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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