1. ## initial value problem

Consider the initial value problem $x''+x+\varepsilon x=0$ with $x(0)=1, x'(0)=0$. I need to find the exact solution to this problem. Thanks.

2. Originally Posted by splash
Consider the initial value problem $x''+x+\varepsilon x=0$ with $x(0)=1, x'(0)=0$. I need to find the exact solution to this problem. Thanks.
I assume $\epsilon$ is constant? Then
$x'' + (1 + \epsilon)x = 0$

The characteristic equation is $m^2 + (1 + \epsilon) = 0$, so
$m = \pm i \sqrt{1 + \epsilon}$. So the general solution is:

$x(t) = Ae^{it\sqrt{1 + \epsilon}} + Be^{-it\sqrt{1 + \epsilon}}$

A physicist would probably prefer the equivalent expression:
$x(t) = Csin(t\sqrt{1 + \epsilon} ) + Dcos(t \sqrt{1 + \epsilon})$

Now use the form of your choice and apply the initial conditions.

-Dan

3. How did you get the characteristic equation? The way I remember is for matricies.

4. Originally Posted by splash
How did you get the characteristic equation? The way I remember is for matricies.
I was not aware that the "characteristic equation" could be used in that regard. I thought it was solely to describe the solutions of:

Characteristic equation (of A):

det(A - LI) = 0, where L is lambda;

or, similarly, det(LI - A) = 0

If not, I learned something new.

5. Originally Posted by AfterShock
I was not aware that the "characteristic equation" could be used in that regard. I thought it was solely to describe the solutions of:

Characteristic equation (of A):

det(A - LI) = 0, where L is lambda;

or, similarly, det(LI - A) = 0

If not, I learned something new.
Isn't A a matrix?

6. Originally Posted by splash
How did you get the characteristic equation? The way I remember is for matricies.
Originally Posted by AfterShock
I was not aware that the "characteristic equation" could be used in that regard. I thought it was solely to describe the solutions of:

Characteristic equation (of A):

det(A - LI) = 0, where L is lambda;

or, similarly, det(LI - A) = 0

If not, I learned something new.

Perhaps I have the wrong term. I was too lazy to look it up.

-Dan

Edit: Ha! My ODE book does call it a characteristic equation, but you might know it better by the term "auxiliary equation" perhaps.

7. Originally Posted by splash
Isn't A a matrix?
Yes, A is a matrix. I meant to quote topsquark and not you.

8. Originally Posted by topsquark
Perhaps I have the wrong term. I was too lazy to look it up.

-Dan
Oh ok. Then how did you get the $m^2 + (1 + \epsilon) = 0$ equation. Sorry to pester, I'm just confused about where that came from. Thanks.

9. Originally Posted by splash
Oh ok. Then how did you get the $m^2 + (1 + \epsilon) = 0$ equation. Sorry to pester, I'm just confused about where that came from. Thanks.
Given the ODE $a_ny^{(n)} + a_{n-1}y^{(n-1)} + ... + a_1y' + a_0 = 0$ (with constant coefficients) you may construct a general solution by solving the polynomial equation:
$a_nm^n + a_{n-1}m^{n-1} + ... + a_1m + a_0 = 0$
(the characteristic or auxiliary equation)

Solutions to the ODE will be:
$y = Ae^{m_1x} + Be^{m_2x} + ... + Ze^{m_nx}$
where the $\{ m_i \}$ are the solutions to the polynomial and A, B, ...Z are constants of integration.

-Dan

10. Thanks a lot tops. Very helpful.

11. Originally Posted by topsquark
I assume $\epsilon$ is constant? Then
$x'' + (1 + \epsilon)x = 0$

The characteristic equation is $m^2 + (1 + \epsilon) = 0$, so
$m = \pm i \sqrt{1 + \epsilon}$. So the general solution is:

$x(t) = Ae^{it\sqrt{1 + \epsilon}} + Be^{-it\sqrt{1 + \epsilon}}$

A physicist would probably prefer the equivalent expression:
$x(t) = Csin(t\sqrt{1 + \epsilon} ) + Dcos(t \sqrt{1 + \epsilon})$

Now use the form of your choice and apply the initial conditions.

-Dan
How do you know that,
$1+\epsilon<0$
And if it is, then it should be,
$\pm i \sqrt{-(1+\epsilon)}$

12. Originally Posted by ThePerfectHacker
How do you know that,
$1+\epsilon<0$
And if it is, then it should be,
$\pm i \sqrt{-(1+\epsilon)}$
Good point! (As usual. )

-Dan