# Going back x 15 2 cost sint ce 2t this is a general

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Going back, x = (1/5) (2 cos(t) + sin(t)) + ce^{-2t} This is a general thing: a linear ODE with p constant and sinusoidal input signal has a sinusoidal solution. If p > 0, any solution differs from this one by transients which die out as t ---> infty. [3] Linear constant coefficient ODEs with exponential input signal Let's try x' + 2x = 4 e^{3t} We could use variation of parameter or integrating factor, but instead let's use the method of optimism, or the inspired guess. The inspiration here is based on the fact that differentiation reproduces exponentials: d -- e^{rt} = r e^{rt} dt Since the right hand side is an exponential, maybe the output signal x will be too: TRY x = A e^{3t} . I don't know what A is yet, but: x' = A 3 e^{3t} 2 x = 2 A e^{3t} ---------------- - 4 e^{3t} = A (3+2) e^{3t} which is OK as long as A = 4/5: x = (4/5) e^{3t} is one solution. The general solution is this plus a transient: x = (.8) e^{3t} + c e^{-2t} . [4] Replacing sinusoidal signals with exponential ones Let'e go back to the original ODE x' + 2x = cos(t) This equation is the real part of a complex valued ODE: z' + 2z = e^{it} This is a different ODE, and I use a different variable name, z(t) . We just saw how to get an exponential solution: k = 2 , r = i : z_p = 1/(i+2) e^{it} To get a solution to the original equation we should take the real part of this! Expand each factor in real and imaginary parts:

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______________________ z_p = ((2-i)/5) ( cos(t) + i sin(t) ) x_p = Re(z_p) = (1/5) (2 cos(t) + sin(t) ) as before!
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