# Damped Spring problem

• November 28th 2012, 06:39 AM
chizmin10
Damped Spring problem
The position y(t) of a damped spring satisfies the differential equation y''+2y'+5y=0. Determine y(t) when the initial position is y(0) and the initial velocity is y'(0)=1?
• November 28th 2012, 08:41 AM
hollywood
Re: Damped Spring problem
This type of differential equation is solved by substituting $y=e^{rt}$ to get the characteristic equation $r^2+2r+5=0$. This equation has two roots that are complex conjugates, $a+bi$ and $a-bi$, so the general solution is $e^{at}(C_1\cos{bt}+C_2\sin{bt})$. You can use the initial conditions to find $C_1$ and $C_2$.

- Hollywood