# 4th oder ODE

1. Oct 6, 2009

### teeoffpoint

how can i solve d4y/dt4 - λ4 y= 0

2. Oct 6, 2009

### HallsofIvy

Staff Emeritus
Since that is a linear equation with constant coefficients, you can immediately write down its "characteristic equation", $x^4- \lambxa^4= 0$. That factors as $(x^2- \lambda^2)(x^2+ \lambda^2)= (x- \lambda)(x+ \lambda)(x- i\lambda)(x+ i\lambda)= 0$ and so has characteristic values $\lambda$, $-\lambda$, $i\lambda$, and $-i\lambda$. Do you know what to do with those?

3. Oct 7, 2009

### teeoffpoint

hmm... i dont really understand. can u show me a clearer working? wat you mean by x^4- \lambxa^4= 0? thanks

Last edited: Oct 7, 2009
4. Oct 7, 2009

### HallsofIvy

Staff Emeritus
What kind of explanation would you understand? Do you know anything at all about linear differential equations? I need to know what you do understand before I can explain much more.

5. Oct 8, 2009

### teeoffpoint

oh how did you solve x^2 + λ^2?

6. Oct 8, 2009

### HallsofIvy

Staff Emeritus
I didn't solve $x^2+ \lamba^2$. That is not an equation. I did solve $x^2+ \lambda^2= 0$ by the obvious method: I subtracted $\lambda^2$ from both sides to get $x^2= -\lambda^2$ and then took the square root of both sides. But now, in addition to my previous questions, which you still haven't answered, why are you trying to do differential equations if you don't know how to solve a simple quadratic equation?

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