Download Exam 3: Solved Problems on Differential Equations and Laplace Transforms - Prof. William A and more Exams Mathematics in PDF only on Docsity! Name: Exam 3 Instructions. Answer each of the questions on your own paper. Put your name on each page of your paper. Be sure to show your work so that partial credit can be adequately assessed. Credit will not be given for answers (even correct ones) without supporting work. A copy of the Table of Laplace transforms from the text will be provided. 1. [16 Points] Solve: 2t2y′′ + 7ty′ − 3y = 0. 2. [16 Points] Solve: y′′ + y′ − 6y = 3e2t. 3. [20 Points] Find the unique solution of the initial value problem y′′ + 2y′ + 5y = 0, y(0) = 3, y′(0) = −1. Is the equation under damped or over damped? Does y(t) = 0 for some t > 0? If so, find the first t > 0 for which y(t) = 0. Sketch the graph of your solution. 4. [16 Points] Find the Laplace transform of the following function: f(t) = { sin 2t if 0 ≤ t < π, 0 if t ≥ π. 5. [16 Points] Find the inverse Laplace transform of the following function: F (s) = (s + 1)e−πs s2 + 2s + 10 6. [16 Points] Solve the following initial value problem: y′′ + 5y′ + 6y = δ(t− 4), y(0) = 0, y′(0) = 0. Recall that δ(t − c) refers to the Dirac delta function which produces a unit impulse at time t = c. In the Table of Laplace transforms, this is referred to as δc(t) (that is δc(t) = δ(t− c)). (See Formula 24, Page 431 of the Laplace Transform Tables). Math 2065 Section 1 April 22, 2005 1 Name: Exam 3 Instructions. Answer each of the questions on your own paper. Put your name on each page of your paper. Be sure to show your work so that partial credit can be adequately assessed. Credit will not be given for answers (even correct ones) without supporting work. A copy of the Table of Laplace transforms from the text will be provided. 1. [16 Points] Solve: 2t2y′′ + 7ty′ − 3y = 0. I Solution. This is a Cauchy-Euler equation with indicial polynomial q(s) = 2s(s − 1) + 7s − 3 = 2s2 + 5s − 3 = (2s − 1)(s + 3), which has the two distinct real roots 1/2 and −3. Hence the general solution is y = c1 |t| 1/2 + c2 |t| −3 . J 2. [16 Points] Solve: y′′ + y′ − 6y = 3e2t. I Solution. This equation could be solved either by the method of undetermined coefficients or variation of parameters, or it could be solved directly by use of Laplace transforms. I will use the method of undetermined coefficients. The characteristic equation is p(s) = s2 + s− 6 = (s− 2)(s + 3) and F (s) = L{3et} = 3/(s − 2) which has denominator Q(s) = s − 2. Thus p(s)Q(s) = (s − 2)2(s + 3) and we compute the fundamental sets corresponding to p(s) and p(s)Q(s) as FSp(s)Q(s) = { e−3t, e2t, te2t } FSp(s) = { e−3t, e2t } . The only element in the first set that is not in the second is te2t, so we conclude that a particular solution of the nonhomogeneous equation has the form yp = Ate 2t for some constant A to be computed by substitution in the differential equation. Since y′p = A(1 + 2t)e 2t and y′′p = A(4 + 4t)e 2t, substitution of yp in the differential equation gives 3e2t = y′′p − 3y ′ p − 4yp = A(4 + 4t)e 2t + A(1 + 2t)e2t − 6Ate2t = 5Ae2t. Hence, 5A = 3, so A = 3/5 and yp = (3/5)te 2t. The general solution is then y = c1e −3t + c2e 2t + 3 5 te2t. J 3. [20 Points] Find the unique solution of the initial value problem y′′ + 2y′ + 5y = 0, y(0) = 3, y′(0) = −1. Is the equation under damped or over damped? Does y(t) = 0 for some t > 0? If so, find the first t > 0 for which y(t) = 0. Sketch the graph of your solution. Math 2065 Section 1 April 22, 2005 1
