ODE2.3

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Solve y''+4y'+3y=0,\,\,\,y(0)=1,\,y'(0)=0\,

Find the characteristic equation.

m^{2}+4m+3=0\,

(m+1)(m+3)=0\,

The roots are m=-1,-3\,

The general solution is y(t)=c_{1}e^{{-t}}+c_{2}e^{{-3t}}\,

Plug in the initial conditions. Notice that two inital conditions (t=0\,) are needed for a unique solution to a second order ODE, and the general solution has two arbitrary constants.

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

y'(0)=-c_{1}-3c_{2}=0,\,\,\,c_{2}=-1/2,\,\,\,c_{1}=3/2\,

So the unique solution is

y(t)={\frac  {3}{2}}e^{{-t}}-{\frac  {1}{2}}e^{{-3t}}\,

Ordinary Differential Equations

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