Modelling with second order ODEs

**acevipa** · September 7th 2010, 09:29 PM

Show that for the overdamped system:

$\dfrac{d^2x}{dt^2}+3\dfrac{dx}{dt}+2x=0$

if $x=0$ and $\dfrac{dx}{dt} \neq 0$ when $t=0$ then x can never equal 0 again.

This is what I've done so far.

The characteristic equation is:

$\lambda^2+3\lambda+2=0$

$(\lambda+2)(\lambda+1)=0$

$\Rightarrow \lambda=-1, -2$

$x=Ae^{-2t}+Be^{-t}$

$\dfrac{dx}{dt}=-2Ae^{-2t}-Be^{-t}$

When $t=0, x=0$

$A+B=0$

When $t=0, \dfrac{dx}{dt}\neq 0$

$-2A-B\neq 0$

Not too sure if I'm doing it right, or what to do next?

**Prove It** · September 7th 2010, 10:14 PM

What you've done so far is fine. You've also shown that since $A+B=0$ then $B = -A$ .

This gives $x = Ae^{-2t} - Ae^{-t}$

$x = Ae^{-t}(e^{-t} - 1)$ .

IF $x = 0$ then $0 = Ae^{-t}(e^{-t} - 1)$

$Ae^{-t} = 0$ or $e^{-t} - 1 = 0$

$e^{-t} = 0$ or $e^{-t} = 1$

The first case is impossible. So the ONLY time that this can happen is $t = 0$ .

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