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Problems #5 pg.202 and #11 pg.232 are graded.

**#5,pg.202:** A mass weighing $2$ pounds stretches a spring $6$ inches. If the mass is pulled down an addition $3$ inches and then released, and if there is no damping, determine the position $u$ of the mass at any time $t$.

**Solution:** Comparing this to the general equation
$$mu''+\gamma u'+ku=F$$
we see that we have $F(t)=0$ (a forcing function is not mentioned) and $\gamma=0$ (no damping). We will proceed using feet as our units. The mass being stretched $6$ inches means that $u_0=6 \mathrm{in}=\dfrac{1}{2} \mathrm{ft}$. Using $g=32 \dfrac{\mathrm{ft}}{\mathrm{s}^2}$ we may solve for the mass by calculating
$m = \dfrac{2}{32} = \dfrac{1}{16}$. To find the spring constant we recall the equation $mg=u_0k$ and solve for $k$ to get $k = \dfrac{2}{\frac{1}{2}}=4$. The initial conditions are $u(0)=3 \mathrm{in} = \dfrac{1}{4} \mathrm{ft}$ and $u'(0)=0$. Thus we have the initial value problem
$$\dfrac{1}{16}u''+4u=0;u(0)=\dfrac{1}{4},u'(0)=0.$$

To solve it, we get the characteristic equation
$$\dfrac{1}{16}r^2 + 4 = 0$$
which has solution $r= \pm 8i$. Hence
$$u(t) = c_1 \cos(8t) + c_2\sin(8t).$$
We see that
$$u'(t)=-8c_1 \sin(8t) + 8c_2\cos(8t).$$
Now apply the initial conditions to get
$$\left\{ \begin{array}{ll}
\dfrac{1}{4}=u(0)=c_1 \\
0=u'(0)=8c_2.
\end{array} \right.$$
We see that $c_2=0$ and $c_1=\dfrac{1}{4}$. Thus the solution is
$$u(t) = \dfrac{1}{4} \cos(8t).$$
**#11,pg.232::** Find the general solution of
$$y'''-y''-y'+y=0.$$
**Solution:** The characteristic equation of this problem is
$$r^3-r^2-r+1=0.$$
The left hand side factors to $(r^2-1)(r-1)$ and hence we have roots $t=1,\pm 1$. Thus the general solution is
$$y(t)=c_1 e^t + c_2 e^{-t} + c_3te^t,$$
because the root $r=1$ is repeated.