# Series solutions of differential equations

• FaNgS

#### FaNgS

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=$$\sum$$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)$$\sum$$x^(2n)/(2n)! + A(1)$$\sum$$x^(2n+1)/(2n+1)!

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

I tried using the series of e^x=$$\sum$$x^n/n! but i couldn't get the solution

help people

THANKS

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:
$y(x)=\sum{A(2k)x^{2k}+A(2k+1)x^{2k+1}} = \sum{\frac{A(0)x^{2k}}{k!}+\frac{A(1)x^{2k+1}}{k!}}$

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.

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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

but how do you go from $y(x)=\sum{A(2k)x^{2k}+A(2k+1)x^{2k+1}} = \sum{\frac{A(0)x^{2k}}{k!}+\frac{A(1)x^{2k+1}}{k!} }$ to an e^x and e^(-x) disregarding the coefficients of the two functions?

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.

Erm, I am sorry I was in too much of a hurry before apparently.
You solved the recurrence correctly getting:
$y(x)=\sum{A(2k)x^{2k}+A(2k+1)x^{2k+1}} = \sum{\frac{A(0)x^{2k}}{(2k)!}+\frac{A(1)x^{2k+1}}{(2k+1)!} }$

With a little manipulation I get:

$$y(x)=\sum_{0}^{\infty}{\frac{A(0)x^{2k}}{(2k)!}}+\sum_{1}^{\infty}{\frac{A(1)x^{2k}}{(2k)!}}$$

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

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

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 $2+2\sum_{1}^{\infty}{\frac{x^{2k}}{(2k)!}}$ 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:
$$y(x)=(A+B)+(A+B)\sum_{1}^{\infty}{\frac{x^{2k}}{(2k)!}}+(A-B)\sum_{1}^{\infty}{\frac{x^{2k+1}}{(2k+1)!}}$$
Now recall the original solution progress:
$$y(x)=\sum_{0}^{\infty}\frac{A(0)x^{2k}}{(2k)!}+\frac{A(1)x^{2k+1}}{(2k+1)!}}$$
Re-write as:
$$y(x)=A(0)+A(1)+\sum_{1}^{\infty}\frac{A(0)x^{2k}}{(2k)!}+\frac{A(1)x^{2k+1}}{(2k+1)!}}$$

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

This doesn't completely agree since 2A=A+B iff A=B, but let's 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.

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= $$\sum$$x^n/n!)...or by any other method

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:
$$y(x)=(A+B)+(A+B)\sum_{1}^{\infty}{\frac{x^{2k}}{(2 k)!}}+(A-B)\sum_{1}^{\infty}{\frac{x^{2k+1}}{(2k+1)!}}$$

Even if you don't see the conversion to exponentials... you should recognize the power series
$$\sum_{x =0}^{+\infty} \frac{x^{2n}}{(2n)!}$$
as the Taylor series for familiar function.

Even if you don't see the conversion to exponentials... you should recognize the power series
$$\sum_{x =0}^{+\infty} \frac{x^{2n}}{(2n)!}$$
as the Taylor series for familiar function.

I'm not sure that many students are really familiar with that particular function.

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.