Find Closed Form of Differential Equation: y''' - y = 0

[fittipaldi]

Hi, everyone, I need some help with the following:

Homework Statement

Given is, that the following power series:
\sum_{n=0}^{\infty} \frac{x^{3n}}{(3n)!}

is the solution to the following differential equation: y''' - y = 0. Find the closed form of the series.

Homework Equations

None

The Attempt at a Solution

Well I tried differentiating the sum, then applying it to the differential equation, but I get something very nasty, so I am sure, I am on a bad way.

Please, help!

 

[Char. Limit]

Hi, everyone, I need some help with the following:

Homework Statement

Given is, that the following power series:
\sum_{n=0}^{\infty} \frac{x^{3n}}{(3n)!}

is the solution to the following differential equation: y''' - y = 0. Find the closed form of the series.

Homework Equations

None

The Attempt at a Solution

Well I tried differentiating the sum, then applying it to the differential equation, but I get something very nasty, so I am sure, I am on a bad way.

Please, help!

Well, you're given that that particular series is a solution to y''' - y = 0. You can also solve this another way. Might I recommend characteristic polynomials?

(also, doesn't that power series look suspiciously close to the power series for e^x?)

 

[fittipaldi]

Well, you're given that that particular series is a solution to y''' - y = 0.

That is true, but I cannot seem to understand how to continue ... a bigger tip maybe?

(also, doesn't that power series look suspiciously close to the power series for e^x?)

Also noticed that, the only difference is the coefficient. C*e^x is also one of the solutions to the given equation, how does this help?

 

[Char. Limit]

That is true, but I cannot seem to understand how to continue ... a bigger tip maybe?
Also noticed that, the only difference is the coefficient. C*e^x is also one of the solutions to the given equation, how does this help?

Well, it's entirely possible that the closed form of your power series is C*e^x for some C. In fact, it looks like that's the case to me.

EDIT: After checking, I can conclude that C*e^x is PART of the solution. However, you'll need to find ALL solutions to the differential equation. C*e^x is just one.

 

[tiny-tim]

hi fittipaldi!

That is true, but I cannot seem to understand how to continue ... a bigger tip maybe?

as Char. Limit says, use the characteristic polynomial