\magnification 1400 \vsize 9.9 truein \voffset -0.1truein \hoffset -0.5truein \hsize 7.5 truein \nopagenumbers \parskip=10pt \parindent= 0pt \def\u{\vskip 0pt} \def\v{\vskip 10pt} \def\w{\vskip 9pt} \def\L{{\cal L}} \input epsf \input graphicx Math 34 Differential Equations Exam \#2 \vskip -10pt October 29, 2010 \hfill SHOW ALL WORK ~~~ %%[5]~~ 1.) Define: The LaPlace transform of $f = {\cal L}(f) = \underline{\hskip 1.6in}$ \u\u [4]~~ 1a.) ${\cal L}(0) = \underline{~~0~~}$ \u [10]~~ 1b.) ${\cal L}^{-1}({2 \over (s-4)^2 + 5}) = \underline{~~{2 \over \sqrt{5}} e^{4t}sin(t\sqrt{5})~~}$ ${\cal L}^{-1}({2 \over (s-4)^2 + 5}) =$ ${2 \over \sqrt{5}} {\cal L}^{-1}({\sqrt{5} \over (s-4)^2 + 5}) =$ ${2 \over \sqrt{5}} e^{4t}sin(t\sqrt{5})$ \vfill [4]~~ 2.) Circle T for True or F for False: \u\u \vskip -4pt Suppose $y = f(t)$ is a solution to $3y'' + 10y = cos(t)$, $y(0) = 0$, $y'(0) = 0$ and suppose $y = g(t)$ is a solution to $3y'' + 10y = cos(t)$, $y(0) = 100$, $y'(0) = -200$. For large values of $t$, $f(t) - g(t)$ is very small. \hfill {\bf Note: no damping} %%\rightline { ~~~~~~~~~F} %%[4]~~ 2b.) The initial conditions have a transient effect on the solution to $y'' + y = cos(t)$, $y(0) = 0$, $y'(0) = 0$. %%\rightline{T~~~~~~~~~~F} \vfill [4]~~ 3a.) Given $2y'' + 5y = cos(wt)$, determine the value $w$ for which undamped resonance occurs: $2r^2 + 5 = 0$. Thus $r = i\sqrt{5 \over 2}$. Hence homogeneous soln is $y(t) = c_1 cos(t\sqrt{5 \over 2}) + c_2 sin(t\sqrt{5 \over 2})$. Hence a potential solution for the non-homogeneous equation $2y'' + 5y = cos(t\sqrt{5 \over 2})$ would be of the form: $t[A cos(t\sqrt{5 \over 2}) + B sin(t\sqrt{5 \over 2})]$. \centerline{Answer $\underline{ w = \sqrt{5 \over 2}~~}$} \u\u [3]~~ 3b.) Briefly describe in words the long-term behaviour of a solution to \break $2y'' + 5y = cos(wt)$ for this value of $w$. The solution oscillates and the pseudo-amplitude gets increasingly larger, approaching infinity. %%Longer answer: Solution is of the form $y(t) = c_1 cos(t\sqrt{5 \over 2}) + c_2 %%sin(t\sqrt{5 \over 2}) + t[A cos(t\sqrt{5 \over 2}) + B sin(t\sqrt{5 \over 2})] $ %%where $A = 0$ and $B =$. Hence for large values of $t$, $t[A cos(t\sqrt{5 \over 2}) %%+ B sin(t\sqrt{5 \over 2})]$ dominates. Thus solution oscillates with a %%pseudo-amplitude which gets increasingly larger, approaching infinity. \v [15]~~ 4.) A mass of 4 kg stretches a spring 5m. The mass is acted on by an external force of $6e^t$ N (newtons) and moves in a medium that imparts a viscous force of 8 N when the speed of the mass is 15 m/sec. The mass is pulled downward 1m below its equilibrium position, and then set in motion in the upward direction with a velocity of 10 m/sec. Formulate the initial value problem describing the motion of the mass. $m = 4$. $F_{viscous}(t) = -\gamma v(t)$, where $v = $velocity. Hence $8 = 15\gamma$ implies $\gamma = {8 \over 15}$. Also, $mg = kL$. Hence $k = {mg \over L} = {4(9.8) \over 5}$. \vfill \centerline{Answer $\underline{~~ 4u''(t) + {8 \over 15}u'(t) + {4(9.8) \over 5} u(t) = 6e^t}$} \eject [20]~~ 5.) Use ch 3 methods to solve the given initial value problem. \centerline{ $y'' + 4y = sin(t), ~~~~ y(0) = 0,~~ y'(0) = 0$} Step 1.) Find the general solution to $y'' + 4y = 0$: Guess $y = e^{rt}$. ~Then $r^2e^{rt} + 4e^{rt} = 0$ implies $ $r^2 + 4 = 0$ which implies $r = \pm 2i$. \vskip 5pt \centerline{homogeneous solution: $y(t) = c_1cos(2t) + c_2 sin(2t)$} Step 2.) Find ONE solution to $y'' + 4y = sin(t)$: Educated guess: $y = Asin(t)$ ~~(since no y' term). $y = Asin(t)$ \hfill $y' = Acos(t)$ \hfill $y'' = -Asin(t)$ $-Asin(t) + 4Asin(t) = sin(t)$. $3Asin(t) = sin(t)$. Hence $3A = 1$ and $A = {1 \over 3}$. The general solution to NON-homogeneous equation is \vskip 10pt \centerline{$c_1cos(2t) + c_2 sin(2t) + {1 \over 3}sin(t)$} %%\vskip 10pt Step 3.) Initial value problem: {Once general solution to problem is known, can solve initial value problem (i.e., use initial conditions to find $c_1, c_2$).} $y(t) = c_1cos(2t) + c_2 sin(2t) + {1 \over 3}sin(t)$ $y'(t) = -2c_1sin(2t) + 2c_2 cos(2t) + {1 \over 3}cos(t)$ $y(0) = 0$: $0 = c_1 $ $y'(0) = 0$: $0 = 2c_2 + {1 \over 3} $. Hence $2c_2 = -{1 \over 3}$ and $c_2 = -{1 \over 6}$ \vfill \centerline{Answer $\underline{~~y(t) = -{1 \over 6} sin(2t) + {1 \over 3}sin(t)~~}$} \eject [25]~~ 6.) Use the LaPlace transform to solve the given initial value problem. \centerline{ $y'' + 4y = sin(t), ~~~~ y(0) = 0,~~ y'(0) = 0$} $\L(y'' + 4y) = \L(sin(t))$ $\L(y'') + 4\L(y) = {1 \over s^2 + 1}$ $s^2\L(y) - sy(0) - y'(0) + 4\L(y) = {1 \over s^2 + 1}$ $s^2\L(y) + 4\L(y) = {1 \over s^2 + 1}$ $\L(y)[s^2 + 4] = {1 \over s^2 + 1}$ $\L(y) = {1 \over (s^2 + 1)(s^2 + 4)}$. Hence $y = \L^{-1}({1 \over (s^2 + 1)(s^2 + 4)})$ Partial Fractions: ${1 \over (s^2 + 1)(s^2 + 4)} = {As + B \over s^2 + 1} + {Cs + D \over s^2 + 4}$ $1 = (As + B)( s^2 + 4) + (Cs + D)(s^2 + 1)$ $1 = As^3 + Bs^2 + 4As + 4B + Cs^3 + Ds^2 + Cs + D$ $1 = (A + C)s^3 + (B + D)s^2 + (4A + C)s + 4B + D$ $A + C = 0$ and $4A + C = 0$. \vskip -5pt Hence $C = -A$ and $4A - A = 0$. Hence $3A = 0$ and $A = 0$, $C = 0$. \vskip -8pt ~~~Alternatively note $A = 0$, $C = 0$ is ``obviously a solution" and you only need one (plus it is ``obvious" that there is only one solution). Note how ``obvious" this is depends on your linear algebra background. $B + D = 0$ and $4B + D = 1$ \vskip -5pt Hence $D = -B$ and $4B - B = 1$. Hence $3B = 1$ and $B = {1 \over 3}$, $D = -{1 \over 3}$ ${1 \over (s^2 + 1)(s^2 + 4)} = {1 \over 3(s^2 + 1)} + {-1 \over 3(s^2 + 4)}$ $y = \L^{-1}({1 \over (s^2 + 1)(s^2 + 4)})$ $= \L^{-1}( {1 \over 3(s^2 + 1)} + {-1 \over 3(s^2 + 4)})$ $= {1 \over 3}\L^{-1}( {1 \over s^2 + 1}) - {1 \over 3}\L^{-1}( {1 \over s^2 + 4})$ $= {1 \over 3}sin(t) - {1 \over 6}\L^{-1}( {2 \over s^2 + 4})$ $= {1 \over 3}sin(t) - {1 \over 6}sin(2t)$ \vfill \centerline{Answer $\underline{~~{1 \over 3}sin(t) - {1 \over 6}sin(2t)~~}$} \eject [15]~~ 7.) Prove that if $F(s) = {\cal L}(f(t))$ exists for $s > a \geq 0$, and if $c$ is a positive constant, then ${\cal L}(u_c(t)f(t-c)) = e^{-cs}{\cal L}(f(t))$ with domain $s > a$. \u\u Hint: $ \int_0^\infty h(t)dt = \int_0^c h(t)dt + \int_c^\infty h(t)dt$ and use $u$-substitution (let $u = t-c$). Proof: If the integral $\int_0^\infty e^{-st} u_c(t)f(t-c)dt$ exists, then ${\cal L}(u_c(t)f(t-c)) = \int_0^\infty e^{-st} u_c(t)f(t-c)dt$ $ = \int_0^c e^{-st} u_c(t)f(t-c)dt + \int_c^\infty e^{-st} u_c(t)f(t-c)dt$ $ = \int_0^c e^{-st} \cdot 0 \cdot f(t-c)dt + \int_c^\infty e^{-st} \cdot 1 \cdotf(t-c)dt$ $ = 0 + \int_c^\infty e^{-st} f(t-c)dt$ Let $u = t-c$, then $du = dt$ and $t = u + c$. When $t = c$, $u = c - c = 0$ $ \int_c^\infty e^{-st} f(t-c)dt$ $ = \int_0^\infty e^{-s(u + c)} f(u)du$ %%$ = \int_0^\infty e^{-su -sc} f(u)du$ $ = \int_0^\infty e^{-su}e^{-sc} f(u)du$ $ = e^{-sc}\int_0^\infty e^{-su} f(u)du$ ~~since $e^{-sc}$ is a constant with respect to $u$. $ = e^{-sc}{\cal L}(f(u))$ $ = e^{-sc}{\cal L}(f(t))$ Note $F(s) = {\cal L}(f(t))$ = $\int_0^\infty e^{-su} f(u)du$ exists for $s > a$. Hence ${\cal L}(u_c(t)f(t-c)) = \int_0^\infty e^{-st} u_c(t)f(t-c)dt$ exists for $s > a$ and \hfil \break ${\cal L}(u_c(t)f(t-c)) =e^{-sc}{\cal L}(f(t))$. \end 32 = 5 + 12 + 15 25 28 15 68 32 97 1.) Circle T for True or F for False: 1a.) If $y = \phi_1(t)$ and $y = \phi_2(t)$ are solutions to a second order differential equation then $c_1\phi_1 + c_2\phi_2$ is also a solution. \rightline{T ~~~~~~~~F} 1b.) If $y = \phi_1(t)$ and $y = \phi_2(t)$ are solutions to a second order linear differential equation then $c_1\phi_1 + c_2\phi_2$ is also a solution. \rightline{T ~~~~~~~~F} 1c.) $ln(t)y'' - {y' \over t} + y\sqrt{t} = e^tcos(t)$ is a second order linear differential equation. \rightline{T ~~~~~~~~F} 1d.) \vfill Choose 4 problems from problems 2 - 6. You may do all the problems for up to 4 pts extra credit. If you do not choose your best 4 problems, I will substitute your extra problem for your lowest scoring problem, but with 3 point penalty. Circle the numbers corresponding to your 4 chosen problems: 2 \hfil 3 \hfil 4 \hfil 5 \hfil 6 Extra credit problem (choose 1 from problems 2 - 6): $ \underline{\hskip 1in}$ \eject [12]~ 2a.) Match the following differential equation to its graph. Indicate all equilibrium solutions (if any) and state whether {\bf stable, unstable or semi-stable.} I.) $y' = 1 - y$ \u %%~~~Equilibrium: $ \underline{\hskip 1in}$ \v II.) $y' = -1 + y$ III.) $y' = y(y + 2)$ IV.) $y' = y^2(2 - y)$ V.) $y' = t + y$ \vskip -1.5in \rightline {A.) \includegraphics[width=18ex]{e3plot2d3} ~~~~ B.) \includegraphics[width=18ex]{e3plot2d4}~~} %%\rightline{A.) Eq'n: $ \underline{\hskip 1in}$ ~~~~~B.) Eq'n: $ \underline{\hskip 1in}$ } \v\v %%\rightline{A.) Equilibrium(s): \hskip 1in ~~~~~B.) Eq'l: \hskip 1in } \v ~~ C.) \includegraphics[width=18ex]{e3plot2d5} \hfil D.) \includegraphics[width=18ex]{e3plot2d2} \hfil E.) \includegraphics[width=18ex]{e3plot2d1} \v\v\v [8]~ 2b.) Match the following differential equation initial value problem to its graph: I.) $y'' + y' + 49y = 0$, $y(0) = 0$, $y'(0) = 5$ II.) $y'' + y' + 49y = 0$, $y(0) = 1$, $y'(0) = 5$ III.) $y'' +10 y' + 49y = 0$, $y(0) = 0$, $y'(0) = 0$ IV.) $y'' +10 y' + 49y = 0$, $y(0) = 0$, $y'(0) = 0$ A.) \includegraphics[width=15ex]{e3plot2d9} \hfil B.) \includegraphics[width=15ex]{e3plot2d7} \hfil C.) \includegraphics[width=15ex]{e3plot2d8} \hfil D.) \includegraphics[width=15ex]{e3plot2d10} \eject 3.) Solve the differential equation %% $2y' + {y \over t} = {1 \over t^2}$ \vfill \centerline{Answer: $ \underline{\hskip 4in}$ } \eject %%4.) Solve $y''e^{y'} - {1 \over y} = 0$ %%4.) Solve ${y'' \over y'} - {ye^{y^2}} = 0$ 4.) Solve ${y'' \over y'} - {1 \over y^2} = 0$, ~$y(2) = 1$, ~$y'(2) = -1$ Let $v = y' = {dy \over dt}$, ${dv \over dy}v ={dv \over dy} {dy \over dt} ={dv \over dt} = v' = y''$ $y'' = v' = {dv \over dt} ={dv \over dy} {dy \over dt} ={dv \over dy}v $ %% ${y'' \over y'} - {ye^{y^2}} = 0$ %% ${{dv \over dy}v \over v} - {ye^{y^2}} = 0$ %%${{dv \over dy}} - {ye^{y^2}} = 0$ %%${{dv \over dy}} = {ye^{y^2}}$ ${y'' \over y'} - {1 \over y^2} = 0$ ${{dv \over dy}v \over v} - {1 \over y^2} = 0$ ${{dv \over dy}} = {1 \over y^2}$ $\int dv = \int {y^{-2}dy}$ $y' = v = -y^{-1} + c_1$ ~~$y(2) = 1$, ~$y'(2) = -1$: when $t = 2$, $-1 = -1 + c_1$. Thus $c_1 = 0$ ${dy \over dt} = -y^{-1}$ $\int y{dy} = \int{-dt}$ ${1 \over 2}y^2 = -t + c_2$ $y^2 = -2t + c_2$ $y = \pm \sqrt{-2t + c_2}$ ~~$y(2) = 1$: $1 = \pm \sqrt{-4 + c_2}$. Thus $c_2 = 5$ ~~~~~~~~~~~ $y = \sqrt{-2t + 5}$ \eject 4.) Solve ${y'' \over y'} - {y} = 0$, ~$y(2) = 1$, ~$y'(2) = -1$ Let $v = y' = {dy \over dt}$, ${dv \over dy}v ={dv \over dy} {dy \over dt} ={dv \over dt} = v' = y''$ $y'' = v' = {dv \over dt} ={dv \over dy} {dy \over dt} ={dv \over dy}v $ ${y'' \over y'} - {y} = 0$ ${{dv \over dy}v \over v} - {y} = 0$ ${{dv \over dy}} = {y}$ $\int dv = \int {y dy}$ $y' = v = {1 \over 2} y^{2} + c_1$ ~~$y(2) = 1$, ~$y'(2) = -1$: when $t = 2$, $-1 = -1 + c_1$. Thus $c_1 = 0$ ${dy \over dt} = -y^{-1}$ ${dy \over dt} = {-1 + c_1y \over y}$ \vfill \centerline{Answer: $ \underline{\hskip 4in}$ } \eject 5.) Suppose a mass weighs 64 lbs stretches a spring 4 ft. If there is no damping and the spring is stretched an additional foot and set in motion with an upward velocity of $\sqrt{8}$ ft/sec, find the equation of motion of the mass. \vfill \centerline{Answer: $ \underline{\hskip 4in}$ } \eject 6a.) Define: A function $f$ is linear if \vskip 20pt 6b.) Show that $L:$ {set of all twice differentiable functions} $\rightarrow$ {set of all functions}, $L(f) = af'' + bf' + cf$ is a linear function. Hint: Calculate $L(rf + tg)$ where $r, t$ are real numbers and $f, g$ are twice differentiable functions. \vfill \vfill\vfill\vfill\vfill \vskip 4.3in If $y = \phi(t)$ is a solution to $af'' + bf' + cf = 0$, then $L(\phi) = \underline{\hskip 1in}$ . If $y = \psi(t)$ is a solution to $af'' + bf' + cf = 0$, then $L(\psi) = \underline{\hskip 1in}$. $L(c_1\phi + c_2\psi) = \underline{\hskip 1in}$. Is $c_1\phi + c_2\psi$ a solution to $af'' + bf' + cf = 0$? $ \underline{\hskip 1in}$ \end [18]~ 1.) Solve the differential equation $2y' + {y \over t} = t^2$ \vfill \centerline{Answer 1.) $\underline{\hskip 5in}$} \eject [24]~ 2.) Solve the differential equation $y'' - 2y' + y = t$, $y(0) = 3$, $y'(0) = 4$. \u \vfill \vfill \vfill \centerline{Answer 2.) $\underline{\hskip 5in}$} \eject [18]~ 3.) Solve the differential equation $(t^2 + t - 2) y' y'' = 1$ \vfill \vfill \vfill \u \centerline{Answer 3.) $\underline{\hskip 5in}$} \eject [14]~ 4.) Draw the direction field for $y' = y(y - 4)$. Find the equilibrium solution(s) and determine if asymtptotically stable, semistable, or unstable. \v \vfill \vfill \vfill \eject [14]~ 5.) A ball with mass 0.3kg is thrown upward with an initial velocity of 98 m/sec from the roof of a building 20m high. If there is no air resistance, find the maximum height above the ground that the ball reaches. \vfill \vfill \vfill \centerline{Answer 5.) $\underline{\hskip 5in}$} \eject [14]~ 6.) Suppose $y = f(t)$ is a solution to $y' + p(t)y = q(t)$. Show that $y = 3f(t)$ is a solution to $y' + p(t)y = 3q(t)$. \end [5]~ 2.) Use Euler's formula to write $e^{2 + 3i}$ is the form of $a + ib$. \vskip 1.7in \centerline{Answer 2.) $\underline{\hskip 3in}$}