REVIEW PROBLEMS FOR MIDTERM II MATH 2373, FALL 2016 ANSWER KEY

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1 REVIEW PROBLEMS FOR MIDTERM II MATH 7, FALL 6 ANSWER KEY This list of problems is not guaranteed to be an absolutely complete review. For completeness you must also make sure that you know how to do all of the homework assigned in the course up to and including that due on November, and all the worksheet problems through week 8, with emphasis on weeks 8. (But of course don t forget basic skills learned in weeks such as using reduced row echelon form to solve linear equations!) To study under typical test conditions, here are the ground rules. You may use matrix addition and multiplication functions on your calculator along with det, and matrix inverse functions, but you must indicate clearly where and how you used these calculator functions otherwise your answers would be considered unjustified and get little credit. Fancier calculator functions, e.g., those finding eigenvalues and eigenvectors, are not allowed to justify your work. All algebra and calculus work (except for the matrix functions specifically allowed above) must be performed by hand and shown clearly. Below we supply mostly full solutions, but occasionally skip a few details if similar details can be seen elsewhere in this answer key. Problem Solve the initial value problem y + 9y, y(), y (), expressing the answer in the form A cos(ωt δ) where A, ω and δ are positive numbers and furthermore δ < π. Give numerical approximations for A, ω and δ accurate to four decimal places. y C cos(7t) + C sin(7t) (general solution) y() C y () 7C C /7. y cos(7t) + 7 sin(7t). A + (/7). and δ arctan( /7 ).78. y. cos(7t.78). Note: In general to find δ one might also have to add π or π depending on the quadrant. Problem A mass of kg is suspended from a high ceiling by a spring. When hanging motionless, the mass stretches the spring meters beyond its natural length. Let y denote the height of the mass above the equilibrium point. Initially the mass is pulled down / meter below the equilibrium point and released with a velocity Date: November, 6.

2 REVIEW FOR MT of m/sec upward. The mass encounters air resistance of. nt/(m/sec). A force of 7 cos(t) nt is applied to the weight. Write out the initial value problem determining the motion of the mass. Is this system underdamped, overdamped or critically damped? Use m/sec as the acceleration of gravity. You do not have to solve the initial value problem. The template is my +by +ky F (t), y() y, y () v. We are given m kg, b. nt/(m/sec), F (t) 7 cos(t) nt, y. m and v m/sec. The weight of the mass is ( kg) ( m/sec ) nt, so k( nt)/( m) nt/m. Thus the initial value problem is y +.y +y 7 cos(t), y()., y (). Because b km. <, the system is UNDERDAMPED. Solve the following IVP s: Problem y + y + 6y, y(), y () y + 6y + 9y, y(), y () y + y + 9y, y(), y () 9 First equation. r + r + 6 (r + )(r + ), r,. y C e t + C e t y() C + C y () [ C C [ 7 y e t 7e t Second equation. r + 6r + 9 (r + ), r, (double root) y C e t + C te t y() C y () [ C + C [ y e t te t Third equation. r + r + 9 (r + ) +, r ± i. y C e t cos(t) + C e t sin(t) y() C y () [ C + C 9[ 9 7 y e t cos(t) + 7e t sin(t)

3 REVIEW FOR MT Problem A crash-dummy weighs lbs (and thus has mass / 7 slugs). The crash-dummy encounters air resistance of. lbs/(ft/sec). The crash-dummy is tossed off a building ft high, initially with a velocity of ft/sec upward. Let y denote the height of the crash-dummy above the ground. Set up and solve the initial value problem for the vertical motion of the crash dummy before it hits the ground. Use your calculator to find the time the crash dummy hits the ground, accurate to four decimal places. Use the method of undetermined coefficients to solve the initial value problem. Also do the problem without air resistance and compare the times of impact. setup: (7 slugs )(y ft/sec ) (. lbs/(ft/sec))(y ft/sec) lbs y() ft, y () ft/sec. after simplification: y +.y, y(), y (). r +.r, r,.. y h C + C e.t y p At (appropriate guess, because there s some trouble ) y p +.y p.a A 6. y 6t + C + C e.t y() C + C y () 6.C [.C + 6 [ 9 y 6t + 9e.t y t.69 (use cursor on graphing calculator) Without air-resistance: setup: (7 slugs )(y ft/sec ) lbs y() ft, y () ft/sec. after simplification: y, y(), y (). solve by freshman calculus y 6t + + t y t. (solve quadratic equation or just use cursor on graphing calculator again) Ignoring air-resistance the crash-dummy hits the ground sooner. Problem Find particular solutions by the method of undetermined coefficients for each of the following equations: y + y + y te t y + y + y te t y + y cos(t)

4 REVIEW FOR MT First equation. y p (At + B)e t y p + y p + y p (A)te t + (9A + B)e t te [ [ t / 9 9/ y p (t/ 9/)e t Second equation. y p (At + B)e t is a nice guess but if you try it does not work. y p (At + Bt)e t works because there is some trouble. y p + y p + y p ( A)te t + (A B)e [ [ t / y p ( t / t)e t Third equation. y p At cos(t) + Bt sin(t) y p + y p B cos(t) cos(t) A and B B / y p (t/) sin(t) Problem 6 Find the eigenvalues and eigenvectors for [ / 9 Given that the eigenvalues are λ,,, find the eigenvectors for. 7 Then use the eigenvectors and eigenvalues you have found to diagonalize the matrices. t / 9 t. t t + (t 6)(t + ) λ 6,. Since eigenvalues for the two-by-two matrix are distinct, diagonalization is guaranteed to be possible. We just need to find some eigenvector for each eigenvalue. [ [ 6 / / [ [ / [ [ ( ) / / [ [ /. 9 ( ) We did not bother to hit the button because in each case the two equations are really just one equation we can solve by inspection. Here is a diagonalization: [ [ [ [ / 6 9

5 REVIEW FOR MT Since eigenvalues for the three-by-three matrix are distinct, diagonalization is guaranteed to be possible. We just need to find some eigenvector for each eigenvalue. /, Here is a diagonalization: 7 Consider the matrix: / /9 9 Problem 7 A [, 9 9. Diagonalize the matrix A, and find the matrix power A n. Multiply out the matrix so that each element has a concrete expression. Diagonalization: [ AP P D, P [, D, P [ (The details of finding the diagonalization are similar to those in the previous problem and therefore omitted.) Matrix Power: [ [ [ A n P D n P n ( ) n / [ [ n ( ) n n ( ) n / [ n + ( ) n n ( ) n n ( ) n n + ( ) n. Problem 8 Find the eigenvalues and *all* eigenvectors for the following matrices: 8 [ 7 8, (We leave you to find the eigenvalues of the two-by-two matrix, but we tell you that the eigenvalues of the three-by-three matrix are,,, i.e., is a double eigenvalue.) Then use the eigenvalues and eigenvectors to diagonalize the given matrices if possible. If not possible, explain why not.

6 6 REVIEW FOR MT Since 6 6 / / (note two nonpivot columns) the eigenvectors for λ are all vectors of the form s + t where not both s and t are zero. Since 6 / (note just one nonpivot column) the eigenvectors for λ are all vectors t for t. We have enough eigenvectors because the total number of nonpivot columns we found in the course of analyzing all the eigenvalues was three, matching the size of the matrix we are diagonalizing, which is three by three. So we get a diagonalization: 8 6 Eigenvalues for the two-by-two are,, i.e., we have a double root. (We skip details of finding and solving the characteristic polynomial since we ve been there.) Since [ 8 8 [ / and there is just one nonpivot column. The eigenvectors for the eigenvalue are [ t for t. The eigenspace is only one-dimensional. The total number of nonpivots discovered in the course of trying to find eigenvectors, namely one, does not match the size of the matrix, which is two-by-two. We can t diagonalize. For more information on what happens when diagonalization is not possible, proceed to an advanced linear algebra course to learn about Jordan canonical form. Problem 9 Approximate the solution of the initial value problem y t y /, y() on the interval t. by using Euler s method with steps.,,

7 REVIEW FOR MT 7 Let F (t, y) t y /. We have step-size h (. )/.. Here is the answer in nice neat tabular form: k t k y k.. + F (, ). + (.7)(.)... + F (.,.).. + (.898)(.) Problem Solve the following initial value problem: y + 6y + y 66 cos(7t), y(), y () 6 Identify the steady-state and transient parts of the solution clearly. Express the steady-state part of the solution in the form A cos(ωt δ), where A >, ω >, δ and δ < π. Numerical approximations to four decimal places of accuracy are requested for A, ω and δ. r + 6r + (r + ) + r ± i. y h C e t cos(t) + C e t sin(t) y p A cos(7t) + B sin(7t) y p + 6y p + y p ( A + B) cos(7t) + ( A B) sin(7t) [ [ 66 y y p + y h cos(7t) + sin(7t) + C e t cos(t) + C e t sin(t) y() + C C y () 7 C + C C 6 C C. y cos(7t) + sin(7t) + e t cos(t) e t sin(t) }{{}}{{} steady state transient A ( ) +.866, δ arctan( ) + π.98 (note that the point (, ) is in quadrant II) y.866 cos(7t.98) + e t cos(t) e t sin(t) }{{}}{{} steady state transient Problem Find an initial value problem of the form y + by + ky A cos(ωt δ), y() y, y () v of which cos(7t) sin(7t) + e t cos(t) e t sin(t) is the solution. Identify the stationary and transient parts of the solution, and determine the constants b, k, A, ω, δ, y and v.

8 8 REVIEW FOR MT derivative: You can find y and v just by plugging into the given function and its y ( cos(7t) sin(7t) + e t cos(t) e t sin(t) ) t v d ( cos(7t) sin(7t) + e t cos(t) e t sin(t) ) dt t Transient part of the given solution: e t cos(t) e t sin(t) We will call this part of the solution y h. To get the indicated transient part we need characteristic roots ± i. These roots are the solutions of the equation (r + ) + r + 6r +. Thus we have b 6, k. We emphasize that the coefficients b and k have been cooked up so that y h + 6y h + y h. Stationary (steady state) part of given solution: cos(7t) sin(7t). We will call this part of the given solution y p. Now plug y p into the left side of the equation we have already figured out for y h. After some algebra we have omitted, we get y p + 6y p + y p cos(7t) + sin(7t). This must be equal to the right hand side of the equation we are looking for. Thus, ω 7, A +.9 (just a wee bit greater than ), ( ) δ arctan + π.968 (just a wee bit less than π). FINAL ANSWER: y +6y +y cos(7t)+ sin(7t). cos(7t.), y(), y (). Problem Solve the initial value problem y + 6y 8 sin(8t), y(), y (). Solution. y h C cos(8t) + C sin(8t) Trouble ahead. y p At cos(8t) + Bt sin(8t) t(a cos(8t) + B sin(8t)) Note that (fg) f g + fg applied twice gives (fg) f g + f g + fg, hence y p + ( 8A sin(8t) + 8B cos(8t)) + t( 6 cos(8t) 6 sin(8t)) y p + 6y p 6B cos(8t) 6A sin(8t) B and 6A 8 A /. y p t cos(8t)/. y y p + y h t cos(8t)/ + C cos(8t) + C sin(8t) y() C

9 REVIEW FOR MT 9 y () / + 8C C /6. y t cos(8t)/ + cos(8t) sin(8t)/6 Are the vectors, Problem, and one of the vectors as a linear combination of the others. linearly independent? If not, express The equations to be analyzed are x + y z If these equation have only the solution (x, y, z) (,, ) then the answer is yes, linearly independent and otherwise the answer is no, linearly dependent. Well, we have / 9/ So indeed no, the vectors are linearly dependent because there is some solution (x, y, z) of the equations above different from (,, ). Reading off one of these solutions different from (,, ) from the reduced row echelon matrix, we find that Finally, we can rearrange the preceding equation to get the equation + 9 which expresses the last of the vectors as a linear combination of the preceding two

10 REVIEW FOR MT Express the vector replaced by Problem as a linear combination of if possible. If not possible, explain why. Then same question with (only a tiny change).,, and To answer the first part of the question we have to solve the equations x + y + z Well, we have + + and thus To answer the second part of the question we have to solve the equations x + y + z. Well, we have.. The last equation here is which can t be solved, so no linear combination of the requested type can be formed.