ODE2.4

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y''+4y'+5y=0,\,\,\,y(0)=1,\,y'(0)=0\,

The characteristic equation is

m^{2}+4m+5=0\,

Use the quadratic formula to get m=-2\pm i\,

The general solution is y(t)=e^{{-2t}}\left[c_{1}\cos(t)+c_{2}\sin(t)\right]\,

Plug in the IC's:

y(0)=c_{1}=1\,

y'(t)=e^{{-2t}}\left[-\sin(t)+c_{2}\cos(t)\right]-2e^{{-2t}}\left[c_{1}\cos(t)+c_{2}\sin(t)\right]\,

y'(0)=c_{2}-2c_{1}=0,\,\,\,c_{2}=2\,

The unique solution is

y(t)=e^{{-2t}}\left[\cos(t)+2\sin(t)\right]\,

Ordinary Differential Equations

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