Math 215 Sec. 201

Second Midterm

March 11, 2002

1.     [ 7 pts ]      The change of the amount of a certain substance in a chemical reaction is modeled by the differential equation

$${dy\over dt} = k (100 - y)(50 - y)$$

where $k$ is a positive constant, $y$ is the amount of the substance, and $t$ is the time. What are the equilibrium values of $y$? Are they stable or unstable? What happens as $t$ increases?

Note that you were not asked to solve this differential equation. The equilibrium values are the $y$ values for which $dy/dt = 0$, namely 50 and 100. By looking at the direction field ($dy/dt > 0$ for $y < 50$ and $y > 100$, while $dy/dt < 0$ for $50 < y < 100$) you see that 50 is stable while 100 is unstable. As $t$ increases, $y \to 50$ if the initial $y$ value is less than 100, and $y \to \infty$ if it is more than 100.

2.     [ 11 pts ]     Solve the initial value problem $$y'' - 3 y' + 2 y = \hbox{e}^{t}$$, $y(0) = 1$, $y'(0) = 2$.

The characteristic polynomial $P(r) = r^2 - 3 r + 2 = (r-1)(r-2)$ has roots 1 and 2, so the homogeneous equation has solutions $$\hbox{e}^t$$ and $$\hbox{e}^{2t}$$. Since 1 is a root, the non-homogeneous equation doesn't have a solution that is a constant multiple of $$\hbox{e}^t$$. Using Exponential Shift, we take $y = \hbox{e}^t u$, and the equation for $u$ is $P(D+1) u = D (D-1) u = 1$. With $u' = v$ this says $v' - v = 1$. A particular solution is the constant $v = -1$, so $u = -t$ and $y = -t \hbox{e}^t$. Adding the solutions of the homogeneous equation, the general solution of the DE is

$$y = -t \hbox{e}^t + c_1 \hbox{e}^t + c_2 \hbox{e}^{2t}$$

From the initial conditions, $y(0) = c_1 + c_2 = 1$ and $y'(0) = -1 + c_1 + 2 c_2 = 2$. Solving for $c_1$ and $c_2$ we get $c_1 = -1$ and $c_2 = 2$, so the answer is

$$y = -t \hbox{e}^t - \hbox{e}^t + 2 \hbox{e}^{2t}$$

3.     [ 11 pts ]     Find a solution of the differential equation $$y'' - 3 y' + 2 y = (12 t+2) \hbox{e}^{-t}$$.

Using Exponential Shift, we write $y = \hbox{e}^{-t} u$. The equation for $u$ is

$$P(D-1) u = (D-2)(D-3) u = (D^2-5D+6) u = 12 t + 2$$

Since $6 \ne 0$, there should be a solution that is a polynomial of the same degree as the right side: $u = a_1 t + a_0$. Then $u' = a_1$ and $u'' = 0$, so the equation becomes

$$-5 a_1 + 6 (a_1 t + a_0) = 12 t + 2$$

Equating the coefficients of each power of $t$, we need $6 a_1 = 12$ and $-5 a_1 + 6 a_0 = 2$. Thus $a_1 = 2$ and $a_0 = 2$, so a solution is $$y = (2t+2) \hbox{e}^{-t}$$.

4.     [ 11 pts ]     A spring-mass system with damping is acted upon by an external force of the form $$F_0 \cos(\omega t)$$. It happens that one solution is $$y = 2 \cos(3 t - \pi/2) + 3 \hbox{e}^{-t} \cos(\mu t)$$ where $\mu$ is a constant. What is the value of $\mu$? Hint: what does the value of $\delta$ say about $\omega$ and $\omega_0$?

The term $2 \cos(3 t - \pi/2)$ must be the steady-state solution $(F_0/\Delta) \cos(\omega t - \delta)$, so $\omega = 3$ and $\delta = \pi/2$. Since $\cos \delta = m(\omega^2 - \omega_0^2)/\Delta$ we have $\omega_0 = \omega = 3$. Since $\omega_0 = \sqrt{k/m}$ this says $k = 9 m$. Now from the term $3 \hbox{e}^{-t} \cos(\mu t)$ we see that $r = -1 \pm i \mu$ are the roots of the characteristic polynomial $P(r) = m r^2 + \gamma r + 9 m$, which are $(-\gamma \pm i \sqrt{36 m^2 - \gamma^2})/(2 m)$. Thus $\gamma/(2m) = 1$, i.e. $\gamma = 2 m$, and then $\mu = \sqrt{36 m^2 - 4 m^2}/(2 m) = \sqrt{32}/2 = \sqrt{8}$.