Obtain the steady periodic solutin $x_{sp}(t)=Asin(\omega t+\phi)$ and the transient equation for the solution t $x''+2x'+26x=82cos(4t)$, where $x(0)=6$ & $x'(0)=0$.
So I'm not sure what's being asked and I'm guessing a little bit. I want to obtain $$x(t)=x_H(t)+x_p(t)$$ so to find homogeneous solution I let $x=e^{mt}$, and find
$$x_H=Acos5t+Bsin5t$$ so then
$$x_p=Ccos4t+Dsin4t$$
So I feel s if I have dne something wrong at this point. Continuing,
$$x_p'=-4Csin4t+4Dcos4t$$
$$x_p''=-16Ccos4t-16Dsin4t$$
so then
$$-16Ccos4t-16Dsin4t-8Csin4t+8Dcos4t+26Ccos4t+26Dsin4t=82cos4t$$
C=5, D=4
Eventally I solve for A and B, is this the right process? and what am I solving for, how do I get to the transient and steady state solutions? Thanks
Best Answer
The homogeneous form of the solution is actually $$X_H=c_1e^{-t}sin(5t)+c_2e^{-t}cos(5t)$$ which exponentially decays, so the homogeneous solution is a transient. The steady state solution is the particular solution, which does not decay.