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Math 246 - Exam 2 Solutions Monday, June 23rd, 2014 (1) In the following problem D = d . dt (a) Give a general solution to the following differential equation D3 (D2 + 9)2 (D2 − 2D − 7)2 (D2 + 2D + 10)y = 0. (b) What is the order of the differential equation in (a)? Solution: (a) The characteristic polynomial is p(z) = z 3 (z 2 + 9)2 (z 2 − 2z − 7)2 (z 2 + 2z + 10) √ √ √ √ which has roots 0, 0, 0, 3i, 3i, −3i, −3i, 1 + 2 2, 1 + 2 2, 1 − 2 2, 1 − 2 2, 1 + 3i, 1 − 3i. Therefore a general solution is given by Y (t) = c1 + c2 t + c3 t2 + c4 cos 3t + c5 sin 3t + c6 t cos 3t + c7 t sin 3t √ √ √ + c8 e(1+2 2)t + c9 e(1−2 2)t + c10 te(1+2 + c12 et cos 3t + c13 et sin 3t. 2)t + c11 te(1−2 √ 2)t (b) The order is 13. (2) Find the natural fundamental set of solutions (associated with t = 0) for each of the following equations: (a) y 00 − 4y 0 + 13y = 0 (b) y 00 + 8y 0 + 16y = 0 Solution: (a) The characteristic polynomial is p(z) = z 2 − 4z + 13, which has roots 2 ± 3i. Therefore a general solution is Y (t) = c1 e2t cos 3t + c2 e2t sin 3t. We apply the general initial conditions Y (0) = y0 , Y 0 (0) = y1 . Which gives ( y 0 = c1 y1 = 2c1 + 3c2 . 1 This has a solution c1 = y0 and c2 = (y1 − 2y0 )/3. Substituting this into the general solution and grouping by y0 and y1 gives 1 2 2t 2t Y (t) = y0 e cos 3t − e sin 3t + y1 e2t sin 3t. 3 3 Therefore the natural fundamental solutions are 1 N0 (t) = e2t cos 3t − e2t sin 3t, 3 1 N1 (t) = e2t sin 3t. 3 (b) The characteristic polynomial is p(z) = z 2 +8z +16 = (z +4)2 , which has a double root at −4. The general solution is given by Y (t) = c1 e−4t + c2 te−4t . We apply the general initial conditions Y (0) = y0 , Y 0 (0) = y1 . Which gives ( y0 = c1 y1 = −4c1 + c2 . This has solutions c1 = y0 and c2 = y1 + 4y0 . Substituting this into the general solutions and grouping by y0 and y1 gives Y (t) = y0 (e−4t + 4te−4t ) + y1 te−4t . The natural fundamental solutions are therefore N0 (t) = e−4t + 4te−4t , N1 (t) = te−4t . (3) Find a particular solution to the following differential equation d2 y + 4y = sec (2t). dt2 Solution 1: Since the forcing is not in characteristic form we can solve this by Green function method. The characteristic polynomial is p(z) = z 2 + 4, with roots ±2i. Therefore the Green function can be written as a general homogeneous solution, g(t) = c1 cos 2t + c2 sin 2t. Apply the conditions g(0) = 0, g 0 (0) = 1, we find ( 0 = c1 1 = 2c2 2 Therefore c1 = 0 and c2 = 1 2 and the Green function is g(t) = 1 sin 2t. 2 A particular solution is then given by Z 1 t Yp (t) = sin 2(t − s) sec 2s ds 2 0 Z t Z t cos 2s sin 2s 1 1 = sin 2t ds − cos 2t ds 2 2 0 cos 2s 0 cos 2s 1 1 = t sin 2t + cos 2t (log (| cos 2t|)) . 2 4 (4) Suppose that a mass of weight 2 kg in a viscous fluid hangs at rest from a spring and streches the spring by 0.2 m due to the force of gravity (you may assume the spring has zero length when unstretched). The viscous fluid also pushes on the mass with a force of 48 N (N = kg·m/s2 ) when the mass is moving 2 m/s. Assume that the spring force is proportional to the displacement and the damping force is proportional to the velocity. There is no external forcing. At t = 0 the mass is then displaced downward from its rest position 1 m and kicked downwards with a velocity of 1.5 m/s. (a) What is the spring constant k and damping coefficient γ associated with this system? Recall that the acceleration due to gravity is g = 9.8 m/s2 . (b) Write down (but do not solve) the initial value problem (IVP) which describes the displacement of the mass from its rest position. (c) Is the motion under-damped, over-damped, or critically damped? State why. Solution: (a) We know that at rest the spring force and the force of gravity are equal and opposite mg = −kyr , where yr = −0.2 is the rest displacement. This gives k=− 2(9.8) mg = = 98 N/m. yr 0.2 The drag force is given by −48 = Fdrag = −γ dy = −2γ. Therefore dt γ= 48 = 24 kg/s. 2 (b) The initial value problem is 2 d2 h dh + 24 + 98h = 0, 2 dt dt with initial conditions h(0) = −1, h0 (0) = −1.5. 3 (c) We see that r γ k µ= = 6, ω0 = =7 2m m Since ω0 > µ, the oscillations dominate the motion and the system is underdamped. (5) Suppose that x and x log x solve the following homogeneous differential equation (you don’t need to check this) x2 d2 y dy + y = 0, −x 2 dx dx for x > 0. (a) Show that these solutions are linearly independent for x > 0. (b) Solve the initial value problem x2 d2 y dy −x + y = x, 2 dx dx y(1) = 1, y 0 (1) = 2. Solution: (a) To see that the solutions are linearly independent, we check the Wronskian, x x log x W [x, x log x](x) = det = x 6= 0, when x > 0. 1 log x + 1 Therefore they are linearly independent for x > 0. (b) Write the equation in normal form d2 y 1 dy 1 1 − + 2y = 2 dx x dx x x To construct a general solution we use variation of parameters and assume that the general solution is given by Y (x) = u1 (x)x + u2 (x)x log x. We impose the conditions ( u01 x + u02 x log x = 0 u01 1 + u02 (log x + 1) = This has solutions u01 (x) = − Therefore Z u1 (x) = − log x , x 1 x u02 (x) = 1 . x log x 1 dx + c1 = − (log x)2 + c1 x 2 4 and Z u2 (x) = 1 dx + c2 = log x + c2 . x Therefore the general solution is 1 Y (t) = c1 x + c2 x log x + x(log x)2 , 2 x > 0. Imposing the initial conditions ( 1 = c1 2 = c1 + c2 , gives c1 = 1, c2 = 1. Therefore the solution to the IVP is 1 Y (t) = x + x log x + x(log x)2 . 2 (6) Consider the following MATLAB code ode = ‘D2y + 3*Dy + 2*y = 2*t + 5’; dsolve(ode, ‘y(0) = 1’, ‘Dy(0) = 2’,‘t’) (a) What is the initial value problem that is being solved? (b) Give the output of the dsolve command. You do not need to write it in MATLAB output format. Solution: (a) The initial value problem being solved is d2 y dy + 3 + 2y = 2t + 5, 2 dt dt y(0) = 1, y 0 (0) = 2. (b) To solve this problem we note that the characteristic polynomial is p(z) = z 2 +3z+ 2, which has roots −1, −2. The forcing is in characteristic form with µ + iν = 0, d = 1, m = 0. We will use key identity evaluations L (1) = p(0) = 2 L (t) = p(0)t + p0 (0) = 2t + 3. We can write the second equations as L (t) + 2 = 2t + 5, 5 However since the first equation tells us that L (1) = 2, we find L (t + 1) = 2t + 5. Therefore a general solutions is Y (t) = c1 e−t + c2 e−2t + t + 1. Applying the initial conditions ( 1 = c1 + c2 + 1 2 = −c1 − 2c2 + 1, which has solution c1 = −1, c2 = 1. Therefore Y (t) = −e−t + e−2t + t + 1 (7) Give a general (real valued) solution to the following differential equation y 00 + 4y 0 + 5y = 2e−2t cos t. Solution: The characteristic polynomial is p(z) = z 2 + 4z + 5, which has roots −2 ± i. The forcing is characteristic with µ + iν = −2 + i, d = 0, m = 1. We use key identity evaluations L(te(−2+i)t ) = p(−2 + i)te(−2+i)t + p0 (−2 + i)e(−2+i)t = 2ie(−2+i)t Muliplying the equation by −i (or equivalently dividing by i) gives L(−ite(−2+i)t ) = 2e(−2+i)t . Taking the Real part of both sides gives L Re −ite(−2+i)t = 2e−2t cos t. Therefore a particular solution is Yp (t) = Re −ite(−2+i)t = te−2t sin t. The general solution is Y (t) = c1 e−2t cos t + c2 e−2t sin t + te−2t sin t. 6 (8) Extra credit Suppose that one hundred ants break away from, their current colony to start a new one. Assume that their population P grows according to a logistic model dP = r(1 − P/K)P. dt The new colony has a carrying capacity of one million ants. Suppose that once the colony is started t = 0 it is measured that the one hundred ants can produce about 10 ants per day. (a) What are the parameters r and K? You can round to the nearest decimal place. (b) Solve the logistic equation with P (0) = 100 for any r and K. (c) What is the population size after 10 weeks for the r and K computed in (a)? Keep only one significant figure. Solution: (a) r = 10/(100 ∗ (1 − 10−4 ) ≈ .1 (days)−1 . K = 1 × 106 ants. (b) This equation is autonomous so we write Z P 1 dz 100 r(1 − z/K)z Z P 1 1 + dz = rz 100 r(K − z) 1 P (K − 100) = log r 100(K − P ) t= We can take the exponential of both sides, P (K − 100) = ert , 100(K − P ) We can then solve for P to obtain 100Kert K + 100(ert − 1) 100K . = −rt Ke + 100(1 − e−rt ) P = (c) 10 weeks = 70 days. So P (70) = e−7 100 ≈ 9.8104 . + 10−4 (1 − e−7 ) 7