Form A Math 4 Common Part of Final Exam December 6, 996 INSTRUCTIONS: Please enter your NAME, ID NUMBER, FORM designation, and INDEX NUMBER on your op scan sheet. The index number should be written in the upper right-hand box labeled Course. In the box labeled Form, write the appropriate test form letter (A, B, or C). Darken the appropriate circles below your ID number and Form designation. Use a # pencil. Mark your answers to the test questions in rows -5 of the op-scan sheet. You have hour to complete this part of the final exam. Your score on this part of the final exam will be the number of correct answers. Please turn in the op scan sheet with your answers and the question sheet at the end of this part of the final exam.. Which of the following is a second-order, linear, homogenous differential equation? dy () + 4y = x dx () sin(x) y + y tan(y)y = 0 (3) x 3 y x y + y = 0 (4) r + 3r + = e x. The differential equation u'' +4u' sin u = 0 can be transformed into the first order system () x' = y y' = 4x +sin y () x' = y y' = sin x 4y (3) x' = sin y y' = 4x (4) x' = y y' = 4x +sin u 3. If r = and r = 3 are solutions to the characteristic equation for a y + b y + cy = 0, where a,b, and c are real numbers, then which of the following is the general solution to a y + b y + cy = 0? (c and c denote arbitrary real constants.) () y = c e x + c e 3 x () y = c sin(x) + c cos( 3x) (3) y = c e ( +3 i ) (4) y = c ( x) + c (3x) 4. If the method of undetermined coefficients is used to solve the equation y y + y = ( + x)e x, which one of these forms should be used for a particular solution? () (A 0 + A x)e x () (A x + A x )e x (3) (A x + A 3 x 3 )e x (4) (A 3 x 3 + A 4 x 4 )e x
5. Find the solution to the initial value problem y x y = x, y() =. () y = x (x +) () y = x 3 5 + 9 5x (3) y = x + x (4) y ln(x) = x 3 3 6. For the solution of the initial value problem x y + y = x, y() =, find the value y(). () 3 () 5 (3) 5 4 (4) 7 4 7. Find so that the solution of the initial value problem y y y = 0, y(0) =, y (0) = approaches zero as x () - () - (3) (4) 3 dy 8. Consider the system = Ay where A is a matrix. dt For which matrix A does every solution converge to the zero vector as t? 0 () 0 () (3) (4) 0 9. Let y(x) be the solution of the initial value problem What is lim x y(x)? y' = y x, y() = 4. () 4 () 3 (3) (4)
0. Let g(x, y) be a continuous function. Let y (x) and y (x) be solutions of the differential equation y' = g(x, y) on an interval that includes the points x=0 and x=. Suppose y (0) = y (0) and y () < y (). Which one of the following statements must be true? () y '(0) = y '(0) () y '(0) < y '(0) (3) y ( ) < y ( ) (4) y '() y '(). Let (x) be the solution to the initial value problem y + y = 0, (/) is ) e / ) /7 3) /5 4) e /4 3 y(0) = /3. Then. Using the Euler method with h = 0., approximate the solution to the initial value problem dy dx + y = x, y() = at x =.. ().7 ().9 (3). (4).3 3. Which one of the following is the graph on the interval [-,] of the solution to the dy initial value problem dx x y = x, y(0) =? ) ) 3) 4) 3
4. Let A be the 3 x 3 matrix 4 3. The system x = Ax has a solution x(t) = e t v, where v is the vector () 3 0 () 5 4 (3) (4) 4 5. A tank with a capacity of 000 gallons originally contains 00 lb. of salt dissolved in 400 gallons of water. Beginning at time t = 0 and ending at time t = 00 minutes, water containing lb. of salt per gallon enters the tank at the rate of 4 gal/min., and the wellmixed solution is drained from the tank at a rate of gal/min. Then a differential equation for the amount of salt in the tank at time t is () ds dt = 4 s 00 () ds dt = 4 s 300 (3) ds dt = 4 s 00 + t (4) ds dt = 4 s 00(t + ) 4
Form B Math 4 Common Part of Final Exam May, 997 INSTRUCTIONS: Please enter your NAME, ID NUMBER, FORM designation, and INDEX NUMBER on your op scan sheet. The index number should be written in the upper right-hand box labeled Course. In the box labeled Form, write the appropriate test form letter (A, B, or C). Darken the appropriate circles below your ID number and Form designation. Use a # pencil. Mark your answers to the test questions in rows -5 of the op-scan sheet. You have hour to complete this part of the final exam. Your score on this part of the final exam will be the number of correct answers. Please turn in the op scan sheet with your answers and the question sheet at the end of this part of the final exam.. y = xe x is a solution of which differential equation? () y 4 y + 4y = 0 () y 4 y + y = 0 (3) y y + 4y = 0 (4) y y + y = 0.. The Wronskian of the functions, x, e x is () () x (3) e x (4) xe x 3. Let y(x) be the solution of the initial value problem y = y (+ x), y(0) =. What is the value y()? () 3 () 3 (3) (4) 4. Let y(x) be the solution of the initial value problem y + 3 x y = 3, y() =. x What is lim x y(x)? () 0 () (3) (4) infinity
5. Let y(x) be the solution of the initial value problem y + 4y = 0; y(0) = 4, y (0) =. What is the value y(π)? () - () 0 (3) (4) 4 6. Let a < x < b denote the longest interval on which the solution of the initial value problem below exists. (x )(x )x y (x + 5)(x )x y (x )y = 0; y( ) =, y ( ) = Then the interval a < x < b is () < x < 0 () 5 < x < (3) 5 < x < (4) < x< 0 7. Suppose that a certain radioactive material decays at a rate proportional to the amount present. If it takes 30 days for this material to be reduced to one third its original amount, then how long will it take for the same material to be reduced to one tenth of its original value? 0 3 () 30 e days () 00 days 30ln0 (3) days ln 3 0 (4) 30ln days 3 3 8. Suppose that y ( x) = x, y ( x) = x are solutions of y + p( x) y + q( x) y= 0 where p(x) and q(x) are continuous for 0<x<5. Let φ( x ) be the solution of the initial value problem: y + p( x) y + q( x) y = 0 y( ) = y ( ) = 4 Then φ( 3 ) is () 36 () 45 (3) 48 (4) 5 B
9. Let y(x) be the solution of the initial value problem y y = ; y(0) = 3, y (0) = β. y(x) Find β so that lim = 0. x x () β = () β = 0 (3) β = (4) no such β 0. The differential equation y (iv ) 4 y + 8 y 8 y + 4y = 0 has characteristic polynomial p(r) = r 4 4r 3 + 8r 8r + 4 = (r ( + i)) (r ( i)). The general solution of the differential equation is t t () e ( A cost + Bsin t) + e ( C cost + Dsin t) t t () e ( A cost + Bsin t) + te ( C cost + Dsin t ) (3) Acost + Bsin t + t( C cost + Dsin t ) t t (4) e ( Acost + B sin t ) + e ( C cos( t ) + Dsin( t)). The differential equation u 3 u 7u = cos t can be transformed into which first order system? x = x () x = 3x + 7x + cost () (3) (4) x = 7x + cost x = 3x x = x + cost x = 7x + 3x x = x x = 7x + 3x + cost. Let A =. One eigenvector for A is w = 4 eigenvalue is () - () - (3) (4) 4. The corresponding 3B
3. Let A= 0. One fundamental matrix for x =Ax is e t e t () e t 0 e t e t () e t 0 e t e t (3) 0 e t e t e t (4) e t e t 6 4. A =. The eigenvalues for A are 5± i. The general solution for the system 5 4 x = Ax is cost () x= + + C t t e C sint 5t t + t e 5t Let cos sin cos sin cos5t () x= + + C t t e C sin 5t t t + t e cos 5 sin5 cos 5 sin5 (3) x= + t 5t C e C e 3 3 (4) x = 7 + 8 t 5t C e C e 3 3 5. We want to apply Euler s method to the initial value problem y = y +t ; y(0) =. If we use a step size of h =0. to estimate y(0.), we find a value y where y is (). (). (3). (4).4 t 4B
Form A Math 4 Common Part of Final Exam December 5, 997 INSTRUCTIONS: Please enter your NAME, ID NUMBER, FORM designation, and INDEX NUMBER on your op-scan sheet. The index number should be written in the upper right-hand box labeled Course. In the box labeled Form, write the appropriate test form letter (A, B, or C). Darken the appropriate circles below your ID number and Form designation. Use a # pencil. Mark your answers to the test questions in rows -4 of the op-scan sheet. You have hour to complete this part of the final exam. Your score on this part of the final exam will be the number of correct answers. Please turn in the op-scan sheet with your answers and this question sheet at the end of this part of the final exam.. Which of the following differential equations is linear? () x + t x + e t x = cost () x + tx =t (3) x x + 3 x + 4x = t (4) x + x + x = cos x. If y + y = te t with y(0) =, then y() equals: () 3 e () (e + e ) (3) e (4) e + 3. Let y (t) = t and y (t) = e t be solutions of the homogeneous equation y + p(t) y + q(t )y = 0. Let y 3 (t) = cost be a particular solution of the inhomogeneous equation Find the solution of the initial value problem y + p(t) y + q(t )y = h(t). y + p(t) y + q(t )y = h(t) y(0) =, y (0) = () t + e t () 3t e t + cos t (3) e t cost (4) t + cost A
4. A tank initially contains 0 gallons of fresh water. Then water containing 0 grams of salt per gallon flows into the tank at a rate of gallons per hour. The well-stirred mixture leaves at the same rate. With Q(t) denoting the amount of salt in the tank at time t in hours, the initial value problem corresponding to this system is: () Q (t) = Q(t) + 0; Q(0) = 0 () Q (t) = Q(t) + 0; Q(0) = 0 5 (3) Q (t) = Q(t) + ; Q(0) = 0 (4) Q (t) = Q(t) + 0; Q(0) = 0 5 5. Consider the differential equation y y + y = e x + tan x. If the method of variation of parameters is used to find a particular solution, then one should look for a particular solution of the form () ue x + ve x () ue x + vxe x (3) ue x + v tan x (4) uxe x + vxtan x 6. The largest interval on which the initial value problem (x ) y + x y + (tan x) 3 y = e x ; y() =, y () = 3 is guaranteed to have a unique twice differentiable solution is: () ( π,) () ( π,) (3) ( π, π ) (4) (0,3) 7. A particular solution for the differential equation y + y = e x + sin x will have the form () Ae x + Bsin x () Ae x + Bxsin x (3) Ae x + Bx cosx + Cx sin x (4) Axe x + Bx cos x + Cxsin x 8. The general solution to y (4) y = 0 is: () y = C + C x + C 3 e x + C 4 xe x () y = C e x + C xe x + C 3 e x + C 4 xe x (3) y = C + C + C 3 e x + C 4 e x (4) y = C + C x + C 3 e x + C 4 e x 9. At each point (x, y) on a curve, the slope of the curve is 3x. If the curve passes y through the point (,3), then the curve is given by: () ln y = 3x + ln3 3 () y = x 3 + 7 (3) y = x 3 3 x (4) y = 3x + 8 3 A
0. The general solution of y + y = e x is: () y = C cos x + C sin x + C 3 e x () y = C e x + C e x + xe x (3) y = C cos x + C sin x + xe x (4) y = C cos x + C sin x + e x. The second order differential equation u + 5 u + 7u = cost is equivalent to the first 0 order system X = AX + where A is the matrix cost 7 5 5 7 () () 0 0 0 (3) (4) 7 7 5 5 0. The general solution of X (t) = X(t) is: 3 () X(t) = C e 4t +C 3 e t () X(t) = C e 4t +C 3 e t 0 (3) X(t) = C e 4t 0 +C 0 e t (4) X(t) = C 0 e 4 t +C e t 3 3. Consider the first order system x = t + x + y y = + x with initial condition x(0) = 0 and y(0) = 0. If we use the Euler method with step size h = 0. to estimate x(0.), the result will be: () 0.0 () 0. (3) 0. (4) 0 0 0 4. The matrix A = 0 has only the eigenvalue λ =. How many linearly 0 independent eignevectors does it have? () 0 () (3) (4) 3 A3
Form A Math 4 Common Part of Final Exam May 4, 998 Instructions: Please enter your NAME, ID NUMBER, FORM designation, and INDEX NUMBER on your op-scan sheet. The index number should be written in the upper right-hand box labeled Course. Do not include the course number. In the box labeled Form, write the appropriate test form letter A. Darken the appropriate circles below your ID number and Form designation. Use a # pencil; machine grading may ignore faintly marked circles. Mark your answers to the test questions in rows - 4 of the op-scan sheet. You have hour to complete this part of the final exam. Your score on this part of the final exam will be the number of correct answers. Please turn in your op-scan sheet and the question sheet at the end of this part of the final exam.. Consider the differential equation dy dx = x y. If y = when x =, then the value of y when x = 0 is: () 3 () (3) 5 (4) 5. Consider the first order differential equation y = x + y + with initial condition y(0) = 0. If we use the Euler method with step size h = 0. to estimate y(0.3), rounding off all calculations to two decimal places, the result will be: () y(0.3) =. () y(0.3) =.36 (3) y(0.3) = 0 (4) y(0.3) =.53 3. The general solution of y y = e t is: () y = C cost + C sint + e t () y = C e t + C e t + e t (3) y = C e t + C e t + tet (4) y = C cost + C sint te t 4. The general solution of y + y + y = 0 is: () y = C e t cos 3 t + C e t sin 3 t () y = C e 3 t + C e 3 t (3) y = C cos 3 t + C sin 3 t (4) y = C cos t + C sin t
5. The eigenvalues of A = 4 3 are λ = 5 and λ = with corresponding eigenvectors for λ = 5 and for λ =. The solution of x = Ax which satisfies x(0) = 0 is: () x = e t + 0 e5t () x = e5t + e t (3) x = /3 /3 e5t + /3 /3 e t (4) x = e t + 3 e5t 6. If the method of undetermined coefficients is used to find a particular solution y p to the differential equation y + 4y = 3t sint +4e 3t, the assumed form of y p should be: () y p = Atsin t + Btcost + Csint + Dcost + Ee 3t () y p = Atsin t + Bsin t + Ce 3t (3) y p = At sint + Bt cost +Ct sin t + Dt cost + Ete 3t (4) y p = At sint + Bt cost +Ct sin t + Dt cost + Ee 3t 7. Consider the differential equation y + y 3y = sint + sect. If the method of variation of parameters is used to find a particular solution, the assumed form for the particular solution should be: () u (t)e 3 t + u (t)te t () u (t)e 3 t + u (t)(sin t + sect)e t (3) u (t)sin t + u (t)sec t (4) u (t)e 3 t + u (t)e t
8. Which of the following differential equations is linear? () y + 3y y + 4y = 0 () y + ty = t (3) y +(t + ) y + e t y = sin t (4) y + y + y = cos y 9. At initial time t = 0, a tank contains 00 gallons of pure water. Then water containing lb. of salt per gallon flows into the tank at the rate of 5 gallons per minute, and the well-stirred mixture leaves the tank at the same rate. Let Q(t) denote the number of pounds of salt in the tank after t minutes.then when t 0, Q satisfies the differential equation and initial condition: () Q = 5 Q ; Q(0) = 0 () 0 Q = 5 Q 00 ; Q(0) = 0 (3) Q = 5Q 5; Q(0) =00 (4) Q = Q ; Q(0) = 00 00 0. The solution of y 4 y + 4y = 0 satisfying y(0) = 0, y (0) =, is: () y = e t () y = te t (3) y = e t e t (4) y = e t te t. Consider the first order linear differential equation t y + y = tan t with initial condition y() =. The largest interval on which a unique solution is guaranteed to exist is: () (, π ) () (, π ) (3) (0,) (4) (0, π ). The general solution of t y + ty =e t is: () y = C e t + t () y = t (e t + C ) (3) y = t e t + C (4) y = C t e t + C t
3. If y (t) = t is one solution to the differential equation t y 4t y + 6y = 0, then the solution y(t) for which y() = 0 and y () = satisfies: () y(0) = 0 () y(0) = (3) y(0) = 8 (4) y(0) = 4. The matrix A = has one eigenvector v = for the repeated eigenvalue 4 6 λ = 4, and it has a generalized eigenvector w = satisfying (A 4I)w = v. The 3 general solution of x = Ax is: () x = C e 4t +C e4t 3 (3) x = C e 4t +C te 4t 3 () x = C e 4t +C e4t t + 3 (4) x = C e 4t +C e4t t 3 + t
Form A Math 4 Common Part of Final Exam December 4, 998 INSTRUCTIONS: Please enter your NAME, ID NUMBER, FORM designation, and INDEX NUMBER on your op-scan sheet. The index number should be written in the upper right-hand box labeled Course. In the box labeled Form, write the appropriate test form letter A. Darken the appropriate circles below your ID number and Form designation. Use a # pencil. Mark your answers to the test questions in rows -4 of the op-scan sheet. You have hour to complete this part of the final exam. Your score on this part of the final exam will be the number of correct answers. Please turn in the op-scan sheet with your answers and this question sheet at the end of this part of the final exam.. The solution of the initial value problem y + 7 y = 0, y(0) = is () e 7t () e 7t (3) 7e t (4) 7e t. Suppose that the third order differential equation e t u + (cost) u + e t u = 0 is equivalent to the system dx = Ax. Find A. dt () 0 0 0 0 e t cost 0 () 0 0 0 0 e t cost 0 0 0 (3) 0 0 e t cost 0 (4) 0 0 0 0 0 e t cost
- - 3. Let y(t) be the solution of the initial value problem y = y, y(0) = 0. By examining the direction field or otherwise, determine which one of the following equals y(4π). () - () 0 (3) (4) 4. Consider the differential equations and (NH) y + p(x) y + q(x)y = g(x) (H) y + p(x) y + q(x)y = 0. If y (x) = x and y (x) = x are solutions of (H), and ϕ (x) = x is a solution of (NH), then the solution of the initial value problem is y + p(x) y + q(x)y = g(x), y() = 0, y () = () x x () x + x (3) 5x + x + x (4) 3x + x 3 5. For which values of the real constant α does the solution pair y (x) = α e x + e x, y (x) = e x + α e x form a fundamental set of solutions for the differential equation y y = 0? () Every α () Every α except α = (3) Every α except α = and α = (4) Only for α = and α =
- 3-6. Let A be a x constant matrix with real entries. Suppose that λ = is a repeated eigenvalue of A with eigenvector. Assuming that (A I) 0 =, find the solution of the following initial value problem: dx dt = Ax, x(0) =. () (3) e t e t () e t + te t e t + te t (4) e t + te t e t e t + te t e t + te t 7. For the initial value problem y = t y, y() = 3, use one step of the Euler method to calculate an approximation of y(.). () -.5 ().5 (3).9 (4) 3. 8. The general solution of the equation is y + 3 y 4y = 0 () Ae x + Be x + Cxe x () Ae x + Be x (3) Ae x + Bcos x + Csinx (4) Ae x + Be x + Cxe x
- 4-9. Newton's Law of Cooling says dθ dt = k(θ T), where θ (t) is the temperature ( F) of a given object at time t (in minutes), T is the temperature of the surrounding environment, and k is a constant of proportionality. A thermometer reads 60 F. It is taken outdoors, where the temperature is 0 F. One minute later, the thermometer reading is 50 F. Then the temperature reading minutes after the thermometer was taken outdoors is () 60 ln(6/5) () 60 (3) 40 (4) 5 [ln(6/5)] 3 0. If the method of undetermined coefficients is used to find a particular solution y p to the differential equation y y + y = xe x + x, then the assumed form of y p should be () y p = Ax e x + Bxe x + Cx + D () y p = Ax e x + Be x + Cx + D (3) y p = Axe x + Bx (4) y p = Axe x + Be x + Cx + D. Let A be a x constant matrix with real entries. Suppose that A has eigenvalue + i with eigenvector. Find the solution of the initial value problem i dx dt = Ax, x(0) =. cos(t) + () e t 3 sin(t) cos(t) + 4sin( t) () e t cos(t) sin(t) cos(t) sin(t) cos(t) + (3) e t 4 sin(t) cos(t) + (4) e t sin(t) cos(t) + sin(t) cos(t) 4 sin(t)
- 5 -. Find the general solution of the DE dy dt + y = cos(3t). (The symbols C, C and C denote arbitrary constants.) Hint: Examine the four choices. () C e t + C sin(3t) () 3 e t + C cos(3t) + C sin(3t) (3) Ce t + 3 0 sin(3t) + t cos(3t) (4) Ce + 3 0 0 sin(3t) + 0 cos(3t) 3. If y(t) is the solution of the initial value problem then y() is equal to y t y = 8t, y() = 8, () 9 () 0 (3) 38 (4) 40 4. Given that A is a x constant matrix with real entries, which of the following functions can be a solution of the system dx dt = Ax? () e t e 3t e t () e t + e t te t + e t e t e t (3) (4) 5e t + 5e t e t e t e t + e t
INSTRUCTIONS: Form A Math 4 Common Part of Final Exam May 0, 999 Please enter your NAME, ID NUMBER, FORM designation, and INDEX NUMBER on your op-scan sheet. The index number should be written in the upper right-hand box labeled Course. In the box labeled Form, write the appropriate test form letter (A, B, or C). Darken the appropriate circles below your ID number and Form designation. Use a # pencil. Mark your answers to the test questions in rows - of the op-scan sheet. You have hour to complete this part of the final exam. Your score on this part of the final exam will be the number of correct answers. Please turn in the op-scan sheet with your answers and this question sheet at the end of this part of the final exam. NOTE WELL: Throughout the exam arbitrary constants are denoted by c, c or c. Your signature is required on this form: Signature
. If y + y + = 0 and y ( 0) = 0, then y() = () + (/ e) () (/ e) (3) ( / e) (4) e. Given two solutions y ( t ) and y ( t) of a homogeneous linear differential equation, consider the following statements: (a.) y t) + y ( ) must also be a solution to the differential equation. ( t (b.) y ( t) must also be a solution to the differential equation. (c.) y( t) y ( t) must also be a solution to the differential equation. (d.) y ( t) y ( ) must also be a solution to the differential equation. 6 t Which of the statements are always true? () (a.) only () (a.) and (b.) only (3) (a.), (b.) and (c.) only (4) (a.), (b.) and (d.) only (5) all are true 3. If y π = y sin( t), and y ( 0) =, then y = () π () y is undefined at. (3) (4) / 4. At time t = 0, a tank contains 500 gallons of well-mixed salt water containing 00 lbs of salt. Pure water enters the tank at 0 gallons per minute and the well-stirred mixture leaves at the same rate. Let Q(t) be the number of pounds of salt in the tank after t minutes. Then () () (3) (4) Q( t) = e t / 50 Q( t) = + 98e Q( t) = 00e t /50 t / 50 Q( t) = 500 400e t / 50
5. A third order linear homogeneous constant coefficient differential equation has characteristic equation (r ) (r 3) = 0. The general solution of the differential equation has the form () y = c t t e + ce + c e 3t 3 () y = c (+ t)e t + c e 3t (3) y = c e t + c te t + c 3 e 3t (4) y = c t + c t + c 3 e 3t 6. The general solution of y y +7y = 0 has the form () y = c cos(4t) + c sin(4t) () y = c e t cos(4t) + c e t sin( 4t) (3) y = c t cos(4t) + c t sin(4t) (4) y = c e 4t cos(t) + c e 4t sin(t) 7. Given that the homogeneous differntial equation y 9 y + 0 y = 0 has general 5t 4t solution c e + ce, find the solution y to the initial value problem y 9 y + 0y = 4e t, y(0) = 3, y (0) = 6 is () y = e 4t + e t () y = 9e 4t 6e 5t + e t (3) y = 9e 4t 6e 5t (4) y = 3e t 8. The second order linear homogeneous constant coefficient differential equation y + y 6y = 0 has general solution y = c e t + c e 3t. Which of the following statements describes the behavior of the solution of the initial value problem y + y 6y = 0, y(0) = α, y (0) = 3? () For every value of α, the solution does not approach 0 as t. () For every value of α, the solution approaches 0 as t. (3) The solution approaches 0 as t if and only if α = 3. (4) The solution approaches 0 as t if and only if α =.
9. An equation of the form u + au + bu + cu = 0 is transformed to a system of first order equations of the form x = Ax where the matrix A is given by Then () a = 5; b = 6 ; c = 4 () a = 5; b = 6; c = 4 (3) a = 4 ; b = 6; c = 5 (4) a = 4; b = 6 ; c = 5 0 A = 0 4 0 6 0 5 0. Consider an initial value problem of the form x = Ax, x (0) =. For which matrix A does lim x (t) = 0? 3 () A = 0 3 3 () A = 0 3 3 0 (3) A = 0 0 (4) A = t. A matrix A has eigenvalues λ = 3 and λ = 7. An eigenvector corresponding to λ = 3 is. An eigenvector corresponding to λ = 7 is. The general 5 solution of x = Ax is given by () () (3) (4) c e 3t c e c e 4 3 c e 7 3t + 3t 3t 7t + e 5 7t ce 4 + c e 0 + c e 3 7t 7t
. Let A =. The vector w = 4 is an eigenvector for A. A corresponding eigenvalue is () λ = () λ = (3) λ = (4) λ = 4
Form A Math 4 Common Part of Final Exam Dec 3, 999 INSTRUCTIONS: Please enter your NAME, ID NUMBER, FORM designation, and INDEX NUMBER on your op-scan sheet. The index number should be written in the upper right-hand box labeled "Course". In the box labeled "Form", write the appropriate test form letter A. Darken the appropriate circles below you ID number and Form designation. Use a # pencil. Mark your answers to the test questions in rows -4 of the op-scan sheet. You have hour to complete this part of the final exam. Your score on this part of the final exam will be the number of correct answers. Please turn in the op-scan with your answers and this question sheet at the end of this part of the final exam.. Four solutions of y' ''' y=0 are? (a) e t,te t,t e t,t 3 e t (b) e t cost,e t sin t,te t cost,te t sint (c) e t,te t,e t,te t (d) e t,e t,sin t,cos t. The general solution of y'' +5y' +4y = 0 is? (a) y = c e 4t + c e t (b) y = c e 5t + c e 4t (c) y = c e 4 t + c e t (d) y = c e 5 t + c e 4 t 3. A particular solution for the differential equation y'' +5y = 6sin t would be (a) y p = c cos5t + c sin5t (b) y p = 4 sin t (c) y p = 5 cost (d) y p = 5 cost + 4 sint (e) y p = 4 cos5t + 5 sin5t
4. The third order differential equation y'''+0y' ' 9y'+3y=cost can be rewritten as the first order system. x ' = 3x (a) x ' = 0 9 x 3 x 3 ' = 0x 3 + 9x 3x + cost x ' = x (b) x ' = x 3 x 3 ' = 0x 3 + 9x 3x + cost x ' = x (c) x ' = x 3 x 3 ' = 0x + 9x 3x 3 + cost x ' = x (d) x ' = x 3 x 3 ' = 3x 9x + 0x 3 + cost 5. A linear system of two first order differential equations with real coefficients has the i complex solution X(t) = e ( + 3i)t. Then two real-valued solutions are given by sin3t (a) e t cos3t cos3t,et sin3t sint (b) e 3t cost cos t,e3t sint sin3t (c) cos3t, cos3t sin3t cos3t sin3t (d) cos3t + sin3t, sin3t cos3t sin3t cos3t 6. Use Euler's Method with a step size of h = 0.5 to approximate the value of y() given the initial value problem y' = 3t y, y(0) = 0. (a) y() 0 (b) y() 0.375 (c) y() 0.5 (d) y() 0.75
7. Given that y = t and y = t are solutions of the homogeneous differential equationt y'' +ty' y = 0 and the function y p = t is a particular solution of the nonhomogeneous differential equation t y'' +ty' y = 3t. The general solution of the nonhomogeneous differential equation is (a) y = t + t + t (b) y = c t + c t + c 3t (c) y = c t + c t + 3t (d) y = c t + c t + t 8. The general solution of the system X' (t) = 4 X is (a) c e 3t + c e t (b) c e 3t + c e t (c) c e 3t + c e t (d) c e 3t + c e t 9. The solution of the initial value problem y'' y = e t with y(0) =, y'(0)= is (a) 5 4 et + 3 4 e t + tet (b) 3 e t + e t + tet (c) cost+sint + tet (d) e t 3 e t + tet 0. Which of the following differential equations is not linear? (a) y'''+ty' +(cos t)y = t 3 (b) y'''+3y''+4y' 8y =0 (c) (+ x )y' '+e x y'+y=e x (d) u' '( t)+u'(t)+5u(t)=3cost
. Find an implicit solution of the differential equation y' = cost 5y. 4 (a) y 5 = sin t + C (b) y 5 + sin t = C (c) y 5 = Csint (d) y = sint + C y 5. The solution of the initial value problem ty' y = t3, y()= 0 is t + (a) t ln(t +) ( ln)t (b) t t tan t + ( 4 )t (c) t ln(t +) t lnt + ( ln) t (d) t tan t 4 t 3. The decay constant for thorium-34 is 0.088 days. The differential equation which governs the decay of thorium-34 when the isotope is being replenished at a rate of mg/day is (a) dq dt (b) dq dt (c) dq dt (d) dq dt = 0.088Q(t) = 0.088 + Q(t) = 0.088Q(t) = 0.088 Q(t) 4. Matrix A = 4 6 has only one eigenvector v = for the repeated eigenvalue = 4. Also, a solution of [ A I ] = v is = 3. Then general solution of x' = Ax is (a) x = c e 4t + c e 4t 3 (b) x = c e 4t + c te 4 t 3 (c) x = c e 4t + c e 4t t + 3
(d) x = c e 4t + c e 4t + t 3
Math 4 Spring 000 Common Final Exam, Spring 000. The differential equation e t y + ysint = is (a) nonlinear (b) homogeneous (c) linear (d) separable. A tank initially contains 00 liters of pure water. A mixture containing a concentration of g per liter of salt enters the tank at a rate of liters per min, and the well stirred mixture leaves the tank at the same rate. Let Q(t) denote the amount of salt in the tank at time t. Then Q(t) is a solution of the initial value problem (a) Q (t)= Q(t), Q(0)=00 (b) Q (t)= Q(t)/00, Q(0)=0 (c) Q (t)=4 Q(t)/00, Q(0)=0 (d) Q (t)= Q(t)/00, Q(0)=0 3. Suppose that the functions e t and e 3t form a fundamental set of solutions for the differential equation y + a y + a 0 y = 0 where a 0,a are constants. Then the coefficient a 0 is (a) (b) 3 (c) (d) 4. Let Q(t) denote the amount of a certain radioactive material (measured in mg) at time t measured in years. If Q(0)=8 and Q(0)=8, find Q(5). Assume that the decay occurs naturally, and no new material is added. (a) 0 (b) 7 (c) 4 (d) 36 5. The general solution of the differential equation ty + y = t is (a) t 4 /4 +C/t (b) t /3 +C (c) t 0 e lns ds+c (d) t /3 +C/t 6. The solution of the initial value problem y = e t /y, y(0)=, is (a) y = e t lny + (implicit form) (b) e t (c) e t + 3 (d) e t/ +
7. The differential equation y 3y + 4y = e 5t can be converted to which first order system: ( ) ( )( ) x (a) (t) 0 x (t) x (t) = + ( ) 0 3/ x (t) e 5t ( ) ( )( ) x (t) 0 x (t) x (t) = + ( ) e 5t 3/ x (t) 0 ( ) ( )( ) x (t) 0 x (t) x (t) = + ( ) e 5t 4 3 x (t) 0 ( ) ( )( ) x (t) 0 x (t) x (t) = + ( ) 0 3 4 x (t) e 5t 8. Suppose that a differential equation of the form ay + by + cy = 0 where a,b,c are real constants, has the solution y(t)=e t cost. Then a second linearly independent solution is given by (a) e t cost (b) e t cost (c) 3e t sint + e t cost (d) e t sint 9. Consider the initial value problem y = t y 3, y()=. Use the Euler method taking one step to calculate an approximation of y(.3). The approximate value of y(.3) is (a).8 (b). (c).4 (d). 0. The Wronskian of the two solutions y (t)=e t and y = te t of the differential equation y 4y + 4y = 0is (a) e t (b) 4te 4t (c) e 4t + 4te 4t (d) t e 4t / te 4t /. Determine (without solving the differential equation) an interval on which the solution to the initial value problem y +ty/(t )=t, y(/)= is certain to exist: (a) (,) (b) (, ) (c) (,) (d) (, )
. The general real valued solution to y y + y = 0is (a) c + c e t + c 3 te t (b) c + c e t + c 3 te t (c) c t + c e t + c 3 te t (d) c t + c e t + c 3 te t 3. The solution to the initial value problem ( ) x = x, x(0)= ( ) 0 is ) ) ( (a) e t ) ( + e 3t ) ( ( (b) e t + e 3t ) ) ( (c) e t ( ) (d) e t ( + e 3t ( ) + e 3t ( ) 4. If x(t)=c e t + c 0 e t( ( ) ( ) 0 ) t + is the general solution of the system x 0 / = Ax, where A isa matrix with real constant coefficients, then (a) A has two distinct real eigenvalues (b) A has a repeated real eigenvalue (c) A has a pair of distinct complex conjugate eigenvalues (d) A has a repeated nonreal (complex) eigenvalue 5. A suitable form for a particular solution of the differential equation y + 3y + y = e t + e t is (a) Ae t + Bte t (b) Ate t + Be t (c) Ae t + Be t (d) Ate t + Bte t
FORM A Math 4 Common Part of Final Exam Dec 000 Instructions: Please enter your NAME, ID NUMBER, FORM designation, and CRN NUMBER on your op-scan sheet. The crn number should be written in the upper righthand box labeled Course. Do not include the course number. In the box labeled Form, write the appropriate test form letter A. Darken the appropriate circles below your ID number and Form designation. Use a # pencil; machine grading may ignore faintly marked circles. Mark your answers to the test question in row - of the op-scan sheet. You have hour to complete this part of the final exam. Your score on this part of the final exam will be the number of correct answers. Please turn in your op-scan sheet and the question sheet at the end of this part of the final exam.. The differential equation ty'' + (6cos t) y' + y= is: a) Third order. b) Constant coefficient. c) Linear. d) Homogeneous.. Consider the initial value problem y' = y t, with y () =. Using Euler s method with step size h = 0.5, it is found that y(.5) is approximately: a).5 b).5 c) 3.5 d) 8.5 3. Let yt () be the solution of a) 3 b) 35 c) 80 d) 88 ty' 4y t 5 =, with () 4 y =. Then y() equals: t + 4. Let yt () be the solution of y ' =, with y (0) =. Then y() equals: y a) b) c) /3 (8) /3 (4) /3 (6)
d) /3 (3) 5. A 500 gallon tank initially contains 300 gallons of water in which 5 pounds of salt is dissolved. Water containing /3 pounds of salt per gallon then flows into the tank at a rate of 6 gallons per minute while the well mixed solution drains at a rate of 4 gallons per minute. If Qt () represents the amount of salt in the tank at any time t before the tank overflows, then Qt () satisfies the initial value problem: 4Q a) Q' = 6, Q(0) = 5 500 4Q b) Q' =, Q(0) = 300 300 4t 4Q c) Q' =, Q(0) = 5 300+ t 4Q d) Q' = 6, Q(0) = 500 300 6. The general solution of y'' + 6 y' + 5y= 0 is: a) b) c) d) yt () = ce cos8t+ ce sin8t 3 t 3t 4 t 4t 3 t 3t 3 t 3t yt () = ce cos3t+ ce sin4t yt () = ce cos4t+ ce sin4t yt () = ce cos4t+ ce sin4t 7. A particular solution of the differential equation 3t a) yp() t = e 6 3t b) y () t = e c) d) p y () t = te p 3t yp() t = te 6 3t y'' 7 y ' 6y e 3t + + = is: 3 4 8. Without solving the initial value problem y'' + y' y t + t 5 =, with t y (3) = 0, tell which of the following is the largest interval on which the solution is certain to exist. a) (,5) b) (0, ) c) (, ) d) (,5)
9. The general solution of y (4) y = 0 is: yt () = ce + ce + c cost+ c sint a) t t 3 4 b) yt () = ct+ ce + ce cos( 3/) t + ce sin( 3/) t t t/ t/ 3 4 yt () = ce + cte + ce + cte c) t t t t 3 4 yt () = ce + ce + ce cost + ce sint t t t t d) 3 4 0. The second order equation y'' + 5 y ' + 4y = costcan be rewritten as the system: a) b) c) d) x' 0 x x ' = 4 5 x x' 0 x 0 x ' = + 5 4 x cost x' 0 x cost x ' = + 4 5 x 0 x' 0 x 0 x ' = + 4 5 x cost x' x. The general solution of x' = 4 x 3t t a) xt () = ce + ce 3t t b) xt () = ce + ce 3t t c) xt () = ce + ce 3t t d) xt () = ce + ce is:
. Consider the system x' = Ax where the eigenvalues of A are both 5. Let v = be the solution of ( A 5) Iv= 0 and w = 0 be the solution of ( A 5) Iw= v. Then the general solution of x' = Ax is: 5t 5t a) xt () ce = + ce 0 5t 5t b) xt () ce = + cte 5t 5t c) xt () = ce ce t + + 0 5t 5t d) xt () = ce ce t + + 0
Math 4 Spring 00 Common Final Exam, Spring 00. The solution to the initial value problem ty = y +t +t 3, y()=is (a) y = t /0 +t /4 +t 3 /5 (b) y = t / +t 3 +t 4 / (c) y = t +t 3 +t lnt (d) y = t +t 3 +t lnt. Consider the initial value problem y = ty+ 3, y() =4. If we use Euler s method with step size to find a very crude approximation for y(4), we obtain (a) y(4) 64 (b) y(4) 8 (c) y(4) 0 (d) y(4) 0 3. The general solution to the differential equation y = y 3 /( + 4t) has the form (a) y = ±(C + ln( + 4t)) /4 (b) y = ±/ C ln( + 4t) (c) y = ±/ C ln( + 4t) (d) y = ±/ C ln( + 4t) 4 4. The differential equation y + y = 5t + 3e t has general solution (a) y = C +C e t 5t + 5t / + 3te t (b) y = C +C e t + 5t + 3e t / (c) y = C +C e t 5t + 5t / + 3e t / (d) y = C +C e t + 5t / + 3e t 5. The general solution to y + 4y + y = t e t is (a) y = C e t +C te t e t lnt (b) y = C e t +C te t t e t / (c) y = C e t +C te t e t /t (d) y = C e t +C te t e t lnt
6. The initial value problem x = 3x (t) x (t) x = x (t) x (t) with x (0)=0, x (0)= has solution x(t)= ( e (a) /3 e ) /3 4e /3 e /3 ( e (b) /3 + e ) /3 e /3 e /3 ( 4e (c) /3 + e ) /3 e /3 + 4e /3 ( e (d) /3 e ) /3 e /3 e /3 ( ) x (t). The value of x() is equal to x (t) 7. Suppose the polynomial equation a r + a r + + a r + a 0 = 0 has solutions 0,0,,,, 3,±i, 3 ± i, 3 ± i. Then the general solution to the differential equation a y () + a y () + + a y + a 0 y = 0is (a) y = C +C t +C 3 e t +C 4 te t +C 5 t e t +C 6 e 3t +C 7 cost +C 8 sint +C 9 e 3t cost +C 0 e 3t sint +C te 3t cost +C te 3t sint (b) y = C +C t +C 3 e t +C 4 te t +C 5 t e t +C 6 e 3t +C 7 cost +C 8 sint +C 9 t cost +C 0 t sint +C te 3t +C te 3t (c) y = C +C t +C 3 e t +C 4 te t +C 5 t e t +C 6 e 3t +C 7 cost +C 8 sint +C 9 e t cos3t +C 0 e t sin3t +C te t cost +C te t sint (d) y = C +C t +C 3 e t +C 4 te t +C 5 t e t +C 6 e 3t +C 7 cost +C 8 sint +C 9 e 3t cost +C 0 e 3t sint +C e 3t cost +C e 3t sint 8. The differential equation y + y + 5y = 0 has general solution (a) y = C +C e t cost +C 3 e t sint (b) y = C +C te t cost +C 3 te t sint (c) y = C +C e t +C 3 te t (d) y = C +C e t cost +C 3 e t sint
9. Suppose A isa matrix with real entries. ( ) Suppose one eigenvalue of A is + i, and that a corresponding eigenvector is. The solution to the initial value problem i dx dt = Ax with x(0)= is ( ) (a) e t cost + 4 sint cost sint ( ) (b) e t cost + 4sint cost ( sint ) e t cost + 4 sint cost + sint ( ) (c) e t cost + sint cost sint 0. A tank initially contains 0 gallons of pure water. A salt solution containing / lb salt per gallon is then poured into the tank at 4 gallons per minute, while the well mixed solution is simultaneously drained at 3 gallons per minute. The amount of salt in the tank after 0 minutes is (a) lb (b) 63 4 lb (c) 93 8 lb (d) 5 lb. The largest interval containing the point t = 3 on which there is guaranteed to be a unique solution to the initial value problem 4(t + t 5)y + y +(t 7)y/(t )=3t cost, y( 3)=, y ( 3)=6 is the interval (a) ( 7,0) (b) ( 5,) (c) ( 5,3) (d) ( 7,5)
Math 4 Fall 00 Common Final Exam, Fall 00. Which of the following differential equations is of second order and linear? (a) y y +ty +t y = 0 (b) y +t siny + y = te t (c) y +(lnt)(y + ) t = 0 (d) (y ) y(y + )=0. The general solution of the homogeneous differential equation y y + y = 0is y h = c e x + c xe x. If a particular solution of the inhomogeneous differential equation y y + y = e x /x is sought of the form y p = ue x + vxe x and u is found to be u = x, then v can be (a) /x (b) x/e x (c) lnx (d) 3. Suppose A is a matrix with repeating eigenvalue λ = 5,5. Suppose u and v are vectors satisfying (A 5I) u = 0 and (A 5I) v = u. Then the general solution of the system differential equations Y = AY is: (a) Y = c e 5t v + c e 5t (t v + u) (b) Y = c e 5t v + c e 5t (t u + v) (c) Y = c e 5t u + c e 5t (t v + u) (d) Y = c e 5t u + c e 5t (t u + v) 4. Let y h denote the general solution of the homogeneous equation y + y y = 0 and let y p denote a particular solution to the inhomogeneous equation y + y y = e t sint. Then the general solution of the inhomogeneous equation is (a) y = y h + t + (b) y = y h + y p + t + (c) y = y h + y p + t (d) y = y h +t + y + y y = e t sint 4t
5. A solution of the initial value problem ty + y t 3 exist in the interval = tant, y() =4 is guaranteed to (a) (,4) (b) (0,3) (c) (π/,3π/) (d) (0,π/) 6. Suppose you wish a numerical approximation to the solution of the initial value problem y = y + 0t + 0, y(0)=5. If you use h = 0. and the Euler method to approximate y(0.4), the result will be (a) 8 (b) (c) 7. (d) 4.4 7. Consider the differential equation y + ysint = 0 with y(π/)=. Then y(0) is (a) e (b) e (c) e (d) e 8. The general solution to the differential equation y (4) y = 0is (a) c + c t + c 3 sint + c 4 cost (b) c ( +t)+c sint + c 3 cost (c) c + c t + c 3 e t + c 4 e t (d) c ( +t)+c e t + c 3 e t 9. Determine the coefficients b and c such that e t sint and e t cost are solutions of the differential equation y + by + cy = 0. (a) b =, c = (b) b =, c = (c) b =, c = (d) b =, c = 0. The eigenvalues of the matrix A = [ ] satisfying Y (0)= is [ ] e 4t (a) e 4t [ e 4t (b) e 4t [ ] 3 are λ =,4. The solution of Y = AY ] [ 4 e (c) 4t + 3 ] 4 e t 3 e4t 3 e t [ ] e 4t (d) e t
[ ] [ ] 3 5 5. The matrix A = has eigenvalues and eigenvectors λ = +i, v = [ ] i 5 and λ = i, v =. Then a real-valued solution of the system of differential + i equations Y = AY is [ 5e (a) t ] cost e t cost + e t sint [ ] 5e t (b) e t [ 5e (c) t ] cost e t sint [ 5e (d) t ] (cost + sint) e t (cost sint). Suppose an unusual decay law were given by dn/dt = λn, rather than the usual linear decay law. If the initial amount was N(0)=N 0, then what would be a formula for the time t for which N(t)=N 0 /? (a) t = (b) t = λln (c) t = ln (d) t = N 0 λn 0 N 0 λ
Form A Math 4 Common Part of Final Exam May 6, 00 Instructions: Please enter your NAME, ID NUMBER, Form designation, and INDEX NUMBER on your op-scan sheet. The index number should be written in the upper right-hand box labeled Course. Do not include the course number. In the box labeled Form write the appropriate test form letter A. Darken the appropriate circles below your ID number and Form designation. Use a # pencil; machine grading may ignore faintly marked circles. Mark your answers to the test questions in rows of the op-scan sheet. You have hour to complete this part of the final exam. Your score on this part of the final exam will be the number of correct answers. Please turn in your op-scan sheet and the question sheet at the end of this part of the final exam.. If y(t) satisfiesty + y = t with y() =, then y() equals (a) 5 4 (b) (c) e (d) e. If y ty 3 =0withy() =, then y satisfies (a) y = 9 t (b) y = t 3 t (c) y 3 = 4 t (d) y = t ( ) 3. Suppose A has eigenvectors and Then a possible solution to Y = AY is ( ) e (a) Y = t +e t e t +4e t ( ) 5e t (b) Y = 3e t ( ) e (c) Y = t +4te t e t + te t ( ) e (d) Y = t +4e t e t + e t ( ) 4 satisfying A ( ) = ( ), A 4 ( ) 4 = ( ) 8.
4. The general solution of y + y =0is (a) y = c + c e t + c 3 e t + c 4 te t + c 5 t e t (b) y = c + c t + c 3 cos t + c 4 sin t (c) y = c + c t + c 3 t + c 4 cos t + c 5 sin t (d) y = c + c t + c 3 t + c 4 e t + c 5 te t 5. Consider the inhomogeneous differential equation y 3y +y = +e t.ifyoutryto find a particular solution of the form y p = ue t + ve t,thenu and v may satisfy: (a) u = e t +e t, v = e t +e t (b) u = et +e t, v = et +e t (c) u = +et, v = +et e 3t e 3t (d) u = ( + e t )e 3t, v = ( + e t )e 3t 6. A particular solution of y 4y +4y = te t + t may be found of the following form (where the K i have yet to be determined): (a) y p = K t + K t + K 3 t 3 e t + K 4 t e t (b) y p = K t + K t e t + K 3 te t (c) y p = K + K t + K 3 e t + K 4 te t (d) y p = K t + K te t + K 3 e t
7. The differential equation u +u u +3u =0 can be transformed to the first order linearsystemoftheformx = AX, where the matrix A is given by (a) 0 0 0 0 3 (b) 0 0 0 0 3 3 (c) 0 0 0 0 (d) 0 0 0 0 3 8. Which of the following differential equations are linear? (A) y y =0 (B) ( x )yy = x (C) y e x y +(+x )y =tan x (D) y y +3( y )=x + y (a) Only (C) (b) Only (A) and (C) (c) Only (C) and (D) (d) None of the differential equations 3
9. The largest interval on which the initial value problem y = t y unique solution is the interval with y() = has a (a) (, ) (b) (0, ) (c) ( 3, ) (d) (, 3) ( ) 9 0. Let A =. You are given that λ =4 is an eigenvalue for A. Then a corresponding eigenvector is ( ) ( ) ( ) ( ) 3 (a) (b) (c) (d) 3 3 9. Suppose N grams of a radioactive material decays according to the usual decay law N = kn, wherek is the decay rate. If N = N() is the number of grams of undecayed material after time unit, and N 0 = N(0) is the number of grams of undecayed material after 0 time units, then a formula for the decay rate k is (a) k = ln 0 (b) k = ln N N 0 (c) k = 9+ln N 0 N (d) k = 9 ln( N 0 N ). Suppose you decide to approximate the solution of the initial value problem y = y +4t, y(0) =, using the Improved Euler Method with step size h =. Then your approximate value for y() will be (a) y() 3 (b) y() 5 (c) y() 9 (d) y() 47 4 4
Math 4 Fall 00 Common Final Exam, Fall 00. Suppose that y(t)=te t is a solution of y ty +ty = αe t. Then α is equal to () 0 () (3) (4). Let y(t) be a solution of y + 3y = e t, y(0)=0. Find the value of y(). () 4 e 4 e 3 () 4 e 4 e /3 (3) 3 e 3 e (4) 4 e3 4 e 3 3. Let y(x) be the solution of y = xy 3, y()=. Find the value of y(0). () 3 () / 3 (3) (4) 3 3 4. A tank initially contains 00 gal of water with 0 lb of salt. Fresh water is entering the tank at a rate of gal/min, and the well-stirred mixture is leaving the tank at the same rate. How many minutes will pass before the total amount of salt in the tank reduces to 5 lb? 5. Let y(t) be the solution of Then as t, () 5 () 5e.0 (3) 50ln (4) 00ln y + y 6y = 0, y(0)=, y (0)=. () y(t). () y(t) 0. (3) y(t). (4) the limit of y(t) can not be determined from the information given. 6. Find the largest interval in which the following initial value problem is certain to have a unique solution. (x + )y + xy + ytanx = 0, y(3)=π, y (3)=. () (,π) () (,π/) (3) (,π/) (4) (π/,3π/)
7. Noting that y (t)=e t is a solution of ty + ( t)y +(t )y = 0, if you try to find a second solution y in the form y = e t v(t), which of the following equations must be satisfied by v? () v +tv + v = 0 () tv + ( t)v +(t )v = 0 (3) tv + v = 0 (4) tv + v = 0 8. According to the method of undetermined coefficients, find the correct form of a particular solution of y y + y = e t + e t sint. () At e t + Be t sint +Ce t cost () (A 0 t + A t + A )e t + Be t sint +Ce t cost (3) (A 0 t + A t + A )e t +(B 0 + B t)e t sint +(C 0 +C t)e t cost (4) At e t + Be t sint 9. Find the general solution of y + y + y = 0. () C +C e t +C 3 te t () C t 3 e t +C t e t +C 3 te t +C 4 e t (3) C t 3 +C t +C 3 t +C 4 (4) C +C t +C 3 e t +C 4 te t 0. Let A bea ( matrix ) with real entries. Suppose that + i is an eigenvalue of A with i an eigenvector. Find the solution of ( e () t sint + e t ) cost e t cost e t sint ( e () t sint + e t ) cost e t cost + e t sint ( e (3) t sint + e t ) cost e t cost e t sint ( e (4) t cost e t ) sint e t cost dx dt = Ax, x(0)= ( ).
. Let A bea matrix and let I bea identity matrix. Suppose that ( ) ( ) ( ) ( ) 0 0 (A I) = (A I) =. 0 0 0 Find the general solution of dx dt = Ax. ( ) () C e t ( ( ) ( ) +C 0 te t + e 0 t 0 ) ( ) () C e t ( ( ) ( ) +C 0 te t + e 0 t ) 0 ) ( ) ) ) +C ( (3) C e t 0 ( ) (4) C e t 0 ( te t 0 ) +C e t ( 0 + e t ( 0
Math 4 Spring 003 Common Final Exam, Spring 003. The function y = C e t +C te t +C 3 t e t is the general solution of (a) y y = 0 (b) y y + y = e t (c) y + y = 0 (d) y 3y + 3y y = 0. The general solution of y 4y = e t is (a) e t /4 +C e 4t +C (b) te t /4 +C e t +C e t (c) e t /4 +C t +C (d) C e t +C e 4t +C 3 0 0 0 3. Let A = 0 0. Then the number of linearly independent eigenvectors for the 4 0 0 eigenvalue 0 is (a) 0 (b) (c) (d) 3 4. For the initial value problem y = ty+ t, y() =, take two steps with the Euler method and step size 0.. The resulting approximation for y(.) is (a).674 (b).0 (c).3 (d).766
5. The solution of the initial value problem (t )y + e t y +(tant)y = 0, y()=3, y ()=4 is guaranteed to exist on the interval (a) (,π/) (b) (, ) (c) (,π) (d) (π/, 3π/) 6. The third order equation y y = t corresponds to the system (a) y = y +t, y = y, y 3 = y (b) y = y, y = y 3, y 3 = y +t (c) y = y, y = y 3, y 3 = y 3 +t (d) y = y 3 +t, y = y 3, y 3 = y 7. Let y(t) be the solution of t y ty = t, y()=. Then y(3) is equal to (a) 8 (b) /3 (c) (d) 5 8. Two water tanks hold 00 gallons each. Water flows into tank at a rate of 3 gallons per minute, and water drains from tank into tank at a rate of gallons per minute and into the environment at a rate of gallon per minute. Tank is fed only from tank and it drains into the environment at the rate of gallons per minute. Both tanks hold initially fresh water, and the water entering tank contains salt at a concentration of 5 ounces per gallon. With Q and Q denoting the salt content of tanks and in ounces, the differential equation modeling the system is (a) Q = 5 3Q, Q = Q Q, Q (0)=Q (0)=0. (b) Q = 5 0.03Q, Q = 0.0Q 0.0Q, Q (0)=Q (0)=0. (c) Q = 0.03Q, Q = 0.0Q 0.0Q, Q (0)=5, Q (0)=0. (d) Q = 5 0.0Q, Q = Q 0.0Q, Q (0)=Q (0)=0.
9. The general solution of t y + ty y = 0isy = C t +C /t. A particular solution of t y + ty y = /t is (a) t/9 +t lnt/3 (b) t 3 / (c) t +t (d) t / 0. Which of the following is not a linear equation: (a) y + y + y = 0 (b) y e t y = sint (c) y + yy + y = 0 (d) y + y = ty. The general solution of y = y t has the form (a) y = ± e t3 /3+C (b) y = y 3 t /3 +C (c) y = 3/(t 3 +C) (d) y = C/t ( ) ( )( ) x 5 x. The general solution of the system y = is 5 y ( ) ( ) ( ) x (a) = C y e 5t 0 +C e t 0 ( ) x ( ( ) ( ) 5 ) ( ) 5 (b) = C y t + +C 0 ( ) ( ) ( ) x 5 (c) = C y +C e 7t ( ) ( ) ( ) x 5 (d) = C y +C e 7t 5
FORM A Math 4 Common Part of Final Exam Fall 003 Instruction: Please enter your NAME, ID NUMBER, FORM designation, and CRN NUMBER on your op-scan sheet. The CRN NUMBER should be written in the upper right-hand box labeled Course. Do not include the course number. In the box labeled Form, write the appropriate test form letter A. Darken the appropriate circles below your ID number and Form designation. Use a # pencil; machine grading may ignore faintly marked circles. Mark your answers to the test question in row - of the op-scan sheet. You have hour to complete this part of the final exam. Your score on this part of the final exam will be the number of correct answers. Please turn in your op-scan sheet and the question sheet at the end of this part of the final exam.. Consider the initial value problem y tan( t) y ln( t), y(). Find the largest interval in which a solution is guaranteed to exist. (a) (, ) (b) ( 0,) (c) ( /, 3 / ) (d) (,3 / ). Let A be a ( ) constant matrix such that possible solution of the system y Ay is: 4 8 A and A 4. One t t t t 4e e e e (a) (b) (c) t t t t e e 4e 3e e t t e (d) e 8te 4e te t t t t 3. Let yt () solve the initial value problem yy, t y(0). What is the value y()? (a) y() 3 (b) y () 5 (c) y() e (d) y() e 4. A tank initially contains 000 gal of water in which is dissolved 0 lb of salt. A valve is opened at time t = 0 and water containing 0. lb of salt per gallon flows into the tank at a rate of 5 gal/min. The mixture in the tank is well stirred and drains from the tank at the same rate of 5 gal/min. Determine the time t, in minutes, that the tank contains 80 lb of salt. 80/ 00 00/80 (a) t e (b) t e (c) t 00ln 9 (d) t ln(80 / 00)
5. Let yt () solve the initial value problem 4 6, () ty y t y (a) y() 4 (b) y () e (c) y() e (d) y() ln. What is y()? 6. Suppose you decide to approximate the solution of ty y 4 t, y() 4 using Euler s method with step size h /. Your approximate value for y() is: (a) 5 (b) 6 7 (c) 6 5 3 (d) 3 7. Let A be a ( ) constant matrix with real entries such that The solution of 5 y Ay, y (0) is: i i A (3 i). 3t 5cost 5sint (a) e (b) e cost 3sint 3t 5cost sint cost 3sint 3t 5cost 5sint (c) e cost sint 3t 5cost sint (d) e cost sint 8. The solution of the following initial value problem is y'''' y''' 3 y'' 0, y(0), y'(0), y''(0) 0, y'''(0) 0 3t (a) t e e t t (b) e e 3t (c) t (d) 3t t (/ 6) (/) e (3/ 4) e 9. One solution of the nd order linear differential equation is y t t y'' ty' y 0, ( t 0),. Which of the following is the general solution? 3 (a) y ct c t (b) y ct c t (c) y ct c t (d) y ct c t
0. Given the nd order nonhomogeneous differential equation t t y'' y' 5y e sin( t) te 4t, the suitable form for a particular solution Y() t, using the method of undetermined coefficients, is (a) (b) (c) (d) t t Ae sin t Bte Ct t t t Ae cos t Be sin t ( C Dt) e Et Ft G t t t Ate cos t Bte sin t ( C Dt) e Et Ft G t t t Ae cos t Be sin t t( C Dt) e ( Et Ft G). The general solution of the nd order homogeneous equation t y '' y 0 is y ct ct. A particular solution Y() t for the corresponding nonhomogeneous equation t y'' y t is (a) Yt () ( /3) t ln t (/9) t (b) Yt () ( /5)t (c) Yt () 4t (d) Yt ( ) (/ 3) t (/ 8) t (/ 6) 4. Given an initial value problem 9 y'' y 0, y(0) 6, y'(0), find so that the solution approaches zero as t. (a) (b) (c) (d) / 3