FINAL EXAM SOLUTIONS, MATH 23. Find the eigenvalues of the matrix ( 9 4 3 ) So λ = or 6. = λ 9 4 3 λ = ( λ)( 3 λ) + 36 = λ 2 7λ + 6 = (λ 6)(λ ) 2. Compute the matrix inverse: ( ) 3 3 = 3 4 ( 4/3 ) 3. Let A = ( 3 5 6 ) Given that the eigenvalues of A are λ =.98 and λ 2 = 8.98, find an eigenvector for λ 2. [ ] 3 Solution: v λ = so putting λ = λ λ 2 we get [ ] 3 v = 7.98
2 FINAL EXAM SOLUTIONS, MATH 23 4. Find the eigenvalues of the matrix 3 45 2 34 5 Since it s a triangular matrix, the eigenvalues are the diagonal elements, 2,5. 5. Let A = 2 6 5 2 5 2 8 Given that 2.483 is an eigenvalue of A, find a corresponding eigenvector. Give your answer with components in decimal format. Two digits of accuracy is enough. Solution: Consider λ 2 6 5 2 λ 5 2 8 λ Expanding on the third row we have ( ) 2 6 v = 2 λ 5, λ 6 5 5, λ 2 5 2 λ 2 + 6λ = 25 + 5λ ( λ)(2 λ) 2 + 6(2.5) = 25 + 5(2.5) ( 2.5)(2 2.5).9 = 35.75.5.5 =.9 35.75 a good answer 9.83 = 36 also acceptable accuracy
FINAL EXAM SOLUTIONS, MATH 23 3 You didn t have Octave during the exam, but you could still check your answer by performing the matrix multiplication Av = 2 6 5 2 5 36 2 8 = 23 77 22 which is pretty close to 2v = (22, 72, ); since the eigenvalue 2.5 is within % of 2 this should be within % of 2v, which it is. I do have Octave while preparing these solutions. The following Octave session confirms that I didn t make any mistake: octave-3.4.:32> A = [,2,6;5,2,5;2,,8] A = 2 6 5 2 5 2 8 octave-3.4.:33> [v,d] = eig(a) v = d = -.49872 -.64362 -.2886 -.62833.76352 -.9256 -.59756.5378.254426 Diagonal Matrix.79 -.8673 2.483 The third column of v is the desired eigenvector. It should be a scalar multiple of the one I found by hand. If I multiply Octave s eigenvector by 38.636 I get (.89, 35.74, 9.83), confirming my answer.
4 FINAL EXAM SOLUTIONS, MATH 23 6. Consider the differential equation y = sin y on π x π. (a) What are the equilibrium solutions? sin y = when y = nπ for an integer n. (b) Sketch the direction field, showing the graphs of three illustrative solutions as well as the direction field. (This picture shows many more than 3 solutions, because you can t see the direction field very well at this resolution.) 7. Consider the following differential equations. Classify each one as linear or nonlinear. Indicate your answer by circling the letter of the linear ones. Answer. (a) and(c) are linear, (b) is not. (a) (b) (c) t 5 d2 y dt 2 + (t2 + ) dy dt + 2y = t dy dt + t(y2 + y + ) = d 3 y dt 3 + tdy dt + ty = t 2 + 5
FINAL EXAM SOLUTIONS, MATH 23 5 8. (a) To solve the differential equation y y = 2te 3t by the method of integrating factors, what would be the integrating factor? Answer : e t (b) Use your answer to (a) to solve the initial value problem y() = 2 for that equation. e t y ye t = 2te 2t (ye t ) = 2te 2t ye t = Now integrate by parts: = t t 2te 2t dt + y()e 2te 2t dt + 2/e since y() = 2 t ye t = te 2t e 2 e 2t dt + 2/e ( ) e = te 2t e 2 2t 2 e2 + 2/e 2 = te 2t e2t 2 e2 2 + 2/e y = te 3t e3t 2 et+2 2 + 2et That s the answer. Now let s check it. Putting t = we get y = 2, so the initial condition is satisfied. Checking the equation we have y y = d ) (te 3t e3t dt 2 et+2 2 + 2et ) (te 3t e3t 2 et+2 2 + 2et as desired. So it checks! = e 3t + 3te 3t 3 2 e3t et+2 2 + 2et ) (te 3t e3t 2 et+2 2 + 2et = 2te 3t
6 FINAL EXAM SOLUTIONS, MATH 23 9. Find the general solution of the differential equation y + y 2 sin x = Solution: It is separable: y = y 2 sin x dy y 2 = sin x dx dy y 2 = sin x dx y = cos x + C y = C + cos x Since C is an arbitrary constant, we can also write this as y = C cos x (in which the original C has been replaced by C).. For the equation y + 7y + 2y = (a) Find the general solution. The characteristic equation is λ 2 + 7λ + 2 = = (λ + 3)(λ + 4). The roots are λ = 3 and λ = 4. The general solution is y = Ae 3t + Be 4t (b) Solve the initial value problem y() = 2, y () = 3. We have 2 = A + B 3 = 3A 4B Solving the equations by Cramer s rule (or some more tedious way) we have A = 2 3 4 = B = 2 3 3 = 9 Hence the desired solution is y = e 3t 9e 4t We can check directly that the initial conditions are satisfied.. For the equation y + 9y = (a) Find the general solution
FINAL EXAM SOLUTIONS, MATH 23 7 The characteristic equation is λ 2 +9 = so λ = ±3i. The general solution is A cos 3t + B sin 3t. (b) Solve the initial value problem y() =, y () =. A = y() = and y () = 3B so B = 3. The solution is y = cos 3t + sin 3t 3 2. Convert the following linear differential equation to a system of linear equations: y + 5y 7y 4y = Write your answer in matrix form. v = 4 7 5 3. For the equation y 2y + y = (a) Find the general solution. The characteristic equation is λ 2 2λ + = = (λ ) 2, so the general solution is y = Ae t + Bte t v (b) Solve the initial value problem y() = 2, y () =. We have A = y() = 2 and y () = A + B = 2 + B =, so B = and the solution is y = 2e t te t To check the solution we compute y = 2e t te t e t = e t te t, so y () = as it should. Then y = e t te t e t = te t, so So it is correct. y 2y + y = te t 2(e t te t ) + (2e t te t ) = 4. Let A = 2 3 3 2 2 2 2.
8 FINAL EXAM SOLUTIONS, MATH 23 Given that the eigenvalues of A are,, and 2, and the corresponding eigenvectors are,, and (a) What is the general solution of the differential equation Y = AY? Hint: No calculations at all are required to solve this part of the problem. Y = Ae t B Ce 2t (b) Please solve the initial value problem From part (a) we have Y = AY with Y () = Y () = A + C B C A B + C = Adding the last two equations we have A =. Then from the first equation C = and from the second equation B =. So the solution is e t Y = e t One can put t = and check directly that Y () = (,, ) as required. There is no point checking that the equation is satisfied since we did no computation at all to find it the only thing we would be checking is whether the given eigenvalues and eigenvectors are correct..
FINAL EXAM SOLUTIONS, MATH 23 9 By the way, the eigenvalues and eigenvectors given in the problem are correct: octave-3.4.:> A = [2,-3,-3; 2,-2,-2;-2,,]; octave-3.4.:3> [v,d] = eig(a) v = d = -.57735..77 -.57735.77..57735 -.77.77 Diagonal Matrix 2.e+ 7.264e-6 -.e+
FINAL EXAM SOLUTIONS, MATH 23 5. Consider the Laplace transform of te at. (a) Give the definition of this Laplace transform. e st te at dt (b) Compute the answer directly from the definition. Integrate by parts: te (a s)t dt = te(a s)t e (a s)t a s a s dt Technically the first term should be expressed as a limit as the upper limit of integration goes to infinity. The exponential in the numerator kills the denominator so the upper term is zero (again, technically one can use L Hospital s rule). The term when t = gives because of the factor t that is multiplied in. So the first term is zero, and we have That s the answer. e (a s)t te (a s)t dt = a s dt = e(a s)t (a s) 2 = (a s) 2
FINAL EXAM SOLUTIONS, MATH 23 6. Consider the equation y y 6y = with initial conditions y() = 2, y () =. Let y(t) be the solution of this initial-value problem. Let Y (s) be the Laplace transform of y. What is Y (s) in this case? The answer must not contain the letter y, i.e. it must be a specific function of s. Taking the Laplace transform of the equation we have = L(y ) L(y ) 6L(y) = s 2 Y 2sy() y () (sy 2y()) 6Y = s 2 Y 2s + sy + 2 6Y = Y (s 2 s 6) 2s + 3 Y = 2s 3 s 2 s 6 7. Find the inverse Laplace transform of Y (s) = 2 s s 2 s 6 Y = = 2 s s 2 s 6 2 s (s 3)(s + 2) Now we expand that in partial fractions: Y = 5(s 3) 4 5(s + 2) Thus the answer is y = e3t 5 4e 2t 5 8. Evaluate δ(x π/2) sin x dx, where δ is the Dirac delta function. The answer is the value of sin(x) at π/2, namely.
2 FINAL EXAM SOLUTIONS, MATH 23 9. A mass of 5 kg. stretches a spring cm. The mass is acted on by an external force F (t) = sin(t/2) newtons and moves in a medium that imparts a viscous force of 2 newtons when the speed of the mass is 4 cm/sec. The mass is set in motion from its equilibrium position with an initial velocity of 3 cm/sec. (a) Write the differential equation describing the motion of the mass using the letters m, γ, k, and F. my + γy + ky = F (t) (b) What are the units and numerical values of m, γ, and k? What are the units of F (t)? m = 5 kg. is given. γ is 2 newtons/(.4 m/sec), which is 5 newtonseconds/m, or better, 5 kg/sec. That s the right units such that when multiplied by a velocity (m/sec) we get kg m /sec 2. Finally, k is such that k.meters = 5g = 49 kg m/sec 2, so k = 49 kg/sec 2 (c ) Formulate the initial value problem describing the motion of the mass, using the numerical values you just calculated. An answer containing any letters other than y and t will be wrong. Put in the numbers. Don t forget the initial conditions. 5y + 5y + 49y = sin(t/2) with y() = and y () =.3. 2. Consider a situation like that in the previous problem, with the same external force F (t) = sin(t/2), and the same mass (5 kg), but a different number instead of cm, such that k in the equation comes out to 7 (in the units appropriate for the meter-kilogram-second system). (a) Find the frequency of the steady-state solution in radians per second. That will be the same as the driving frequency. sin(ωt) corresponds to ω radians per second, so the frequency required here is /2 radian per second. (b) Find the natural frequency ω of this system in radians per second. ω = 7 k/m = 5 = 4 =.8 radians/second (c) This system is far from resonance because ω is far from ω.