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## Homework Statement

The solution to the Initial value problem, x''+2x'+5x=0, is the sum of the steady periodic solution x_sp and x_tr. Find both.

## Homework Equations

## The Attempt at a Solution

I already found x_sp ( the particular solution). It is (-44/533)cos(7t)+(14/533)sin(7t).

r^2+2r+5=0

-1=+- 2i

x_tr = e^(-t)(Acos(2t)+Bcos(2t))

x(t) = e^(-t)(Acos(2t)+Bsin(2t)) + (-44/533)cos(7t)+(14/533)sin(7t).

x'(t) = e^(-t)(-2Asin(2t)+2Bcos(2t)) - e^(-t)(Acos(2t)+Bsin(2t)) +

(308/533)sin(7t)+(98/533)cos(7t).

x(0)=1(A+0)-(44/553) ==> A = 44/533

x'(0)= 1(2B)-1(A)+(98/533) ==> B= -27/533

So, x_tr=(44/533)cos(2t)-(27/533)sin(2t)

this is not right however. Can someone see where I might have messed up?

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