Higher Order ODE - Multiple Complex Roots?

[kape]

Hello, I have two questions about this problem:

$$(D^4 + 5D^2 + 4)y = 0$$

y(0) = 10
y'(0) = 10
y''(0) = 6
y'''(0) = 8

$$\lambda^4 - 5\lambda^2 + 4 = 0$$

$$(\lambda^2 + 4) (\lambda^2 + 1)$$

Until here I am fairly sure that I didn't mess it up..
But I'm not sure if I have the roots correct. Are the roots:

Root1:
$$\lambda^2 = -4$$
$$\lambda = \pm\sqrt-4$$
$$\lambda = \pm\sqrt-1 \cdot \pm\sqrt4$$
$$\lambda = \pm2i$$

Root2:
$$\lambda^2 = -1$$
$$\lambda = \pm\sqrt-1$$
$$\lambda = \pm1i$$

So this is a complex double root (I think?) and the equation should be:

$$y = e^{\gamma x} [ (A_1 + A_2x) cos\omegax + (B_1 + B_2x) sin\omegax ]$$

Making it:

$$y = e^{(0)x} [ (A_1 + A_2x) cos2x + (B_1 + B_2x) sinx ]$$



* [Question 1]: Is this the right way to proceed?



Because, I finished the problem and it turned out to be wrong... (and it took a long time to differentiate it 3 times too.. and even longer to check it again.. twice.. *sigh*)


Also, I'm not sure exactly how I am supposed to find $$\inline \gamma$$ in $$\inline e^{\gammax}$$...

Is it right to think that if the root is $$\inline \lambda = 5 \pm 3i$$ then $$\inline \gamma$$ is 5 and $$\inline \omega$$ is 3? (making the equation:)

$$y = e^{5x} [ (A_1 + A_2x) cos3x + (B_1 + B_2x) sin\omegax ]$$

But this must be wrong because what if the other root has a value added to the multiple of the i value too?

* [Question 2] How do you find $$\inline \gamma$$ ?

 

[Hurkyl]

So this is a complex double root (I think?)

Nope -- you have four distinct roots, none of them are a double root.

 

[schattenjaeger]

Just checking, I'm pretty sure it's just a typo but your indicial equation(with a lambda)has a minus where there should be a plus, but you still factor it the way it should be, but just fyi

yah, you've got 4 roots, remember, complex roots always come in pairs like that('cuz of the +- nature of the quadratic equation), you're never going to get a root that's JUST +2i, it's always going to be + or - 2i, so you don't need to worry about that double root jazz

 

[Hurkyl]

Sometimes you do. For example, his differential operator could have been:

(D² + 1)²

which does have double roots.

 

[J77]

Jyah, you've got 4 roots, remember, complex roots always come in pairs like that('cuz of the +- nature of the quadratic equation), you're never going to get a root that's JUST +2i, it's always going to be + or - 2i, so you don't need to worry about that double root jazz

...for real coefficients.

How's about:

$$\lambda^2-4i\lambda-4=0$$?

 

[HallsofIvy]

We were talking about differential equations with real coefficients so a characteristic equation like that would never occur.

In order to have double complex roots, since in an equation with real coefficients roots are always in complex conjugate pairs, your equation would have to be of at least 4th order as Hurkyl said.

 

[kape]

Oh I see.. four distinct roots! Thank you. I managed to solve the problem without any difficulties!

I guess I was muddled with the definitions of roots...

I've been looking through my textbook but I'm still not exactly sure of the definitions..

-----------------
Am I right to think:

- Double Root: Where two "roots" are the same; $$\inine (a - 1)^2, a_1 = 1, a_2 = 1$$

- Complex Conjugate Roots: Where the roots are $$\inline \pm \omegai$$ (does conjugate mean +/- ?)

- Complex Conjugate Double Roots: Like... (D^2 + 1)^2

- Multiple Complex Conjugate Roots: actually, I don't quite understand what this is. What IS this? And how do you find gamma??

 

[J77]

We were talking about differential equations with real coefficients so a characteristic equation like that would never occur.

Yeah - but you can have differential equations with complex coefficients, eg. vibroelastic stiffness. It's not completely correct to suggest that the quadratic equation throws up conjugate roots always.

It's worth pointing this out to the OP.