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Ordinary Differential Equations Math 22B-002, Spring 2017 Final Exam: Solutions 1. [15 Pts.] Solve the Initial Value Problem

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Ordinary Differential Equations Math 22B-002, Spring 2017 Final Exam: Solutions 1. [15 Pts.] Solve the Initial Value Problem Ordinary Differential Equations Math 22B-002, Spring 2017 Final Exam: Solutions 1. [15 pts.] Solve the initial value problem 6y00 − y0 − y = 0; y(0) = 10; y0(0) = 0: Solution. • The characteristic equation is 6r2 −r −1 = 0 with roots r = 1=2; −1=3, so the general solution is x=2 −x=3 y(x) = c1e + c2e : • The initial conditions are satisfied if 1 1 c + c = 10; c − c = 0 1 2 2 1 3 2 whose solution is c1 = 4, c2 = 6, so y(x) = 4ex=2 + 6e−x=3: 1 2. [20 pts.] (a) Let y0 is an arbitrary constant. Find the solution y(x) of the initial-value problem 0 y − 2xy = x; y(0) = y0: (b) For what initial value y0 does the solution remain bounded as x ! +1? What is the solution in that case? Solution. • The ODE is first-order, linear, and nohomogeneous, so we use the in- tegrating factor method. • The integrating factor with coefficient p(x) = −2x is given by Z µ(x) = exp − 2x dx = e−x2 : • Multiplying the ODE by the integrating factor and writing the left- hand side as an exact derivative, we get that 2 0 2 e−x y = xe−x ; so 2 Z 2 1 2 e−x y(x) = xe−x dx + c = − ex + c; 2 and the general solution is 1 2 y(x) = − + cex : 2 • The initial condition gives y0 = −1=2 + c, so c = y0 + 1=2 and 1 1 2 y(x) = − + y + ex : 2 0 2 • (b) The solution remains bounded as x ! 1 if c = 0 and y0 = −1=2. In that case, the solution is y(x) = −1=2. Remark. Alternatively, we could use variation of parameters: the solution of the homogeneous equation is y(x) = cex2 , so write y(x) = u(x)ex2 and solve for u(x); or we could observe that yp(x) = −1=2 is a particular solution, and x2 write y(x) = yp(x) + ce . 2 3. [20 pts.] (a) Find the general solution of the ODE y00 − 8y0 + 25y = 0: (b) Find the general solution of the ODE y00 + 25y = 0: Solution. • (a) The characteristic equation is r2 − 8r + 25 = 0 with roots r = 4 ± 3i It follows that the general solution is 4x 4x y(x) = c1e cos 3x + c2e sin 3x where c1, c2 are arbitrary constants. • (b) The characteristic equation is r2 + 25 = 0 with roots r = ±5i: It follows that the general solution is y(x) = c1 cos 5x + c2 sin 5x where c1, c2 are arbitrary constants. 3 4. [20 pts.] (a) Find a particular solution of the ODE y00 + y = ex + x2: (b) Find a particular solution of the ODE y00 − y = ex + x2: Solution. • (a) We look for a particular solution of the form x 2 yp(x) = Ae + Bx + C: • Then 00 x 2 yp + yp = 2Ae + Bx + 2B + C; so we take A = 1=2, B = 1, and 2B + C = 0. It follows that C = −2 and 1 y (x) = ex + x2 − 2: p 2 (b) The function ex is a solution of the homogeneous equation, so we look for a particular solution of the form x 2 yp(x) = Axe + Bx + C: • Then 00 x 2 yp − yp = 2Ae − Bx + 2B − C; so we take A = 1=2, B = −1, and 2B − C = 0. It follows that C = −2 and 1 y (x) = xex − x2 − 2: p 2 4 5. [25 pts.] Suppose that the 2 × 2 matrix A has the following eigenvalues λk and eigenvectors ~rk: 1 2 λ = −2; ~r = ; λ = 1; ~r = : 1 1 3 2 2 −1 (a) Write down the general solution for (x(t); y(t))T of the system x 0 x = A : y y (b) Sketch the phase plane of this system. Include in your sketch the trajec- tory that passes through the point (0; 2). (c) Sketch the graphs of x(t) and y(t) versus t for the solution of this system that satisfies the initial condition x(0) = 0, y(0) = 2. Solution. • (a) The general solution is x 1 2 = c e−2t + c et y 1 3 2 −1 where c1, c2 are arbitrary constants. • (b) The origin is a saddle point, The stable direction, in which (x; y) ! (0; 0) as t ! +1, is (1; 3); the unstable direction, in which (x; y) ! (0; 0) as t ! −∞, is (2; −1). • A numerical plot of the phase plane (using the matlab script pplane8.m) is shown in Figure 1. • (c) Graphs of x(t) and y(t) versus t are shown in Figure 2. 5 Figure 1: Phase plane for 5(b). Figure 2: Graphs for 5(c). 6 Figure 3: Phase plane for 6(b). 6. [25 pts.] (a) Find the general solution of the 2 × 2 system x 0 1 2 x = : y −5 −1 y Write your answer in complex form. (b) Sketch the phase plane of this system. Solution. • The characteristic polynomial of the matrix is 1 − λ 2 2 = λ + 9; −5 −1 − λ so the eigenvalues satisfy λ2 + 9 = 0, and λ = ±3i. • If λ = 3i, then the eigenvector ~r = (p; q)T satisfies (A − 3iI)~r = 0, or 1 − 3i 2 p 0 = : −5 −1 − 3i q 0 7 One solution is p 2 = : q −1 + 3i Any non-zero (complex) scalar multiple of this vector would do just as well. • The eigenvector for λ = −3i is the conjugate of the eigenvector for λ = 3i, so the general solution is x 2 2 = c e3it + c e−3it y 1 −1 + 3i 2 −1 − 3i where c1, c2 are arbitrary (complex) constants. To get real-valued solutions, we take c1 and c2 to be complex conjugates. • The origin is a center with closed (periodic) orbits. Since x0 = x + 2y is positive when x; y > 0 and negative when x; y < 0, the direction of trajectories is clockwise. A numerical plot of the phase plane is shown in Figure 3. 8 2 7. [25 pts.] (a) Let m; c; k; F0 > 0 be positive constants with c > 4mk. Find the general solution for y(t) of the ODE 00 0 my + cy + ky = F0: (b) How does y(t) behave as t ! +1? (c) Give a physical interpretation of this ODE and the behavior of its solu- tions. Solution. • A particular solution of the ODE is the constant yp = F0=k. • The characteristic equation of the homogeneous ODE is mr2+cr+k = 0 with roots p −c ± c2 − 4mk r = = −α; −β: 2 2 pBoth of these roots are negative real numbers, since c − 4mk > 0 and c2 − 4mk < c. • It follows that the general solution of the ODE (particular solution + general solution of the homogeneous equation) is given by F y(t) = 0 + c e−αt + c e−βt k 1 2 where c1, c2 are arbitrary constants. • (b) We have y(t) ! F0=k as t ! +1. • (c) This ODE describes a mass m on a spring with spring constant k and damping proportional to velocity with damping constant c. The mass is subject to a constant external force F0. • The displacement yp = F0=k corresponds to the equilibrium location of the mass, when the external force F0 is balanced by the restoring force −kyp of the spring. As t ! 1, the mass relaxes to its equilibrium state due to damping. 9 8. [25 pts.] Suppose that a hot metal bar with absolute temperature T (t) at time t cools by emitting thermal radiation. According to the Stefan- Boltzmann law, the bar radiates heat energy at rate proportional to T 4. If the initial temperature of the bar is T0 > 0, then T (t) satisfies the initial value problem dT = −kT 4;T (0) = T ; dt 0 where k > 0 is a constant. (a) Solve for T (t). (b) Let t1=2 be the time it takes for the bar to cool to half its initial temper- ature. Find an expression for t1=2 in terms of k and T0. (c) Does t1=2 increase, decrease, or stay the same as T0 increases? How does this compare with the behavior of t1=2 for an object that cools according to Newton's law of cooling (dT=dt = −kT )? Solution. • (a) The ODE is nonlinear and separable. Separating variables, we get Z dT Z = − k; dt + c T 4 Integration gives 1 − = −kt + c: 3T 3(t) 3 The initial condition gives c = −1=3T0 , and solving for T , we find that T T (t) = 0 : 3 1=3 (1 + 3T0 kt) • (b) The temperature of the bar is half the initial temperature when 3 1=3 1 + 3T0 kt1=2 = 2: Solving for t1=2, we get that 7 t1=2 = 3 : 3T0 k • (c) The time t1=2 decreases as T0 increases (the cooling rate is larger at −kt higher temperatures). For Newton's law of cooling, T (t) = T0e and the decay time t1=2 = ln 2=k is independent of T0. 10 9. [25 pts.] Consider the following ODE for y(x): y0 + sin2(x + y) = 0: (1) (a) What is the order of this ODE? Is it linear or nonlinear? Separable or non-separable? (b) What can you say from general theorems about the existence and unique- ness of solutions of the initial value problem for (1) with the initial condition y(x0) = y0? (c) Show that the substitution u(x) = x + y(x) gives the following ODE: u0 = cos2 u: (2) (d) Solve (2) for u(x) and write down the general solution of (1) for y(x).
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