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

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 → +∞? What is the solution in that case?

Solution. • The ODE is ﬁrst-order, linear, and nohomogeneous, so we use the integrating 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 → ∞ 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 → +∞, 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 → +∞? (c) Give a physical interpretation of this ODE and the behavior of its solutions.

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 √ −c ± c2 − 4mk r = = −α, −β. 2 2 Both√ 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 → +∞. • (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 → ∞, 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 temperature. 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). For what x-values do your solutions exist? (e) What are the solutions of (1) with the initial conditions : (i) y(0) = 0; (ii) y(0) = π/2?

Solution. • (a) The ODE is first-order, nonlinear, and non-separable. • (b) The function sin2(x + y) is a continuous function of (x, y) and continuously differentiable with respect to y for all (x, y) ∈ R2, so a unique solution y(x) of the initial value problem exists for every initial condition y(x0) = y0. The solution is defined in some open x-interval that contains x0. • (c) Writing y = u − x and y0 = u0 − 1, we get that u0 − 1 + sin2 u = 0. Since 1 − sin2 u = cos2 u, it follows that u0 = cos2 u. • (d) Separating variable in the u-ODE, and assuming that cos u 6= 0, we get that Z du Z = dx + c. cos2 u Since 1/ cos2 u = sec2 u and Z sec2 u du = tan u + c,

11 it follows that tan u = x + c, so

u(x) = tan−1(x + c)

for some branch of the (multi-valued) inverse tangent.

• When cos u = 0, we also get the constant solutions π u(x) = + nπ, n = 0, ±1, ±2,... 2

• The general solution of the ODE for y is therefore π y(x) = tan−1(x + c) − x, or y(x) = + nπ − x 2 where c is an arbitrary (real) constant and n is an arbitrary integer. These solutions exist for all −∞ < x < ∞.

• (e) (i) The solution with y(0) = 0 is

y(x) = tan−1 x − x

where we take the standard branch of the inverse tangent with

−π/2 < tan−1 x < π/2.

• (ii) The solution with y(0) = π/2 is π y(x) = − x. 2