9. $\displaystyle \frac{(2-s)-(2+s)e^{-s}}{s^2(1-e^{-s})} $
11. $\displaystyle
y(t)=\frac15e^{-t}+\frac{1}{10}\sin(2t)-\frac15\cos(2t).$
13. $y(t) = -(3/80)\cos(t) + (1/80)\sin(t) + (3/1040)\cos(3t) +
(7/1040)\sin(3t) -(1/65)\exp(-2t) + (1/20)\exp(-t),$
or, if you use a
calculator and get answers in decimal format (to the nearest 0.0001),
$y(t) = -0.0375\cos(t) + 0.0125\sin(t) + 0.0029\cos(3t) + 0.0067\sin(3t)
-0.0154\exp(-2t) + 0.05\exp(-t).$
15. The Laplace transform of the solution is
$$
\frac{1-\exp(-s)}{s(1+\exp(-s))(s^2+\pi^2)}
$$
At $s=\pm i\pi$ both the $s^2+\pi^2$ term and the $1+\exp(-s)$ term are
zero, so there is a double pole. Resonance is reasonable because the
equation is undamped with natural (circular) frequency $\pi,$ hence the
natural period is 2, and the square wave has period 2.
17. $$
\frac{e^x}{2}\int_0^x e^{-t^2-t}dt -
\frac{e^{-x}}{2}\int_0^x e^{-t^2+t}dt
$$
19. The convolution integral is
$$ \begin{align}
\int_0^x \sin(x-u)\sec(u) du
&= \int_0^x (\sin(x)\cos(u)-\cos(x)\sin(u))\sec(u) du \\
&= \sin(x)\int_0^x \cos(u)\sec(u) du - \cos(x)\int_0^x \sin(u)\sec(u) du
\\
&= \sin(x)\int_0^x 1 du - \cos(x)\int_0^x \tan(u) du \\
&=x\sin(x)+\cos(x)\ln|\cos(x)|
\end{align}
$$
21. $\displaystyle
y(x) = \frac{e^x}{3} \int_0^x \frac{te^t}{(t-1)^2} dt
-\frac{x}{3} \int_0^x \frac{e^{2t}}{(t-1)^2} dt.$
Laplace transforms aren't helpful here since the problem is not
constant-coefficient, but variation of parameters still works.
23.
$\displaystyle
x(t) = e^{-t}+\frac12 u(t-2\pi)e^{2\pi-t}\sin(t)
+\frac12 u(t-4\pi)e^{4\pi-t}\sin(t).
$
25. $$ \begin{align}
x(t) &= \exp(2t)(\cos(3t)+\frac13\sin(3t)) \\
y(t) &= \frac53\exp(2t)\sin(3t)
\end{align}
$$
27. $$
\begin{align}
x(t) &= \frac12\exp(3t)+\frac12\exp(3t) &= \frac12 \cosh(3t) \\
y(t) &= \frac32\exp(3t) + \frac32\exp(-3t) - \frac12\exp(2t)
-\frac12\exp(-2t) &= \frac32\cosh(3t) - \frac12 \sinh(2t)
\end{align}
$$
29.
The solution curves spiral in to the equilibrium at $(-1,-1).$
The equilibrium at $(1,1)$ is a saddle point.
31.
The equilibrium at $(1,2)$ is a saddle point.
The solution curves go directly out from the equilibrium at
$(2,1).$
The solution curves go directly in to the equilibrium at
$(-2,-1).$
The equilibrium at $(-1,-2)$ is a saddle point.
33.
The solution curves go directly in to the equilibrium at
$(1,1).$
