General solution to third order differential equation

  • #1
188
0
yIII+yII-yI-y = 0

I used the characteristic equation and got:

r3+r2-r = 0

r (r2+r-1) = 0

Which means that r = 0 is one root,

And the other factors from the polynomial are (-1-Sqrt(5))/2 and (-1+Sqrt(5))/2

This means that the final answer would be:

y = C1 Exp(0x) + C2 Exp((-1-Sqrt(5))/2) + C3Exp((-1+Sqrt(5))/2)

Then I'd simplify from there, but I checked my answer in the back of the book, and it says that I'm wrong. It says the answer is

y = C1 Exp(x) + C2 Exp(-x) + C3x Exp(-x)

I also solved the equation using Mathematica and also got the answer that the book states.
Where did I go wrong?
 

Answers and Replies

  • #2
HallsofIvy
Science Advisor
Homework Helper
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yIII+yII-yI-y = 0

I used the characteristic equation and got:

r3+r2-r = 0
No, that's NOT the characteristic equation. That would be the characteristic equation for y'''+ y''- y'= 0. You seem to have forgotten the "- y" at the end.

The characteristic equation for y'''+ y''- y'- y= 0 is
[itex]r^3+ r^2- r- 1= 0[/itex]
It should be easy to see that r= 1 is a root.

r (r2+r-1) = 0

Which means that r = 0 is one root,

And the other factors from the polynomial are (-1-Sqrt(5))/2 and (-1+Sqrt(5))/2

This means that the final answer would be:

y = C1 Exp(0x) + C2 Exp((-1-Sqrt(5))/2) + C3Exp((-1+Sqrt(5))/2)

Then I'd simplify from there, but I checked my answer in the back of the book, and it says that I'm wrong. It says the answer is

y = C1 Exp(x) + C2 Exp(-x) + C3x Exp(-x)

I also solved the equation using Mathematica and also got the answer that the book states.
Where did I go wrong?
 

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