General solution to third order differential equation

In summary, the conversation is about solving a third-order linear differential equation using the characteristic equation method. The individual used the incorrect characteristic equation and obtained a different answer from the book. After reevaluating the equation, they realized their mistake and obtained the correct answer.
  • #1
warfreak131
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?
 
Physics news on Phys.org
  • #2
warfreak131 said:
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?
 

FAQ: General solution to third order differential equation

What is a third order differential equation?

A third order differential equation is a mathematical equation that involves a function and its third derivative. It is written in the form y'''(x) = f(x,y,y',y''), where y denotes the dependent variable and x denotes the independent variable.

What is the general solution to a third order differential equation?

The general solution to a third order differential equation is a family of functions that satisfies the equation for all possible values of the independent variable. It contains three arbitrary constants, which can be determined by applying initial or boundary conditions.

How do you solve a third order differential equation?

To solve a third order differential equation, you can use a variety of methods such as separation of variables, variation of parameters, or the method of undetermined coefficients. The specific method used will depend on the form of the equation and any given initial or boundary conditions.

Can a third order differential equation have multiple solutions?

Yes, a third order differential equation can have multiple solutions. This is because the general solution contains three arbitrary constants, which can result in an infinite number of possible solutions depending on the given initial or boundary conditions.

What are the applications of third order differential equations?

Third order differential equations have various applications in physics, engineering, and other fields of science. They are commonly used to model physical systems that involve acceleration, such as motion of objects under the influence of external forces or electrical circuits with inductors.

Similar threads

Replies
3
Views
2K
Replies
52
Views
3K
Replies
7
Views
3K
Replies
12
Views
2K
Replies
8
Views
2K
Replies
2
Views
2K
Replies
4
Views
4K
Replies
4
Views
1K
Back
Top