#### Example 4.5.1.

Suppose that \(k=2\text{,}\) and \(m=1\text{.}\) The units are again the mks units (meters-kilograms-seconds). There is a jetpack strapped to the mass, which fires with a force of 1 newton for 1 second and then is off for 1 second, and so on. We want to find the steady periodic solution.

The equation is, therefore,

\begin{equation*}
x'' + 2 x = F(t) ,
\end{equation*}

where \(F(t)\) is the step function

\begin{equation*}
F(t) =
\begin{cases}
0 & \text{if } \; {-1} < t < 0 , \\
1 & \text{if } \; \phantom{-}0 < t < 1 ,
\end{cases}
\end{equation*}

extended periodically. We write

\begin{equation*}
F(t) = \frac{c_0}{2} + \sum_{n=1}^\infty
c_n \cos (n \pi t) +
d_n \sin (n \pi t) .
\end{equation*}

We compute

\begin{equation*}
\begin{aligned}
c_n & = \int_{-1}^1 F(t) \cos (n \pi t) \, dt =
\int_{0}^1 \cos (n \pi t) \, dt = 0 \qquad \text{for } \; n \geq 1,
\\
c_0 & = \int_{-1}^1 F(t) \, dt =
\int_{0}^1 \, dt = 1 ,
\\
d_n & = \int_{-1}^1 F(t) \sin (n \pi t) \, dt
\\
& = \int_{0}^1 \sin (n \pi t) \, dt
\\
& = \left[ \frac{-\cos (n \pi t)}{n \pi} \right]_{t=0}^1
\\
& = \frac{1-{(-1)}^n}{\pi n} =
\begin{cases}
\frac{2}{\pi n} & \text{if } n \text{ odd} , \\
0 & \text{if } n \text{ even} .
\end{cases}
\end{aligned}
\end{equation*}

So

\begin{equation*}
F(t) = \frac{1}{2} + \sum_{\substack{n=1 \\ n \text{ odd}}}^\infty
\frac{2}{\pi n} \sin (n \pi t) .
\end{equation*}

We want to try

\begin{equation*}
x(t) = \frac{a_0}{2} + \sum_{n=1}^\infty
a_n \cos (n \pi t) +
b_n \sin (n \pi t) .
\end{equation*}

Once we plug \(x\) into the differential equation \(x''+2x = F(t)\text{,}\) it is clear that \(a_n = 0\) for \(n \geq 1\) as there are no corresponding terms in the series for \(F(t)\text{.}\) Similarly \(b_n = 0\) for \(n\) even. Hence we try

\begin{equation*}
x(t) = \frac{a_0}{2} +
\sum_{\substack{n=1 \\ n \text{ odd}}}^\infty
b_n \sin (n \pi t) .
\end{equation*}

We plug into the differential equation and obtain

\begin{equation*}
\begin{split}
x'' + 2 x & =
\sum_{\substack{n=1 \\ n \text{ odd}}}^\infty
\Bigl[ - b_n n^2 \pi^2 \sin (n \pi t) \Bigr] +
a_0 +
2
\sum_{\substack{n=1 \\ n \text{ odd}}}^\infty
\Bigl[ b_n \sin (n \pi t) \Bigr]
\\
& =
a_0 +
\sum_{\substack{n=1 \\ n \text{ odd}}}^\infty
b_n (2 - n^2 \pi^2 ) \sin (n \pi t)
\\
& =
F(t) = \frac{1}{2} + \sum_{\substack{n=1 \\ n \text{ odd}}}^\infty
\frac{2}{\pi n} \sin (n \pi t) .
\end{split}
\end{equation*}

So \(a_0 = \frac{1}{2}\text{,}\) \(b_n = 0\) for even \(n\text{,}\) and for odd \(n\) we get

\begin{equation*}
b_n =
\frac{2}{\pi n (2 - n^2 \pi^2 )} .
\end{equation*}

The steady periodic solution has the Fourier series

\begin{equation*}
x_{sp}(t) = \frac{1}{4} + \sum_{\substack{n=1 \\ n \text{ odd}}}^\infty
\frac{2}{\pi n (2 - n^2 \pi^2 )}
\sin (n \pi t) .
\end{equation*}

We know this is the steady periodic solution as it contains no terms of the complementary solution and it is periodic with the same period as \(F(t)\) itself. See Figure 4.14 for the plot of this solution.