Answers and Hints to Review Questions for Test 3 (a) Find the general solution to the linear system of differential equations [ dy 3 Y 3 [ (b) Find the specific solution that satisfies Y (0) = (c) What is the natural period of the solutions? Answer (a) The eigenvalues are found by solving λ 3 3 λ = 0 λ + 4λ + 3 = 0 and so λ = ± 6 5 = ± 3i Using λ = + 3i we solve [ 3i 3 3 3i [ v v = One such eigenvector is when v = i and v = Thus the complex solution is [ i Y (t) = e t [ + i [ [ = e t + i e t [ 0 0 Therefore, Therefore, the general (real) solution is Y (t) = c [ e t + c [ e t (b) For the specific solution, one can see that c = and c =, and so [ [ Y (t) = e t + e t (c) π 3 (a) Find the general solution to the linear system of differential equations [ dy Y 4 [ (b) Find the specific solution that satisfies Y (0) =
Answer (a) The eigenvalues are found by solving λ 4 λ = λ 6λ + 9 = 0 Thus λ = 3 is a repeated eigenvalue The general solution is then Y (t) = [ x0 y 0 [ e 3t + [ x0 y 0 te 3t or, simplifying, the general solution becomes [ [ x0 Y (t) = e 3t x0 + y + 0 x 0 + y 0 y 0 te 3t (b) For the given initial condition, x 0 = and y 0 =, thus the specific solution is [ [ 3 Y (t) = e 3t + te 3t 3 3 Sketch the phase portraits for the systems in Questions and, and sketch the specific solutions on those phase portraits Answer Check your answers using HPG System solver 4 Sketch the phase portraits for the following systems of differential equations with the help of the given information about their eigenvalues and/or eigenvectors Also sketch the solution curve with the initial condition Y (0) = (, 0) (a) dy (b) dy (c) dy [ 4 [ 4 [ 4 Y Eigenvalues: λ = 3 repeated; eigenvector: [ Y Eigenvalues: λ = 0, 5 eigenvectors: Y Eigenvalues: λ = 0, 5 eigenvectors: [ [, [, Answer (a) See text Section 35#3 (b) See text Section 35#9 (c) Looks like (b) with arrows reversed [ 5 Write the general solution to the system in 4(b) in both component form, and in vector form Answer In vector form: Y (t) = c [ [ + c e 5t In component form, x(t) = c + c e 5t, y(t) = c + c e 5t
6 Given the following systems of differential equations, and the corresponding eigenvalues, for each system (a) determine if the origin is a spiral sink, spiral source, or a center, (b) determine the natural period of the oscillations, (c) determine whether the solutions go in a clockwise or counterclockwise direction around the origin, (d) sketch the xy-phase portrait for the system, and the x(t)- and y(t)- graphs for solutions with the indicated initial conditions (i) dy [ 0 Y ; initial condition Y 0 0 = (, 0); the eigenvalues are λ = ±i (ii) dy [ Y ; initial condition Y 0 = (0, ); the eigenvalues are λ = ± i (iii) dy [ 6 Y ; initial condition Y 0 = (, ); the eigenvalues are λ = (3 ± i 47)/ Answer (i) See answer to 34#3 (ii) See answer to 34#5 (iii) See answer to 34#7 7 Find a system with the same eigenvalues as in 6(i), but whose solutions travel in the opposite direction [ dy 0 Answer Just consider the negative of the equations in 6(i): Y 0 [ a b 8 Find equations involving a, b, c and d for the matrix A = c d (a) A has 0 as an eigenvalue; (b) A has a repeated eigenvalue so that: Answer First, the eigenvalues of A satisfy the equation λ (a + d)λ + (ad bc) = 0 (a) If ad bc = 0, then λ = 0 is a solution to the previous equation (b) If the discriminant in the quadratic equation is 0, then there is a repeated real root, ie, if (a + d) 4(ad bc) = 0 9 Find the general solutions to the spring mass system where b,m and k are as given below Classify each system as undamped, underdamped, critically damped, or overdamped, and find the period of the oscillations (if applicable) (a) m =, b = 0 and k = 8 (b) m = 3, b = 6 and k = 5 (c) m =, b = 0 and k = 5
(d) m =, b = 7 and k = 0 Answer (a) y = c + c, and the system is undamped with period π 3 (b) y = c e t cos(t) + c e t sin(t), and the system is underdamped with period π (c) y = c e 5t + c te 5t, and the system is critically damped, nonperiodic (d) y = c e t + c e 5t and the system is overdamped, nonperiodic 0 Solve the equation in 9(d) subject to the initial conditions (a) y(0) = 0 and y (0) = 0, and (b) y(0) =, y (0) = 0 Answer (a) y(t) = 0 (b) y(t) = 0 3 e t 4 3 e 5t Do Section 37, Exercise # 7 Find the general solutions to the following homogeneous differential equations Use Math- CAD or some other form of technology to help you find the roots of the characteristic equations (a) y (4) + y (3) 7y 0y y = 0 (b) y (4) + 8y (3) + 4y + 04y + 69y = 0 (c) y (5) 5y (4) + y (3) 0y + y 5y = 0 (d) y (7) + 7y (6) + 7y (5) + 5y (4) 5y (3) 9y 3y 3y = 0 Answer The zeros of the corresponding characteristic equation were found using MathCAD and are listed with each answer (a) The zeros are, 3,,, thus the general solution is y = c e t + c te t + c 3 e t + c 4 e 3t (b) The zeros are ± 3i, ± 3i, thus the general solution is y = c e t + c e t + c 3 te t + c 4 te t (c) The zeros are ±i, ±i, 5, thus the general solution is y = c cos(t) + c sin(t) + c 3 t cos(t) + c 4 t sin(t) + c 5 e 5t (d) The zeros are (multiplicity 5), and 3, thus the general solution is y = c e t + c te t + c 3 t e t + c 4 t 3 e t + c 5 t 4 e t + c 6 e t + c 7 e 3t
3 Find the form of the particular solution necessary for solving the following nonhomogeneous differential equations Do not find the particular solution (a) y 6y + 9y = e t (b) y 5y + 6y = e t (c) y 4y + 4y = e t (d) y 4y + 4y = t (e) y 6y + 9y = t + e t (f) y 4y + 4y = sin(6t) (g) y 4y + 4y = t cos(6t) (h) y + 36y = 0 sin(6t) Answer (a) y p = Ae t (b) y p = Ate t (c) y p = At e t (d) y p = At + Bt + C (e) y p = At + Bt + C + De t (f) y p = A sin(6t) + B sin(6t) (g) y p = A sin(6t) + B sin(6t) + At sin(6t) + Bt sin(6t) (h) y p = A sin(6t) + B sin(6t) + At sin(6t) + Bt sin(6t) 4 Find the general solutions to the following differential equations (a) y + 4y + 3y = e t (b) y + 4y + 3y = t + (c) y + 5y + 6y = e t + 4 (d) y + 4y + 3y = 3 cos(t) Answer (a) y = c e 3t + c e t e t (b) y = c e 3t + c e t + 3 t 9 (c) y = c e 3t + c e t + e t + 3 (d) y = c e t + c e t + 7 4 cos(t) + sin(t) 45 45 5 Find the solutions to 4(c) subject to the initial condition y(0) = 0 and y (0) = 0 Answer y(t) = 6 e 3t 3e t + e t + 3 As t, y(t) 3
6 For what values of k will pure resonance occur in the following undamped spring systems? (a) y + ky = (b) y + ky = 5 cos(6t) Answer (a) k = 9 (b) k = 36 7 Compute the solution to y + 9y = subject to y(0) = and y (0) = 9 Answer See Section 43 #3 8 Suppose the suspension system of the average car can be fairly well modeled by an underdamped harmonic oscillator with a natural period of seconds How far apart should speed bumps be placed so that a car traveling at 0 miles per hour over several bumps will bounce more and more violently with each bump? Answer Place the speed bumps so that the car will hit them every two seconds (the natural period of the car) In seconds, the car travels 76/6 feet, or 9 feet, 4 inches 9 (a) (A Superposition Principle) Suppose y is a solution to ay +by +cy = f(t) and y is a solution to ay + by + cy = g(t) where the a, b and c are the same constants in each equation, but f and g may be different Show that y = αy + βy is a solution to where α and β are constants ay + by + cy = αf(t) + βg(t) (b) Given that y (t) = is a solution to 5 and that y (t) = 3 t sin(t) is a solution to 4 find the general solution to y + 4y = y + 4y = 3 cos(t), y + 4y = 5 + 9 cos(t) Answer (a) Use the linearity property of derivatives and plug in the results: ay + by + cy = a(αy + βy ) + b(αy + βy ) + c(αy + βy ) = α(ay + by + cy ) + β(ay + by + cy ) = αf(t) + βg(t) (b) y = c cos(t) + c sin(t) + + 9 4 t sin(t)
0 (a) Show that y(t) = k cos(ωt)+k sin(ωt) can be written in the form y(t) = A cos(ωt φ) Find formulas for A and φ (b) Convert y(t) = 3 cos(4t) sin(4t) to the form y(t) = A cos(ωt φ) (c) Convert y(t) = cos(4t) + 3 sin(4t) to the form y(t) = A cos(ωt φ) Answer Using the identity cos(α β) = cos α cos β + sin α sin β one obtains A cos(ωt φ) = A[cos(ωt) cos(φ) + sin(ωt) sin(φ) = k cos(ωt) + k sin(ωt) Therefore, k = A cos φ and k = A sin φ, consequently, () A = k + k, cos φ = k A, and sin φ = k A (b) First A = 3 + = Thus sin φ = and cos φ = 3 Therefore, one choice for φ is φ = π/6 Thus y(t) = cos(4t + π/6) (c) First A = + 3 = Thus sin φ = 3 and cos φ = Therefore, one choice for φ is φ = π Thus y(t) = cos(4t π) 3 3 As a double check in each of (b) and (c), you can graph the original function along with your answer and see if the graphs are the same