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.

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

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

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

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

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 ${\cal L}^{-1}\{1/((s-1)^2(s-3))\}$.

   

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

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

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

   (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
   $$dx/dt= 3y + 6, x(0) = 1,$$
   $$dy/dt = -3x, y(0) = 0.$$

10. (a)

    Find the Fourier sine series of the function

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

    (b)

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

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

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

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

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