TITLE: Differential Equations and resonance
QUESTION [0 upvotes]: An approximate solution to $$x''+bx'+{w_0}^2x={w_0}^2 \cos {w_0t};\mbox{ } x(0)=x'(0)=0 $$
is of the form $x(t)\approx \frac{w_0T}{2}B_0 \sin {w_0}t$ such that $T=\frac{2}{b}$ and $B_0(t)=\begin{cases}\\\frac{t}{T} \mbox{ } 0\leq t \leq T\\ 1 \mbox{ } T \leq t\end{cases} $
We set $N=\frac{w_0}{b}=\frac{w_0T}{2}$
I need to find the solution to the first equation of the form $x(t)=N[\sin {w_0}t-C_1(t) \sin {w_1}t-C_2(t) \cos {w_1}t]$ such that $w_1$ is the pseudo frequency, evaluating $C_1(t)$ and $C_2(t)$ as functions of $\frac{t}{T}$ and $\frac{w_0}{w_1}$
Thank you

REPLY [0 votes]: The simplest approach to finding the exact solution to this is to break the solution up into homogeneaous and inhomogeneous solutions $x_H$ and $x_I$, respectively.  $x_H$ satisfies:
$$x_H''+b x_H'+w_0^2 x_H = 0$$
which has solution
$$x_H(t) = e^{-b t/2} [A \cos{w_1 t} + B \sin{w_1 t}]$$
where $w_1 = \sqrt{w_0^2 - b^2/4}$.  The inhomogeneous solution, on the other hand, takes the form $x_I = C \cos{w_0 t} + D \sin{w_0 t}$. One plugs in this form directly into the above equation and get
$$x_I(t) = \frac{w_0}{b} \sin{w_0 t}$$
Using the initial conditions $x(0)=x'(0)=0$, we get the exact solution:
$$x(t) = \frac{w_0}{b} \sin{w_0 t} - \frac{w_0^2}{b w_1} e^{-b t/2} \sin{w_1 t}$$
You should be able to deduce your $N$, $C_1$ and $C_2$ from this solution.
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