show that using characteristic equation and a power series both give the same answer
 
  • #4
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but how do you go from [itex]y(x)=\sum{A(2k)x^{2k}+A(2k+1)x^{2k+1}} = \sum{\frac{A(0)x^{2k}}{k!}+\frac{A(1)x^{2k+1}}{k!} }[/itex] to an e^x and e^(-x) disregarding the coefficients of the two functions?
 
  • #5
Dick
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exk there are no initial conditions the question asks for a general solution, i'm sorry for not specifying that before.
the question is basically trying to show that using characteristic equation and a power series both give the same answer
If the problem asks for a general solution and doesn't give initial conditions then e^(x)+e^(-x) isn't the general solution. A*e^x+B*e^(-x) is as you know from the characteristic equation. So the series solution should also have two undetermined constants in it.
 
  • #6
exk
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Erm, I am sorry I was in too much of a hurry before apparently.
You solved the recurrence correctly getting:
[itex]
y(x)=\sum{A(2k)x^{2k}+A(2k+1)x^{2k+1}} = \sum{\frac{A(0)x^{2k}}{(2k)!}+\frac{A(1)x^{2k+1}}{(2k+1)!} }
[/itex]

With a little manipulation I get:

[tex]y(x)=\sum_{0}^{\infty}{\frac{A(0)x^{2k}}{(2k)!}}+\sum_{1}^{\infty}{\frac{A(1)x^{2k}}{(2k)!}}[/tex]

Now we can expand both summations to get:
[tex]y(x)=A(0)[1+x^2/2+x^4/4....]+A(1)[x^2/2+x^4/4...][/tex]

This can be re-written as:
[tex]y(x)=A(0)+(A(0)+A(1))*(x^2/2+x^4/4...)[/tex]
or:
[tex]y(x)=A(0)+(A(0)+A(1))\sum_{1}^{\infty}{\frac{x^{2k}}{(2k)!}}[/tex]

Now suppose you know before hand that the solution should be e^x+e^-x (lets assume the undetermined coefficients are 1s for now).

ok now expand that solution as a series (the only difference between the two being that e^-x produces a negative term for odd powers of x). Add them together to get:
1+1+2*x^2/2+2*x^4/2+....etc

rewrite that as [itex]2+2\sum_{1}^{\infty}{\frac{x^{2k}}{(2k)!}}[/itex] which looks very similar to our y(x) above if A(0)=2 and A(1)=0.

However your general solution is A*e^x+B*e^-x as Dick pointed out. So going back to addition of the two expansions you get:
[tex]y(x)=(A+B)+(A+B)\sum_{1}^{\infty}{\frac{x^{2k}}{(2k)!}}+(A-B)\sum_{1}^{\infty}{\frac{x^{2k+1}}{(2k+1)!}}[/tex]
Now recall the original solution progress:
[tex]y(x)=\sum_{0}^{\infty}\frac{A(0)x^{2k}}{(2k)!}+\frac{A(1)x^{2k+1}}{(2k+1)!}}[/tex]
Re-write as:
[tex]y(x)=A(0)+A(1)+\sum_{1}^{\infty}\frac{A(0)x^{2k}}{(2k)!}+\frac{A(1)x^{2k+1}}{(2k+1)!}}[/tex]

Finally we can see that if A(0)=A+B and A(1)=A-B you get:
[tex]y(x)=2A+(A+B)\sum_{1}^{\infty}{\frac{x^{2k}}{(2k)!}}+(A-B)\sum_{1}^{\infty}{\frac{x^{2k+1}}{(2k+1)!}}[/tex]

This doesn't completely agree since 2A=A+B iff A=B, but lets say A could be = to B. Then you are back with A(0)=2*A and A(1)=0, which corresponds to what I showed above.

I hope I didn't make a mistake again, all these summations are confusing. However, I certainly have no idea what would prompt me to do the above steps without actually knowing the end solution. Personally I would have probably left the solution in terms of summations.
 
  • #7
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the sums all look correct and are quite similar to what i have on paper but i'm still stuck from going from the sums to e^x
i have no clue on how to manipulate the sums to get it to e^x by using the e^x series (e^x= [tex]\sum[/tex]x^n/n!)....or by any other method
 
  • #8
exk
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Take a second look at what I did above. The conversion to e^x can be seen fairly clearly in tex lines 3-5. The actual solution conversion is seen in this:
[tex]
y(x)=(A+B)+(A+B)\sum_{1}^{\infty}{\frac{x^{2k}}{(2 k)!}}+(A-B)\sum_{1}^{\infty}{\frac{x^{2k+1}}{(2k+1)!}}
[/tex]
 
  • #9
Hurkyl
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Even if you don't see the conversion to exponentials... you should recognize the power series
[tex]\sum_{x =0}^{+\infty} \frac{x^{2n}}{(2n)!}[/tex]
as the Taylor series for familiar function.
 
  • #10
HallsofIvy
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Even if you don't see the conversion to exponentials... you should recognize the power series
[tex]\sum_{x =0}^{+\infty} \frac{x^{2n}}{(2n)!}[/tex]
as the Taylor series for familiar function.
I'm not sure that many students are really familiar with that particular function.
 
  • #11
Hurkyl
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I'm not sure that many students are really familiar with that particular function.
They should at least recognize the other function it's similar to -- and then they could derive the needed change of variable.
 

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