SM 212 Final Examination
11 December 1996

  1. (a)
    Find the general solution to

    (i) $ x^2dy/dx = 4 - 3xy$,

    (ii) $ (1 - x\cos(xy))dy/dx = 1 + y\cos(xy)$,

    (b)
    Solve the initial value problem $ dy/dx = 2x\cos^2(y),  y(0)=\pi/4$.

  2. (a)
    A tank contains $ 100$ gallons of water with $ 100$ lbs. of salt dissolved in it. Saltwater with a concentration of $ 4$ lb/gal is pumped into the tank at a rate of $ 1$ gal/min. The well-stirred mixture is pumped out at the rate of $ 2$ gal/min. At what time does the tank contain the largest amount of salt? How much salt is in the tank at that time?

    (b)
    Use the improved Euler method with a step size of $ h = 0.1$ to approximate $ y(0.1)$ for the initial value problem.

    $\displaystyle dy/dx = x - y^2,    y(0)= -1.
$

  3. Find the general solution to:

    (a)
    $ y'' + 2y' - 8y = 0$,

    (b)
    $ y'' + 8y' + 25y = 0$,

    (c)
    $ y^{(4)} + 8y'' + 16y = 0$.

  4. Use undetermined coefficients (or annihilators) to solve the initial value problem

    $\displaystyle y''-3y' + 2y = 6e^x,   y(0) = 1,    y'(0) = -2
$

  5. (a)
    An $ 8$ lb weight stretches a spring $ 2$ ft upon coming to rest at equilibrium. From equilibrium the weight is raised $ 1$ ft and released from rest. The motion of the weight is resisted by a damping force that in numerically equal to $ 2$ times the weight's instantaneous velocity.

    (i) Find the position of the weight as a function of time.

    (ii) What type of damping does this mass-spring system possess?

    (b)
    The position of a weight in a mass-spring system subject to an external force is given by $ x(t) = e^{-t}\cos(3t) + e^{-t}\sin(3t) + 6\cos(2t) + 4\sin(2t)$.

    (i) What are the amplitude and period of the steady-state part of the solution?

    (ii) Write the transient part of the solution in the form $ Ae^{-t}\sin(3t + \phi)$.

    (iii) Find the time past which the magnitude of the transient part of the solution is less than one-percent of that of the steady-state part of the solution.

  6. (a)
    Find:

    (i) $ {\cal L}\{ 4t^3 - 4e^{-t}\sin(2t) \}$,

    (ii) $ {\cal L}\{t^2U(t-1) \}$,

    (iii) $ {\cal L}\{ t\cos t\}$.

    (b)
    Find $ {\cal L}\{f(t)\}$ for the periodic function given over one period by

    \begin{displaymath}
f(t) =
\left\{
\begin{array}{cc}
1,& 0< t \leq 2,\\
0,& 2< t <3.
\end{array}\right.
\end{displaymath}

  7. (a)
    Find

    (i) $ {\cal L}^{-1}\{1/(s^2 + 9) - e^{-3s}/(s-4)^2\}$,

    (ii) $ {\cal L}^{-1}\{(s+4)/(s^2 + 6s + 25)\}$.

    (b)
    Use the convolution theorem to find $ {\cal L}^{-1}\{1/((s-1)^2(s-3))\}$.

    \includegraphics[height=5.5cm,width=9cm]{/home/wdj/teaching/sm212/sm212_circuit_fall1996.eps}
  8. (a)
    For the circuit in the circuit above show that the charge $ q$ on the capacitor and the current $ i_3$ in the right branch satisfy the system of differential equations

    $\displaystyle q' + (1/RC)q + i_3 = 0,
$

    $\displaystyle i_3' - (1/LC)q = 0.
$

    (b)
    When the switch in the circuit is closed at time $ t=0$, the current $ i_3$ is 0 amps and the charge on the capacitor is $ 5$ coulombs. With $ R = 2$, $ L = 3$, $ C = 1/6$ use Laplace transforms to find the charge $ q(t)$ on the capacitor.

  9. Use eigenvalues/eigenvectors to solve the initial value problem

    $\displaystyle dx/dt= 3y + 6,    x(0) = 1,
$

    $\displaystyle dy/dt = -3x, y(0) = 0.
$

  10. (a)
    Find the Fourier sine series of the function

    \begin{displaymath}
f(x) =
\left\{
\begin{array}{cc}
0, &0\leq x \leq 2, \\
50x, & 2<x \leq 4.
\end{array}\right.
\end{displaymath}

    (b)
    The temperature $ u(x,t)$ of a thin bar of length $ 4$ satisfies the following conditions

    $\displaystyle du/dt = 3d^2u/dx^2,
$

    $\displaystyle u(0,t) = u(4,t) = 0,
$

    $\displaystyle u(x,0) = f(x)
$

    where $ f(x)$ is given in (10.a).

    (i) Use your answer in (10.a) to find $ u(x,t)$. Show all steps of the separation of variables process clearly. Write your answer in summation notation.

    (ii) Use the first two non-zero terms of your answer in ((10)(b)1) to approximate $ u(1, .5)$.

Answers for SM 212 Final Examination
11 December 1996

1. a. i. $ x^2(dy/dx) = 4 - 3xy$, $ dy/dx + (3/x)y = (4/x^2)$, $ y = (\int (e^{\int (3/x)dx}*4/x^2 + C)/e^{\int(3/x)dx}$, $ y=(2x^2 + C)/x^3$.

1. a. ii. $ (1 - x\cos(xy))y' = 1 + y\cos(xy)$, $ (1+y\cos(xy))dx - (1-x\cos(xy))dy= 0$, $ M_y=N_x$, $ \sin(xy) + x  \vert  -y + \sin(yx)/x$, $ \sin(xy) + x - y = C$

1.b. $ dy/dx = 2x\cos^2(y)$, $ y(0)=\pi/4$, $ \int(dy/\cos^2y) = \int(2x dx)$, $ \tan y= x^2 + C$, $ \tan y = x^2 + 1$, $ y = arctan (x^2 + 1)$.

2. $ A'(t) = (flow in)(Cin) - (A(t)flow out)/volume$, $ A'(t) + 2A(t)/(100-t) = 4$, $ A(t) = (4*\int(e^{\int(2/(100-t))}) + C)/(e^{\int(2/(100-t))})$, gives $ A(t) = C*(100-t)^2 + 4*(100-t)$. The IC $ A(0) = 100$, implies $ A(t) = 100 + 2t - (3/100)t^2.$

2.b. $ dy/dx = x - y^2$, $ y(0) = -1$, $ y_{new}=y+(h/2)(f(x,y)+f(x+h,y+hf(x,y)))$, $ h= .1$, $ y_{new}=-2211/2000$

3.a. $ y'' + 2y' - 8y = 0$, $ (r+4)(r-2) = 0$, $ y=C_1e^{-4t} + C_2e^{2t}$

3.b. $ y'' + 8y' + 25 = 0$, $ r^2 + 8r + 25 = 0$< $ r= -4 \pm 3i$, $ y= c_1e^{-4x}\cos 3x + c_2e^{-4x}\sin 3x$,

3.c. $ y^{(4)} + 8y'' + 16y = 0$, $ (x^2+4)(x^2+4) = 0$, $ r= \pm 2i$ (double roots), $ y= c_1\cos 2x + c_2\sin 2x + c_3x\cos 2x + c_4x \sin 2x$,

4. $ y''- 3y' + 2y = 6e^x$, $ y(0)=1$, $ y'(0)=-2$ $ (r-2)(r-1)=0$, $ y_h= c_1e^{2x} + c_2e^x$, $ y_p= Axe^x$, $ y_p'= Axe^x + Ae^x$, $ y_p'' = Axe^x + 2Ae^x$, $ Axe^x + 2Ae^x - 3(Axe^x + Ae^x) + 2 Axe^x = 6e^x$, $ A=-6$, $ y= y_h + y_p = c_1e^2x + c_2e^x -6xe^x$, $ y(0)=1$, $ c_1 + c_2 = 1$, $ y'(0) =-2 = 2c_1 + c_2$, $ c_1= 3$, $ c_2 = -2$, $ y= 3e^2x -2e^x -6xe^x$

5.a. 1. $ mx'' + bx'+ kx = 0$, $ k= 8/2 = 4$, $ m = 8/32$ slugs, $ b=2$, $ r'' + 8r'+16r=0$, $ r= -4,-4$, $ x=c_1e^{-4t} + c_2te^{-4t}$, $ x(0) =-1, x'(0)=0$, $ x=-e^{-4t} -4te^{-4t}$,

5.a.2. critically damped.

5.b.1. Amplitude: $ \sqrt{c_1^2 + c_2^2} = \sqrt{36 + 16} = \sqrt{52}$, Period = $ \pi$,

5.b.2. $ \phi = arctan (c_2/c_1)$, $ x(t) = (\sqrt{2})(e^{-t})\sin(3t + -\pi/4)$,

5.b.3. $ \sqrt{2}e^{-t}<(1/10)\sqrt{6^2+4^2}$, $ e^{-t}<\sqrt{26}/10$, $ t>\log (10/\sqrt{26}$.

6.a.1. $ 24/s^4 - 8/(((s+1)^2 + 4)$,

6.a.2 $ (e^{-s}){\cal L}\{f(t-1)\}$, $ e^{-s}(\frac{1}{s}+\frac{2}{s^2}+\frac{2}{s^3})$

6.a.3 $ {\cal L}\{t\cos t\} = (-1+s^2)/(s^2+1)^2$

6.b.1 $ u(t-a) -> e^{-as}/s -1u(t-2)$, $ F(s) = (-2^{-3s})/s$

7.a.1 $ u_c(t)f(t-c) = {\cal L}{^-1}\{e^{-cs}F(s)\}$, $ 1/3\sin 3t - {\cal L}{^-1}\{e^{-3s}(1/(s-2)^4)\}$, $ f(t) = t^3e^{2t}$, $ f(t-c) = (t-3)^3e^{2t -6}=1/3\sin 3t - 1/6u3(t)(t-3)^3e^{2t-6}$,

7.a.2 $ {\cal L}^{-1}\{(s+4)/(s^2 + 6s + 25)\}$, $ {\cal L}^{-1}\{(s+3)/((s+3)^2+16) + (1/((s+3)^2+16))\}$, $ e^{-3t}\cos (4t) + 1/4e^{-3t}\sin (4t)$

7.b. Use the convolution theorem to find $ {\cal L}^{-1}\{1/((s-1)^2(s-3)\}$, $ {\cal L}^{-1}\{(1/(s-1)^2)(1/(s-3))\}
=e^{-3t}/4+(-t/2-1/4)e^{t}$.

8.a. $ Li_3'- (1/C)q = 0$, $ i_1R + (1/C)q=0$

8.b. Substituting one DE into the other gives $ Lq'' + Rq' + (1/C)q = E(t)$, so $ q'' + (2/3)q' + 2q = 0$. Taking LTs: $ s^2Q(s) - sq(0) - q'(0) + 2/3Q(s) - q(0) + 2Q(s) = 0$, $ Q(s) (s^2 + (2/3)s + 2) = 5s + 5$, $ Q(s) = (5s + 5)/ (s^2 + (2/3)s + 2)$. Now take inverse LTs to get:
$ ={\frac {5}{17}} {e^{-1/3 t}} \left( 17 \cos \left( t/3 \sqrt {17}
\right) +2 \sqrt {17}\sin \left( t/3 \sqrt {17} \right) \right)$.

9. $ x(t) = \cos(3t)+2\sin(3t)$, $ y(t) = -\sin(3t)+2\cos(3t)-2$

10.a. $ b_n=(2/L)\int_0^L f(x)\sin(nx\pi/L) dx$, $ =\int_2^4 50x\sin(nx\pi/4) dx$ $ =200(-2\sin(n\pi/2)+n\pi\cos(n\pi/2)-2n\pi\cos(n\pi))/(n^2\pi^2)$, $ f(x) = \sum_n b_n \sin(n\pi x/4)$

b. $ X(x)T'(t) = X'(x)T(t)$, $ T'(t)/T(t) = X'(x)/ X(x) = -K$, $ u(x,t) = \sum_n b_n\sin(xn\pi/4)e^{-n^2\pi^2t/16}$.



David Joyner 2003-08-10, last modified 11-2-2004