\documentclass[12pt]{article} \setlength{\topmargin}{-1in} \setlength{\oddsidemargin}{-.3in} \setlength{\textheight}{9in} \setlength{\textwidth}{7.2in} \newcommand{\sol}{{\bf Solution: }} \newcommand{\dx}{\ensuremath{\; dx}} \newcommand{\dy}{\ensuremath{\; dy}} \newcommand{\ra}{\rightarrow} \begin{document} {\bf Key for Test 1 Math 2280 - 7} \vspace{.1in} \begin{enumerate} \item (12 points) For each of the initial value problems below, determine \begin{enumerate} \item[(i)] If there is a solution to the problem, and if there is the interval the solution will be valid. \item[(ii)] If this is the unique solution. \item[(iii)] Tell me your answers in the previous parts is correct. \item[(iv)] If you are unable to determine the above, then explain why not. \end{enumerate} \begin{enumerate} \item \[ \frac{dy}{dx}=\sqrt{2x-y} \quad \quad y(1)=2 \] \sol Here we try to apply theorem 1 on page 23 of the text (this is the only existence and uniqueness theorem that applies to this problem). Unfortunately, $f(x,y)=\sqrt{2x-y}$ is not continuous on a rectangle containing $(1,2)$ therefore the theorem doesn't say anything. So we are unable to determine existence and uniqueness for this function. \item \[ \frac{dy}{dx}=\frac{x^4y+1}{2x^2-77x+488} \quad \quad y(10)=6 \] \sol Again we try to apply theorem 1 on page 23. This time it works because $f(x)$ is continuous everywhere except $x=8$ and $x=\frac{61}{2}$. Therefore a solution exists on an interval contained in $(8,\frac{61}{2})$. Note that the theorem only gives existence on a sub-interval of the this interval. To determine uniqueness, we look at $f_y=\frac{x^4}{2x^2-77x+488}$ and this is continuous on the same interval as above. So, our solution is unique. \item \[ y' = (2x-y)^{2/3} \quad y(1)=-2 \] \sol I meant this to be $y(1)=2$, because then $f_y$ would not be continuous on a rectangle, but as it stands, $f(x,y)$ is continuous in the whole plane, so therefore a solution exists. Also, $f_y=-\frac{2}{3}(2x-y)^{1/3}$ and this is continuous on a rectangle containing $(1,-2)$. Therefore we have a unique solution on an interval. \item \[ y^{(4)}+\sqrt{x+1}y^{(3)}-\frac{x}{x^2-4}y''+6e^xy=0 \quad \quad y(0)=4 \] \sol This is a linear DE, therefore we apply theorem 2 on page 156. This says that linear differential equations have unique solutions on the entire interval that the coefficient functions are continuous. Unfortunately, our initial conditions are not sufficient to get only one solution, therefore we will not have a unique solution. In fact the theorem tells us that we will have a 3 parameter family of solutions (not 4 because we are given one initial condition) on the interval $(-1,2)$ (because this is the largest interval on which the coefficient functions are continuous containing $x=0$). \end{enumerate} \item (10 points) Find the general solution to the differential equation \[ \frac{6}{x^3} \frac{dy}{dx} = y \] \sol Separation of variables: \begin{eqnarray*} \int \frac{dy}{y} & = & \int \frac{x^3}{6} \dx \\ \ln|y| & = & \frac{x^4}{24}+C \\ |y| & = & Ae^{x^4/24} \quad A>0 \\ y & = & Be^{x^4/24} \quad B\neq 0 \end{eqnarray*} \item (10 points) Solve the initial value problem \[ xy' -6y =x^4 \quad y(2) = 24 \] \sol Linear first order, so we find an integrating factor, but we first need to divide by $x$: \begin{eqnarray*} y' -\frac{6}{x}y & = & x^3 \\ \rho & = & e^{\int -6/x \dx} = e^{-6\ln|x|} = x^{-6} \\ x^{-6}y' -6x^{-7}y & = & x^{-3} \\ \frac{d}{dx}\left(x^{-6}y\right) & = & x^{-3} \\ x^{-6}y & = & -\frac{x^{-2}}{2}+C \\ y & = & -\frac{x^4}{2}+Cx^6 \end{eqnarray*} Now we use our initial conditions to get our specific solution: \[ 24=-\frac{2^4}{2}+64C \] And this gives $C=1/2$. So our solution is: \[ y = -\frac{x^4}{2}+\frac{x^6}{2} \] \item (10 points) Find the general solution to the differential equation below: \[ y''-y'-2y=e^{3x} \] Hint: First solve the homogeneous equation, then to find a particular solution, try $Ae^{3x}$ and see if you can find $A$. \sol We first solve the homogeneous equation $y''-y'-2y=0$ by trying $y=e^{rx}$ and finding the roots of the characteristic equation: \begin{eqnarray*} r^2-r-2 & = & 0 \\ (r-2)(r+1) & = & 0 \end{eqnarray*} So we get $r_1=2$ and $r_2=-1$ are the solutions to the characteristic equation. That means that the general solution to the homogeneous equation is \[ y_c = C_1e^{2x}+C_2e^{-x} \] Now to solve the non-homogeneous equation. We have to recall that the general solution to the non-homogeneous equation is given by \[ y=y_c+y_p \] where $y_c$ is the general solution the the homogeneous equation and $y_p$ some solution to the non-homogeneous equation (see page 164). So we use the hint and try $y_p=Ae^{3x}$ and see if we can determine $A$ so that this is a solution. \[ y_p=Ae^{3x} \quad y_p'=3Ae^{3x} \quad y_p''=9Ae^{3x} \] Plugging these into the non-homogeneous equation gives: \begin{eqnarray*} 9Ae^{3x}-3Ae^{3x}-2Ae^{3x} & = & e^{3x} \\ 4Ae^{3x} & = & e^{3x} \\ A & = & \frac{1}{4} \end{eqnarray*} So, a particular solution is $y_p=(1/4)e^{3x}$ and the general solution is \[ y = \frac{1}{4}e^{3x} + C_1e^{2x} + C_2e^{-x} \] \item (10 points) Prove that the functions below are linearly independent \[ f(x)=x \quad g(x)=xe^x \quad h(x)=x^2e^x \] \sol (I think this was a homework problem) To prove this I will compute the Wronskian $W(f,g,h)$. I first need to compute derivatives. \[ f'(x)=1 \quad f''(x)=0 \] \[ g'(x)=(x+1)e^x \quad g''(x)=(x+2)e^x \] \[ h'(x)=(x^2+2x)e^x \quad g''(x)=(x^2+4x+2)e^x \] \[ W(f,g,h) = \left| \begin{array}{ccc} x & xe^x & x^2e^x \\ 1 & (x+1)e^x & (x^2+2x)e^x \\ 0 & (x+2)e^x & (x^2+4x+2)e^x \end{array} \right| = x^3e^{2x} \] This is a nonzero function so therefore $f,g,h$ are linearly independent. \item (15 points) A hovering helicopter's engines suddenly dies 2000 feet above the ground. The helicopter plummets to the ground. \\ Assume that the air resistance is proportional to the velocity of the helicopter with $\rho=.18$ (so that up to sign we have $F_R=m\rho v$). Find the following: \begin{enumerate} \item The terminal velocity of the helicopter, \item The amount of time that it takes the helicopter to hit the ground, \item The speed that the helicopter is going when it hits the ground. \end{enumerate} You MUST set up and solve the differential equations to get full credit (don't use the book's general solution). \sol We have the following: \begin{eqnarray*} \frac{dv}{dt} & = & -\rho v - g \quad \quad v(0)=0 \quad y(0)=2000 \end{eqnarray*} We want to solve this DE for $v$ and $y$ and then answer the questions. We first separate variables and integrate to solve for $v$: \begin{eqnarray*} \int \frac{dv}{.18v+32} & = & \int -dt \\ \frac{1}{.18}\ln|.18v+32| & = & -t + C_1 \\ \ln|.18v+32| & = & -.18t+C_2 \\ .18v+32 & = & C_3e^{-.18t} \\ v & = & \frac{C_3}{.18}e^{-.18t}-\frac{32}{.18} \end{eqnarray*} We now use $v(0)=0$ and solve for $C_3$: \begin{eqnarray*} 0 & = & \frac{C_3}{.18}-\frac{32}{.18} \\ C & = & 32 \end{eqnarray*} So, we have $v=\frac{dy}{dx}$: \[ v(t) = \frac{32}{.18}e^{-.18t}-\frac{32}{.18} \] Now we solve this differential equation for $y$: \begin{eqnarray*} \frac{dy}{dt} & = & \frac{32}{.18}e^{-.18t}-\frac{32}{.18} \quad \quad y(0)=2000 \\ y & = & -\frac{32}{(.18)^2}e^{-.18t}-\frac{32}{.18}t+C \end{eqnarray*} We use the initial conditions ($y(0)=2000$) to solve for $C$: \begin{eqnarray*} 2000 & = & -\frac{32}{(.18)^2}+C \\ C & = & 2000+\frac{32}{(.18)^2} \approx 2987.65 \end{eqnarray*} So, now we have the functions $v(t)$ and $y(t)$, so now we answer the questions: \begin{enumerate} \item The terminal velocity of the helicopter: \[ v_\tau = \lim_{t \ra \infty} v(t) = -\frac{32}{.18} \approx -177.8 ft/sec \approx -121 mph \] The negative of course means the helicopter is traveling down and not up. \item The amount of time it take the helicopter to hit the ground: This is more difficult. To find this we need to solve the equation: \[ 0 = -\frac{32}{(.18)^2}e^{-.18t}-\frac{32}{.18}t+ 2000+\frac{32}{(.18)^2} \] There are several ways to do this and perhaps the easiest is to graph it on your graphing calculator (so that the calculator does all the work). Other options are to use Newton's method to find a zero. I graph it and let Hewlett Packard do the work and I get the following zeros for the above equation \[ t = 16.52,-8.40 \] Clearly the negative one is not the root for us, so it takes 16.52 seconds for the helicopter to hit the ground. \item How fast is the helicopter going when it hits the ground: We now take the 16.52 and plug it for $t$ in the velocity equation and get: (I make sure I keep all the digits when I plug it in so that I minimize my roundoff) \[ v(16.52) = -168.69 ft/sec \approx -115 mph \] \end{enumerate} \item (12 points) Suppose that a certain species of fish in the Deer Creek Reservoir satisfy the differential equation \[ \frac{dP}{dt} = -\frac{1}{2}P^3+\frac{21}{2}P^2-72P+160 \] where the units are in hundreds of fish per year. \begin{enumerate} \item Find all equilibrium solutions and label them as stable, unstable or semi-stable, \item draw the phase diagram, \item draw a graph of possible solution curves for this differential equation. \item If there are initially 100 fish in the reservoir, what happens over time? \item If 20,000 fish were suddenly planted in the reservoir (a one time occurrence) what would happen over time? \end{enumerate} Hint: get your units right! Do you remember how to factor polynomials? \sol I'm not going to draw the graphs on the computer but we first need to factor the polynomial. I do this using a combination of synthetic division, graphing and long division. In any case, we have: \[ \frac{dP}{dt} = -\frac{1}{2}(P-5)(P-8)^2 \] So our equilibrium solutions are \[ P = 5 \quad \quad P = 8 \] After drawing the phase portrait we see that $P=5$ is a stable equilibrium solution and $P=8$ is a semi-stable equilibrium solution. If we started with 100 fish (so that $P(0)=1$) then looking at either the phase portrait or the solution curves we should be able to see that as $t\ra \infty$ we will have $P(t)=5$. In other word, our population should approach 500. If we had 20,000 fish, then we would have $P(t_0)=200$ and in this case, by looking at either the phase portrait or the solution curves, we should have a limiting population of 800 fish (our semi-stable equilibrium). \item (10 points) Consider the initial value problem \[ \frac{dy}{dx} = e^{2y} \cos (xy) \quad y(1)=2 \] Using Runge-Kutta and a step size of $0.5$, find an approximation for $y(0)$. \sol First of all, I picked a bad function to do this with. What I compute for the $k$-values are terrible. Second, both $f(x,y)$ and $f_y$ are continuous in a rectangle containing $(1,2)$. In fact, both $f$ and $f_y$ are continuous in the whole plane. After getting the screwy results below, I used Maple to graph the slope field and the slopes are quite steep. Theoretically we should be able to get some sort of solution though. Here we let \[ f(x,y) = e^{2y} \cos (xy) \] and we let $h=-.5$. We need $h$ to be negative because we actually want to find the $y$-value for and $x$-value to the left of our initial point and not to the right. So we have $x_0=1$, $y_0=2$. We now compute the $k's$ and stuff as in the book (section 2.6): (I used all the decimal places, but only wrote down a few): \[ \begin{array}{|c|c|c|c|c|c|c|} \hline n & x_n & y_n & k_1 & k_2 & k_3 & k_4 \\ \hline \hline 0 & 1 & 2 & & & & \\ \hline & & & -22.7208 & 4060890.757 & 0 & 1 29.499506 \\ \hline 1 & .5 & -676813.68 & & & & \\ \hline & & & 0 & 0 & 0 & 0 \\ \hline 2 & 1 & -676813.68 & & & & \\ \hline \end{array} \] The $k$'s are not actually zero, but they are a large negative power of $e$. Therefore the $k$ values are very close to zero. Clearly this isn't working well. I reduced the step size and used Maple and still got exponentials that were too big. We need to look at this closer, perhaps use a different method. \item (12 points) Suppose we have the differential equation \[ y''+p(x)y'+q(x)y=0 \] and we have two solutions $y_1$ and $y_2$. Make the following definitions: \[ y_3:=y_1+y_2 \quad y_4:=y_1-y_2 \] Prove that if $y_1$ and $y_2$ are linearly independent solutions then $y_3$ and $y_4$ are linearly independent solutions. \sol We first need to show that $y_3$ and $y_4$ are solutions, but this is just the principle of superposition. (The really is part of the problem.) Next, we need to show that they are linearly independent. I suppose that this can be done by using the Wronskian, but that seems a bit more difficult then a more direct proof. Suppose that we have a linear combination of $y_3$ and $y_4$ and this sum is zero: \[ Ay_3 + By_4 = 0 \] We can rewrite this using the definitions of $y_3$ and $y_4$: \begin{eqnarray*} A(y_1+y_2)+B(y_1-y_2) & = & 0 \\ (A+B)y_1 + (A-B)y_2 & = & 0 \end{eqnarray*} Now if one or both of $A$ and $B$ are nonzero, then this implies that one of $A+B$ or $A-B$ is nonzero. In other words, $y_1$ and $y_2$ are linearly independent if and only if $y_3$ and $y_4$ are linearly independent. So, that does it. \end{enumerate} \end{document}