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Problems #1 on pg.336 and #1, pg.343 are graded.
#1,pg.336: Find the solution of the initial value problem
$$y''+y=f(t); y(0)=0,y'(0)=1; f(t) = \left\{ \begin{array}{ll}
1 &; 0 \leq t < 3 \pi \\
0 &; 3\pi \leq t < \infty.
\end{array} \right.$$
Solution: First rewrite $f$ using step functions:
$$f(t)=u_0(t)-u_{3\pi}(t),$$
so that the differential equation becomes
$$y''+y=u_0(t)-u_{3\pi}(t).$$
Taking Laplace transform of the equation yields
$$(s^2\mathscr{L}\{y\}-sy(0)-y'(0))+\mathscr{L}\{y\}=\dfrac{e^{-0s}}{s} - \dfrac{e^{-3\pi s}}{s}$$
or simplified
$$(s^2+1)\mathscr{L}\{y\}=1 + \dfrac{1}{s} - \dfrac{e^{-3\pi s}}{s}.$$
Hence
$$\mathscr{L}\{y\} = \dfrac{1}{s^2+1} + \dfrac{1}{s(s^2+1)} - \dfrac{e^{-3\pi s}}{s(s^2+1)}.$$
Partial fractions shows that
$$\dfrac{1}{s(s^2+1)} = \dfrac{1}{s} - \dfrac{s}{s^2+1}.$$
Using the fact that $u_c(t)f(t-c)=\mathscr{L}^{-1}\{e^{-cs}\mathscr{L}\{f\}(s)\}$ we take inverse laplace transforms to see
$$\begin{array}{ll}
y(t) &= \mathscr{L}^{-1} \left\{ \dfrac{1}{s^2+1} + \dfrac{1}{s} - \dfrac{s}{s^2+1} - \dfrac{e^{-3 \pi s}}{s} + \dfrac{s e^{-3 \pi s}}{s^2+1} \right\} \\
&= \sin(t) + 1 - \cos(t) - u_{3\pi}(t) + u_{3 \pi}(t)\cos(t-3\pi).
\end{array}.$$
We will write the solution in piecewise function form for easy comparison to the solution provided by WolframAlpha. To do this we use trigonometry to notice that $\cos(t-3\pi)=-\cos(t)$ and thus we see
$$y(t) = \left\{ \begin{array}{ll}
\sin(t)+1-\cos(t) &; 0 \leq t < 3 \pi \\
\sin(t)-2\cos(t) &; t \geq 0
\end{array} \right.$$
Note: WolframAlpha uses the function Heaviside(t-a) to represent our step function $u_a(t)$. The solution there also has values for $t<0$ which we have ignored.
#1,pg.343: Find the solution of the initial value problem
$$y''+2y'+2y=\delta(t-\pi); y(0)=1, y'(0)=0.$$
Solution: Use the fact that $\mathscr{L}\{\delta(t-t_0)\}=e^{-t_0s}$ to take Laplace transforms and get
$$(s^2\mathscr{L}\{y\}-sy(0)-y'(0))+2(s\mathscr{L}\{y\}-y(0))+2\mathscr{L}\{y\}=e^{-\pi s},$$
so that
$$(s^2+2s+2)\mathscr{L}\{y\} - s - 2 = e^{- \pi s},$$
and hence
$$\mathscr{L}\{y\} = \dfrac{s+1}{s^2+2s+2}+\dfrac{1}{s^2+2s+2}+\dfrac{e^{-\pi s}}{s^2+2s+2}.$$
Complete the square on the denominator to see
$$\mathscr{L}\{y\} = \dfrac{s+1}{(s+1)^2+1} + \dfrac{1}{(s+1)^2+1} + \dfrac{e^{-\pi s}}{(s+1)^2+1}.$$
If we define the functions $F(s) = \dfrac{1}{s^2+1}$ and $G(s)=\dfrac{s}{s^2+1}$ then it is clear that $\mathscr{L}^{-1}\{F\}=\sin(t)$ and $\mathscr{L}^{-1}\{G\}=\cos(t)$ and may rewrite the formula for $\mathscr{L}\{y\}$ as
$$\mathscr{L}\{y\} = G(s+1) + F(s+1) + e^{-\pi s}F(s+1).$$
Using the formulas $\mathscr{L}^{-1}\{e^{-cs}F(s)\}=u_c(t)f(t-c)$ and $\mathscr{L}^{-1} \{F(s-a)\}=e^{at}f(t)$ we see
$$\begin{array}{ll}
y(t) &= \mathscr{L}^{-1}\{G(s-(-1)) + F(s-(-1)) + e^{-\pi s}F(s-(-1))\} \\
&=e^{-t}\mathscr{L}^{-1}\{G(s)\}(t) + e^{-t}\mathscr{L}^{-1}\{F(s)\}(t) + u_{\pi}(t) \mathscr{L}^{-1}\{F(s-(-1))\}(t-\pi) \\
&= e^{-t} \cos(t) + e^{-t} \sin(t) + u_{\pi}(t) e^{-(t-\pi)}\sin(t-\pi).
\end{array}$$
This answer could be further simplified (NOTE: this is not required) using the relation $\sin(t-\pi)=-\sin(t)$ to get
$$y(t) = e^{-t}\cos(t) + e^{-t}\sin(t) - u_{\pi}(t)e^{\pi-t}\sin(t).$$
