Find the general solution of the differential equation y '' '+ 8y = 0
The characteristic equation of Y '' '+ 8y = 0 is as follows:
λ^3+8=(λ+2)(λ^2 -2λ+4)=0
There are roots: λ 1 = - 2, λ 2 = 1 + I √ 3, λ 3 = 1-I √ 3
So the equation is
y1=e^-2x
y2=e^x*cos√3x
y3=e^x*sin√3x
The general solution of the differential equation y '' '+ 8y = 0:
y=C1e^(-2x)+C2(e^x*cos√3x)+C3(e^x*sin√3x)
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