APPM 23: Final Exam :3am :pm, May, 25. ON THE FRONT OF YOUR BLUEBOOK write: ) your name, 2) your student ID number, 3) lecture section, 4) your instructor s name, and 5) a grading table for eight questions. Text books, class notes, and calculators are NOT permitted. A one-page two-sided crib sheet is allowed. In acknowledgment of attending the final exam, please sign and return your exam paper in your BLUEBOOK. Problem 3 points): i) 5 points) Solve the following ODE ty = t + 2)y + t 3 e t ii) 5 points) Consider the nonlinear ODE y = y2 t 2 yt Using the v-substitution v = y/t) find the first order ODE satisfied by v. DO NOT solve the ODE. i) Normalize y + 2 ) y = t 2 e t t Integrating factors µ = exp + 2/t) = t 2 e t. Thus d t 2 e t yt) ) = yt) = t 2 e t C + t) ii) v = y/t implies y = vt and y = v t + v Thus v t + v = v v = v tv. Problem 2 3 points): Alice is a graduate student in chemistry. In her lab she finds a tank that contains gallons of chemical solution, which she labels as solution A. The chemical concentration of solution A is measured to be 2 pounds per gallon. She also finds plenty of chemical solution with chemical concentration 4 pounds per gallon, which she labels as solution B. To achieve the chemical concentration that she desires, she injects solution B into solution A at a flow rate of 3 gallons per hour, stirs the mixture uniformly, and drains the mixture out at a flow rate of 3 gallons per hour. i) 8 points) Set up the initial value problem for the mass of the chemical in the mixture as a function of time. ii) 2 points) Solve the initial value problem. iii) points) Suppose that instead of draining the mixture at 3 gallons per hour, Alice evaporates the mixture at 3 gallons per hour. In this process, water escapes from the mixture while the chemical remains. In this case, what is the mass of the chemical in the mixture as a function of time? iv) 4 points) To reach a chemical concentration of 3 pounds per gallon, is it more efficient to evaporate the mixture or to drain the mixture? You do not need to justify your answer.
i) Let the chemical mass as a function of time be xt). Then the differential equation is x t) = rate in) rate out) = 4lb/gal) 3gal/hr) x lb/gal) 3gal/hr) = 2 3xt) lb/hr). The initial condition is x) = 2lb/gal) gal) = 2lb). ii) To solve the differential equation, we rewrite it in standard form The homogeneous solution is x t) + 3 xt) = 2. x h t) + 3 x ht) = x h t) = ce 3 t. A particular solution is, by inspection Therefore the general solution is Using the initial condition, we have x p t) = 4. xt) = x h t) + x p t) = ce 3 t + 4. c + 4 = 2 c = 2. Hence the solution to the initial value problem is xt) = 2e 3 t + 4. iii) In this case the concentration of the outgoing flow is, so the initial value problem becomes x t) = 2, x) = 2. The solution is then xt) = 2 + 2t. iv) It is more efficient to evaporate the mixture. This is because the rate at which the chemical mass increases is always greater for evaporation than drainage. Problem 3 3 points): Consider the second order non-homogeneous differential equation: x )y xy + y = x ) 2. Two solutions to the corresponding homogeneous problem are y x) = x and y 2 x) = e x. a) 4 pts) Write the corresponding homogeneous equation. Verify that y x) and y 2 x) solve the homogeneous equation. b) 4 pts) Show that y x) and y 2 x) are linearly independent functions on x R. c) 2 pts) Using the method of variation of parameters, find a particular solution to the non-homogeneous equation. d) 5 pts) Write the general solution to the non-homogeneous equation. e) 5 pts) Obtain the solution to the non-homogeneous equation with the initial condition y) =, y ) =.
a) The homogeneous equation is: ) x )y xy + y =. For the first solution: y x) = x, y =, y =. Plugging into??), we get x + x =. For the second solution, y 2 x) = e x, y 2 = ex, y 2 = ex. Plugging into??), we get x )e x xe x + e x = x )e x + x)e x =. b) To show y x) = x and y 2 x) = e x are linearly independent we compute their Wronskian. Both functions are defined and have continuous derivatives for all x R. It is enough to show that the Wronksian determinant is nonzero at one particular value of x. We take x =. Then: W y, y 2 ) = x ex e x = xex e x = = c) We assume y p x) = v x)y x)+v 2 x)y 2 x). We rewrite ) in standard form assume x, can check at end that x = works in solution): 2) y x x y + y = x ) x Hence, fx) = x ). We already have the Wronskian of y and y 2 which is e x x ). The equations for v and v 2 give using integration by parts for v 2): v = y 2f W y, y 2 ) = ex x ) e x x ) = = v x) = x v 2 = y f xx ) = W y, y 2 ) e x x ) = xe x Hence, the particular solution is: d) The general solution is: = v 2 x) = x + )e x y p x) = x) x x + )e x ) e x = x 2 x 3) yx) = y h x) + y p x) = C x + C 2 e x x 2 x e) We have y x) = C + C 2 e x 2x. The first IC gives y) = C 2 = = C 2 = 2. The second IC gives y ) = C + C 2 = = C = C 2 = 2 =. So the solution is: 4) yx) = x + 2e x x 2 x = 2e x x 2 2x Problem 4 3 points) i) Write down the form of the particular solution for the following differential equations using the method of undetermined coefficients. Do not solve for the coefficients explicitly. i) 5 points) y 3y = 3e 2t ; ii) 5 points) y 2y + y = te t + t; iii) 5 points) y + y = 2 sin t. ii) For certain linear inhomogeneous differential equations in applications, as time goes to infinity, the homogeneous solution decays to zero, and what remains is the particular solution known as the steady state. Determine the steady states of the following problems. i) 5 points) Cooling: T = 2T 5); ii) points) Forced oscillation: 2x + 4x + 2x = 2 cos t.
i) i) Ae 2t ; ii) t 2 At + B)e t + Ct + D iii) ta cos t + B sin t). ii) i) Rewrite in standard form: 5) T + 2T =. MUC guess: ) T p = A. Solve for coefficients: 7) 2A = A = 5. Steady state: 8) T p = 5. ii) Rewrite in standard form: 9) x + 2x + x = cos t. MUC guess: ) x p = A cos t + B sin t. Solve for coefficients: ) A + 2B + A =, B 2A + B = B = 2, A =. Steady state: 2) x p = sin t. 2 Problem 5 35 points): Answer the following questions. Each question is worth 5 points. Only your final answers will be considered no partial credit will be awarded for work. a) For the system of differential equations dx = +y 2 dy ye+x2 = +y 2 xe+x2, i) Find all equilibrium points. ii) Find an implicit representation of the phase-plane trajectories. iii) What do you think your answer to part ii) says about the non-equilibrium solutions of the system? b) The system of differential equations dx = y xx2 + y 2 2) x 2 + y 2 dy = x yx2 + y 2 2) x 2 + y 2 has one equilibrium point at, ) it s unstable) and an attracting limit cycle x 2 + y 2 = 2 you do not have to verify any of this take it as given). i) What is the direction of motion of solution trajectories as they spiral into the limit cycle? ii) Make a rough sketch of the phase plane that includes only the limit cycle and two solution trajectories one that starts inside the limit cycle but not at the origin) and one that starts
outside the limit cycle. You do not have to find nullclines or a direction field. c) i) Find a solution of the IVP y = t y 2, y) = other than equilibrium solution d yt) =. HINT: Recall that du [arcsinu)] = u 2 ii) Does your answer violate Picard s Theorem? Answer YES or NO. iii) Briefly explain your answer to part ii). d) For m a positive constant, find the general solution of y my + m 2 y m 3 y = HINT: factor by grouping. e) TRUE or FALSE? A figure 8 that is, a curve that looks like the infinity symbol ) could be a phase-plane solution trajectory of a system of differential equations dx = fx, y) dy = gx, y) where f and g have continuous partial derivatives with respect to x and y. f) TRUE or FALSE? Every homogeneous linear differential equation with constant coefficients and purely imaginary roots of its characteristic equation must have periodic solutions. g) TRUE OR FALSE? For every k >, the equilibrium solution xt) = of the nonlinear van der Pol equation ẍ + kx 2 )ẋ + x = is unstable. a) i) e +x2 +y 2 is never zero, so, ) is the only equilibrium point. ii) Use dy dx = dy/ dx/ = x/y separable) = x2 + y 2 = C, C >. iii) They re probably periodic since solution trajectories are circles. b) i) Set y = in dy dy and note that for x > we have >. So COUNTERCLOCKWISE since x >, y = is on the positive x-axis in the phase plane and trajectories must be heading roughly upwards there. ii)
c) i) It s separable, so... yt) = sint 2 /2). ii) NO iii) With ft, y) = t y 2, f y is not continuous in any rectangle including y =, so unique solutions aren t guaranteed by Picard s. d) The characteristic equation is r 3 mr 2 + m 2 r m 3 = r 2 r m) + m 2 r m) = r 2 + m 2 )r m) = when r = m, ±im. So the general solution is for c, c 2, c 3 R. yt) = c e mt + c 2 cosmt) + c 3 sinmt) e) FALSE this would violate an existence/uniqueness theorem. f) FALSE consider y 4) + 2y + y =. The characteristic equation is r 2 + ) 2 = which has repeated roots ±i, so the general solution is yt) = c + c 2 t) cost) + c 3 + c 4 t) sint), which will, in general, increase in amplitude as t. g) TRUE. Convert to the system note the Jacobian is ẋ = y ẏ = x kx 2 )y, [ ] J, ) = k with characteristic polynomial λ 2 kλ + and hence eigenvalues λ = k ± k 2 4. When 2 < k < 2 we have complex eigenvalues with positive real part k/2, when k = 2 we have a repeated real eigenvalue k/2, and when k > 2 we have distinct real eigenvalues at least one of which is definitely positive. In all cases the linearized system predicts an unstable equilibrium point for the original system. Problem 35 points): Answer TRUE or FALSE. You do NOT need to justify your answer. Only write TRUE if the statement is always true.
i) Consider the two 5 5 real valued matrices A and B. If the vector v is an eigenvector of A with eigenvalue λ A, as well as an eigenvector of B with eigenvalue λ B. Then matrix C = A + B has an eigenvalue λ A + λ B ) and matrix D = AB has an eigenvalue λ A λ B. ii) Consider the vectors v, v 2, v 3 R 3 and the square matrix A formed by using the vectors as columns. If det A) =, then dim span {v, v 2, v 3 }) < 3 iii) Consider a 4 4 matrix A. If {v, v 2, v 3, v 4 } forms a basis of ColA, then the matrix A is invertible. iv) Consider the two invertible n n matrices A and B. If B A C = I and CB A = I, then the n n matrix C is also invertible. v) Consider the three invertible n n matrices A, B and C, then AB) C) T = C A ) T B ) T. vi) The set of all solutions yt) to the second order differential equation y + sint)y + t 3 y = forms a vector space. vii) Consider the n n real valued upper triangular matrix A, if deta, then all eigenvalues are nonzero. viii) The set of functions {, x, 2x 2, 4x 3 3x } is linearly independent. i) True ii) True iii) True iv) True v) True vi) True vii) True viii) True Problem 7 points): Consider the following system of differential equations: ẋt) = 2 2 xt) i) Find three linearly independent solutions x t), x 2 t), x 3 t) ii) Find the general solution xt) iii) Find the particular solution for the initial condition x) = i) We start by computing the eigenvalues. The characteristic polynomial is given by λ 2 λ 2 5 λ = λ ) λ 2 2 λ = λ )λ2 +2λ 3) = λ ) 2 λ+3) = We obtain a repeated eigenvalue λ,2 = and λ 3 = 3.
We compute the eigenvector for λ,2 : 2 2 2 2 2 2 2 2 2 We get a single eigenvector v,2 =. We compute the eigenvector for λ 3 : 2 2 4 2 2 2 2 2 2 2 We get an eigenvector v 3 =. 2 2 2 2 To get the third linearly independent eigenvalue, we need to compute a generalized eigenvalue u for λ,2. 2 2 2 2 2 2 2 2 We get a generalized eigenvector u = s + We can now write the three solutions: x t) = e t v,2 = e t x 2 t) = e t tv,2 + u) = e t t + x 3 t) = e 3t v 3 = e 3t c ii) The general solution is xt) = c x t) + c 2 x 2 t) + c 3 x 3 t) iii) To find the particular solution, we need to solve the linear system + c 2 = We compute the RREF: 2 + c 3 2
We get the three coefficients: c =, c 2 =, c 3 = and the particular solution is given by: xt) = x 2 t) = e t t + Problem 8 points): Consider the following system of differential equations: ẋ = y ẏ = y + x x 2 i) Find all equilibrium points. ii) Determine the type of all equilibrium points. iii) Sketch the local behavior about the point, ). iv) Using the knowledge gained from ii) and iii). Sketch the phase plane. Vector field is given as [y, y + x x 2 ]. i) Equlibrium points, ),, ) ii) Classify using Jacobian matrix 2x ) ),) ), ) is a saddle node it has T r =, Det = given λ 2 + λ =, ) is a spiral sink it has T r =, Det = given λ 2 + λ + = iii) local vector field about,) is given by [v, u v] along the line v = this this [, u] for a clockwise spiral sink. iv) Spiral sink contains all information required to sketch the phase portrait.,)