Math 2280-2, Practice Midterm Exam 1February 12, 2008Total: 110/100 pointsProblem 1 (25 pts) A certain population P can be modeled by the differential equationdPdt= −P2+ 8P − 15(a) (4 pts) Find the equilibrium solutions of the differential equationWe have −P2+ 8P − 15 = −(P − 3)(P − 5). Thus the equilibrium solutions of thedifferential equation are P = 3 and P = 5.(b) (4 pts) Sketch the phase diagram for the differential equation. Identify the type ofequilibrium solution.The phase diagram for this DE is,−+−P = 3 P = 50 0PP0We see that P = 3 is unstable and P = 5 is stable.(c) (4 pts) Sketch the slope field for the differential equation together with the equilibriumsolutions and a few representative solutions (note: you do not need to find the formulasof the solutions to sketch them).t00.5 1.0 1.5 2.0P(t)K4K224681(d) (9 pts) Solve the differential equation with initial condition P(0) = 2. (Here you doneed to find the formula for P (t) for this initial condition).We rewrite the differential as dP/dt = −(P −3)(P −5) and use separation of variables:ZdP(P − 3)(P − 5)= −Zdt⇒12Z−1P − 3+1P − 5dP = −t + C1⇒ lnP − 5P − 3= −2t + C2⇒P − 5P − 3= C3e−2tsince P (0) = 2 ⇒P − 5P − 3> 0 for t near zero.Using the initial condition in the last equation we get C3= 3. Solving for P gives,P (t) =9e−2t− 53e−2t− 1.This is plotted in the slope field above.(e) (4 pts) Check that your solution is consistent with the slope field, that is study thebehavior of P (t). In particular, find the value T for which limt→TP (t) = −∞.The denominator of P (t) can be zero. This happens for T such that 3e−2T= 1, i.e. forT = ln(3)/2. This is the vertical asymptote plotted in the slope field and is consistentwith the behavior of P (t).Problem 2 (25 pts) Consider a 10 gallon tank, initially full of pure water. A brinewith concentration of 1 pound of salt per gallon enters the tank at the rate of 1 gallon perminute. The mixture at the tank is kept perfectly mixed, and the tank has a hole that letsthe solution escape at 2 gallons per minute. Thus the tank will be empty after exactly 10minutes and the volume of solution in the tank is V (t) = 10 − t.(a) (5 pts) Show that the quantity of salt x(t) (pounds) inside the tank before it is emptysatisfies the differential equation,dxdt= 1 −210 − tx.Using the notation from class we have that ri= 1, ci= 1 and ro= 2. The concentrationof salt in the tank is co= x/V = x/(10 − t). Thus at any time 0 ≤ t ≤ 10, the rate atwhich the quantity of salt in the tank changes isdxdt= rici− roco= 1 −210 − tx.2(b) (15 pts) Use the integrating factor method to solve the above differential equation.We rewrite the differential equation as,dxdt+210 − tx = 1.The integrating factor method tells us to multiply on both sides of the DE byeR2dt/(10−t)= e−2 ln(10−t)=1(10 − t)2, (Note the minus sign in the exponential)to obtainddtx(10 − t)2=dxdt1(10 − t)2+2(10 − t)3x =1(10 − t)2,which by integration givesx(t)(10 − t)2=Zdt(10 − t)2+ C1=110 − t+ C1.Thus the general form for x is,x(t) = (10 − t) + C1(10 − t)2.Since there is no salt in the tank at t = 0 we get the value of the constant C10 = x(0) = 10 + 100C1⇒ C1= −1/10.and thus the answerx(t) = (10 − t) −110(10 − t)2.(c) (5 pts) What is the maximum quantity of salt ever in the tank?The maximum amount salt occurs at T where x0(T ) = 0. Using the expression wehave found for x(t) we getx0(t) = −1 +15(10 − t).Setting x0(T ) = 0 we get T = 5. Evaluating, the maximum quantity of salt in thetank is x(T ) = 5/2 = 2.5 pounds.3Problem 3 (25 pts) Consider the differential operator L(y) = y00+ 2y0+ 2y = 0.(a) (5 pts) Write the characteristic polynomial p(r) of L and find its roots.The characteristic polynomial is p(r) = r2+ 2r + 2 = (r + 1 + i)(r + 1 − i) Its rootsare complex: r1= −1 + i and r2= −1 − i.(b) (5 pts) Find the general form of a solution yHto the homogeneous differential equationL(y) = 0.A general solution has the form yH(x) = Ae(−1+i)x+ Be(−1−i)x= e−x(a cos(x) +b sin(x)).(c) (10 pts) Find a particular solution ypto L(y) = cos(2x), using the method of unde-termined coefficients.Our search space for ypis V = {cos 2x, sin 2x}, that is we seek a particular solutionof the form yp= a cos 2x + b sin 2x. We now find the matrix (L)Bof L in the basisB = {cos 2x, sin 2x} of V . This is the matrix with columns being the images of thebasis vectors:L(cos 2x) = −4 cos 2x − 4 sin 2x + 2 cos 2x = −2 cos 2x − 4 sin 2x, andL(sin 2x) = −4 sin 2x + 4 cos 2x + 2 sin 2x = 4 cos 2x − 2 sin 2x.Then we get the linear system Ac = b, withA = (L)B=−2 4−4 −2and b = (cos 2x)B=10.Solving the system we getc =ab=120−2 −44 −210=110−12,thus a particular solution to L(y) = cos 2x isyp(x) = −110cos 2x +15sin 2x.Problem 4 (25 pts) Consider a mass-spring system with damping, where the mass m = 1,the string constant k = 104 and the damping constant c = 4. Recall that the displacementx(t) from the equilibrium position satisfies the differential equationmx00+ cx0+ kx = 0.(a) (5 pts) Find the characteristic polynomial p(r) of this DE and its roots for the givenvalues of m, k and c.The characteristic polynomial of the DE is p(r) = r2+ 4r + 104, the roots are complexr1= −2 + 10i and r2= −2 − 10i.4(b) (5 pts) What kind of motion is this?Since the discriminant we found above is ∆2= c2− 4k = 16 − 416 = −400 < 0, wehave an underdamped motion.(c) (5 pts) Find the general form of a solution to the DE.A general solution to the DE has the formx(t) = Ae(−2+10i)t+ Be(−2−10i)t, or equivalently x(t) = e−2t(a cos 10t + b sin 10t).(d) (5 pts) Sketch (qualitatively) a typical x(t) for this system.t1 2 3K0.6K0.4K0.200.20.40.60.81.01.2(e) (5 pts) What is the pseudoperiod of the motion?The function e2tx(t) is2π10−periodic. This is also the pseudoperiod of the motion.Problem 5 (10 pts) Let L be a third order linear differential operator. Assume y1, y2,and y3are solutions to the following initial value problems,L(y1) = 0, y1(0) = 1, y01(0) = 0, y001(0) = 0 (1)L(y2) = 0, y2(0) = 0, y02(0) = 1, y002(0) = 0 (2)L(y3) = 0, y3(0) = 0, y03(0) = 0, y003(0) = 1 (3)Let a, b and c be some real constants. Give a solution y to the following IVPL(y) = 0, y(0) = a, y0(0) = b, y00(0) = c.What result from class did you use?By the superposition principle we that y =