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?
This type of differential equation is solved by substituting $\displaystyle y=e^{rt}$ to get the characteristic equation $\displaystyle r^2+2r+5=0$. This equation has two roots that are complex conjugates, $\displaystyle a+bi$ and $\displaystyle a-bi$, so the general solution is $\displaystyle e^{at}(C_1\cos{bt}+C_2\sin{bt})$. You can use the initial conditions to find $\displaystyle C_1$ and $\displaystyle C_2$.
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