Polytechnic Institute of NYU MA Final Practice Answers Fall Studying from past or sample exams is NOT recommended. If you do, it should be only AFTER you know how to do all of the homework and worksheet problems. These are additional practice problems designed to cover the material, but not necessarily specific to the exam. The final is cumulative, covering all material in the course. new material. In general, 75-8% of the exam is Consider the initial value problem P = te t P, y =. a Use the Euler Method with stepsize h =.5 to find the approximate value for the solution of the above problem at x =.5. We will need.5.5 = iterations to get the answer. n t n = t + n h y n ft n, P n = te t P P n+ = P n + h ft n, P n e = P = +.5 =.5.5 e.5 =.66 P = +.5.66 =.665.5.665 n/a n/a b Find the exact solution and determine its value at x =.5. Separable: e P dp = te t dt so P = ln.5 e t + C Using the IVP,, = ln.5 + C =.5 + C C =.5. So, P = ln.5 e t +.5 and P.5 =.8. The approximation is poor due to the relatively large stepsize. Consider the initial value problem x y = xy x, y =. a Use the Euler Method with stepsize h =. to find the approximate value for the solution of the above problem at x =.. n x n y n fx n, y n = y/x x y n+ = y n + h fx n, y n / = 5 y = +.5 =.5 +. =..5 5.66 y =.5 +.5.66 = 4.66. 4.66 5.56 y = 4.66 +.5.56 = 4.577. 4.577 n/a n/a
b Find the exact solution and determine its value at x =.. First order linear: You must reformulate the problem as y = y/x x as in the fx n, y n column above. The integrating factor is e /x dx = e ln x = x. Then x y x y = x. x y = ln x + C y = x ln x + Cx. Using the IVP,, C =, so y = x ln x + x. Then y. = 5.54. Consider the initial value problem xy = ylny/x +, y = e. a Use the Euler Method with stepsize h =. to find the approximate value for the solution of the above problem at x =.. n x n y n fx n, y n = y/x lny/x + y n+ = y n + h fx n, y n e e lne + = e y = e +. e =. e =.69..69 6.888 y =.888..888 n/a n/a b Find the exact solution and determine its value at x =.. Homogeneous: Use y = vx substitution to get v x + v = vlnv +. Note that with the given IVP, x in this problem. Then v = v lnv x dv v lnv = dx x lnlnv = lnx + C lnv = xec = kx for k = e C. So, v = e kx y = xe kx. Using the IVP, e, k =, and y = xe x. Thus, y. =.984. 4 Consider the d.e. y e x + xe y + cosxdx + x e y y e x sinπydy =. a Show that the equation is exact. M y = y e x +xe y and N x = xey +y e x. Since they are equal, the d.e. is exact. b Find the implicit solution to the d.e. F x, y = y e x + xe y + cosx dx = y e x + x e y + sinx + gy F y = y e x + x e y + g y = x e y y e x sinπy, so g y = sinπy gy = /π cosπy F x, y = y e x + x e y + sinx + cosπy/π = C c Solve for the particular solution satisfying y =. F, = + + /pi = π + /π = C, so F x, y = y e x + x e y + sinx + cosπyπ = π + /π
5 Find the explicit solution to the d.e. x y = x y xy, x >. Bernoulli: Use the substitution z = y = y, z = y y. Then z = x + z/x. The integrating factor is e /x dx = e ln x = x, so x z x z = x. x z = ln x + C z = x ln x + Cx y = x ln x + Cx y = x ln x + Cx. 6 Find the general solution to the d.e. y 6y + y = e x + 5x. First solve the homogeneous linear d.e. y 6y + y =. r 6r +r = rr 6r+ = r =, ±i y H = c e x +c e x cosx+c e x sinx = c + c e x cosx + c e x sinx. Since the d.e. is third order, we cannot use the method of variation of parameters unless we convert the d.e. to a system, and use variation of parameters for systems, so we must use the method of Undetermined Coefficients. Then y p = Ax + Bx + Cx + De x. The extra power of x is due to the already used homogeneous solution e x =. y p = Ax + Bx + C + De x ; y p = 6Ax + B + 9De x ; y p = 6A + 7De x Substituting into the original d.e., Ax + B 6Ax + C B + 6A + De x = e x + 5x, leading to: D = D = 4, 6A = 5 A = 5, B 6A = B =, C B + 6A = C = 5. Thus, y p = 5 x + x + 5 x + 4ex. Then the general solution is y = y H + y p = c + c e cosx + c e x sinx + 5 x + x + 5 x + 4ex. 7 Find the general solution to the d.e. d y dx 4 dy dx = e x + x cosx. Solve the homogeneous linear d.e. to get y H = c + c e x + c e x. Then y p = Axe x + Bx + Cx + D cosx + E sinx, as e x and e x = are both solutions to the homogeneous d.e. Substituting into the original d.e., Ae x 8Axe x + D sinx E cosx 4Ae x Axe x +Bx+C D sinx+e cosx = 8Ae x 8Bx 4C + 5D sinx 5E cosx, so A = /8, B = /4, C =, D =, E = /5. Thus, y p =.75xe x.5x +. sinx and y = c +c e x +c e x +.75xe x.5x +. sinx. 8 Find the general solution to the d.e. x y + xy y = x lnx, x >. Solve the homogeneous Cauchy-Euler d.e., x y +xy y =, by solving the auxiliary equation mm +m = to get m = ±. Thus, the homogeneous solution is y H = c x+ c x. In order to use variation of parameters, divide through by x, to get fx = x lnx. Then, with y p = v x + v, applying the v. of p. equations, get: x v x + v x = and v v x = lnx which result in v = x x lnx and v = lnx x. Using the given integration tables, specifically III- n= or integration by parts, get v = x lnx 4 x and using u-substitution, get v = 4 lnx to get the particular solution: y p = 4 x lnx 4 x lnx + 8 x and the general solution y = c x + c x + 4 x lnx 4 x lnx + 8 x x
4 9 Solve the d.e. x y 5xy + 9y =, x >. Cauchy-Euler: auxiliary equation mm 5m + 9 = ; solution y = C x + C x ln x. Consider the 5th order differential equation d 5 y dt 5 d y dt + d y 4y = e t dt a Express the equation as a 5-dimensional system of linear differential equations. First you must divide through by, so the leading coefficient is. Then solve for y 5 = y y + 4 y + e t. u = u u = u u = u 4 u 4 = u 5 u 5 = u 4 u + 4 u + e t b Rewrite the system in the matrix form x = A x + b. u u u u u = 4 u u u 4 + u 4 5 u 5 e t Write down a fundamental set of solutions to each of the following. a d y dx = r = r =,, {, t, t } b d y dx dy dx = r = rr + r = r =,, {, e t, e t } c d y dx d y dx = r r = r r = r =,, {, t, e t } d d y dx y = r = r r + r + = r =, r = ± i {e t, e t/ cos t/, e t/ sin t/} Solve the given initial value problem. X = X, X = 7 5 4 5
5 X = 6 e t + e t e 4t 8 For each of the following, express the general solution of the given system of equations in terms of real-valued functions. a b c X = X 5 cost sint X = C + C cost + sint sint cost X = X X = C e t + C e t cost + C e t sint sint cost X = X 4 X = C e t + C e t + C e t 4 For each of the following, express the general solution of the given system of equations in terms of real-valued functions. a X = X λ = with v = ; λ = with v =. But the algebraic multiplicity of λ = is, so we must find a generalized eigenvector. Using A + I w = v, we find that w =, so our new vector is t v + w. X = C e t + C e t + C e t t + t t
6 b 4 X = 4 X t 5 X = C e 4t + C e 4t + C e t 9 + C 4e t 9 5 Consider the following system of first order equations: x = 6x x x = 5x + x a Find the solutions to the characteristic equation of the matrix of this system. λ = 4 ± i with associated eigenvector v = i b Find formulas for x t and x t. x = e 4t cost cost sint x = e 4t sint sint + cost So, x t = c e 4t cost + c e 4t sint x t = c e 4t cost sint + c e 4t sint + cost c Find a solution of the system of differential equations that satisfies the initial conditions x =, x =. c =, c = X = e 4t cost cost sint 6 Find the general solution to the following system of first order equations: x = x + x x = 4x + x x = c e t + c t + e t
7 7 Solve the given initial value problem. x = x, x = 5 x = e t + t + e t + e t = t e t + e t t 8 A fundamental matrix for a dimensional homogeneous linear system is Φt = et e t e t e t e t. e t e t e t a Find the general solution Xt of the system. Xt = C e t + C e t + C e t b Find the particular solution satisfying X =. so C =, C =, C = and Xt = e t e t + e t 9 A system is of the form x = A x + ft; e t e ϕ = t e t e t, and the particular solution to e t the system is x p = e t. a The general solution to the system is: x = c e t + c e t + e t
8 b If the initial value of the system is x = 4, find the solution to the IVP. Substituting for t, the system of equations becomes: x = c + c + = x = c c + = 4 which leads to and c = 7, c =. Then x = 7e t + e t + e t Consider the system of equations x = A x. If λ = + i is an eigenvalue of A with + i corresponding eigenvector v =, express the solution to the system of equations in terms of real-valued functions. The real part of λ is, and the imaginary part is. The real part of the eigenvector is p = and the imaginary part is q =. [ ] [ ] So the solution is x = c e t cost sint +c e t cost + sint = c e t cost sint + c cost e t cost + sint. sint Consider the system of equations x = x + x x = x. The system has a repeated eigenvalue of, and x = e t is one solution to the system. Find the second linearly independent solution to the system, x. The first solution is x = e λt v where v is the solution to A λi v =. Here, v =. The second solution is x = e λt [t v + w] where w is the solution to A λi w = v. So, solve b the matrix to get w = where b is a free variable corresponding b to the second column of the matrix. If b =, w = t + x = e [t t + = e ] t. t, and
9 Find a formula for x and x. From the previous solution, x = c e t x = c e t + c e t t + x = c e t c te t. t + + c e t. Then, t Find the general solution to x = x = c + c e t + x + t t + t e t + t t + e t Solve the IVP for the system of first order equations, to find formulas for x t and x t. x = x + x + e t, x = x x +, x = 6, x = 6 This is the system x e t = x +, x = 6 6 The homogeneous solution is x = c e t + c e t. The nonhomogeneous solution is te t + e t + te t e t + 4. Then x = c e t + c e t + te t + e t + te t e t + 4. As to the IVP: x = c + c + 5 6 = 6 x = c c + 6 = 6 which leads to and c = c =. Then x = e t e t + te t + e t + te t e t + 4 and x t = e t + te t + x t = e t + e t + te t +
4 Consider the system x = A x + ft where 8 A = 4, ft = 6 e 4t. e t 4e 4t e 4t, find the particular solution to the system. If ϕ = e First find ϕ by finding detϕ = e 6t so ϕ = e 6t 4t 4e 4t e t 4e e t = t e 4t. Next, find ϕ f 6e t 8e t 8e =, then integrate to get t 4e t. Last, left- t multiply the previous vector by ϕ to get x p = 8 4e 4t + 8te 4t te 4t. Write the general solution to the system. 4 8 4e x = c e t + c e 4t + 4t + 8te 4t te 4t. 5 The eigenvalues of the matrix A are and, and two solutions of the homogeneous linear system x = Ax are x = + et and x = + t e t. Fill in the blanks. You do not have to explain your answers. a The algebraic multiplicity of λ = is, its geometric multiplicity is, and the eigenvector is v = x = e t and the generalized eigenvector w = x = e t t +. b The algebraic multiplicity of λ = is, and its geometric multiplicity is, and the eigenvector is e t. c The determinant of the matrix A is =.
6 Determine which of the following functions are solutions of the differential equation The solutions are a yt = t + t y ty + y = and e yt = + t + t 7 Consider the matrix 9 A = 4. Determine which sets of vectors form a fundamental set of solutions for the differential equation x = A x. The solutions are sets c y = e t, y = e t 6, y = e t 6t t and e y = e t, y = e t t, y = e t t.