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Math 286 · Exam 2 Solutions · Fall 2012 · Dr. Greene Name: Please show supporting work or an explanation of your reasoning. These elements are considered part of the answer and will be graded. Good luck! 1. Solve the initial value problem y 00 + y 0 − 6y = 0, y(0) = 0, y 0 (0) = 1. (10 points) Solution: The auxillary equation m2 + m − 6 = (m + 3)(m − 2) = 0 has roots m = −3, 2. The general solution of the differential equation is y = c1 e−2x + c2 e3 x which implies y 0 = −2c1 e−2x + 3c2 e3x . The initial conditions y(0) = 0 and y 0 (0) = 0 imply c1 + c2 = 0 −2c1 + 3c2 = 1 The first equation implies c2 = −c1 , which after substituting in the second equation yields −5c1 = 1. Therefore, c1 = −1/5 and c2 = 1/5. The solution to the initial–value problem is 1 1 y = − e−2x + e3x . 5 5 1 2. Use the method of undetermined coefficients to solve y 00 − 4y 0 + 4y = x2 + 2. (10 points) Solution: To solve the homogeneous differential equation y 00 − 4y 0 + 4y = 0, observe that the auxillary equation m2 − 4m + 4 = (m − 2)2 has solution m = 2 with multiplicity 2. Thus, the homogeneous equation has solution y = c1 e2 x + c2 xe2 x. The form of the right side of the original nonhomogeneous differential equations suggests a particular solution of the form y = Ax2 + Bx + C. By taking derivatives, we see y 0 = 2Ax + B y 00 = 2A. Substituting these expressions in the differential equation gives y 00 − 4y 0 + 4y = 2A − 4(2Ax + B) + 4(Ax2 + Bx + C) = 4Ax2 + (4B − 8A)x + 2A − 4B + 4C = x2 + 0x + 2 Comparing the coefficients of x2 , we see A = 1/4. Comparing the coefficients of x, we see that 0 = 4B − 8A = 4B − 8/4 = 4B − 2. Thus, B = 1/2. Finally, the remaining coefficient gives 2 = 2A − 4B + 4C = 2/4 − 4/2 + 4C = 1/2 − 2 + 4C. Thus, C = 7/8. Consequently, the general solution of the differential equation is 1 7 1 y = c1 e2 x + c2 xe2 x + x2 + x + . 4 2 8 2 3. Use variation of parameters to solve y 00 + y = sec x. (10 points) Solution: To solve the homogeneous equation y 00 + y = 0, observe that the auxillary equation m2 + 1 = 0 has solutions m = 0 ± i. Thus, the complmentary solution is y = c1 cos x + c2 sin x. A particular solution to the nonhomogeneous equation is u1 cos x + u2 sin x where u1 and u2 are to be determined. Solve for u1 : 0 sin x sec x cos x 0 · cos x − sin x sec x 0 = u1 = = − sin x sec x. 2 x − (− sin2 x) cos cos x sin x − sin x cos x Z u1 = Z − sin x sec xdx = − sin x dx = cos x Z du = ln | cos x| + C u Solve for u2 : cos x 0 − sin x sec x cos x sec x − 0 · (− sin x) 0 = u2 = = 1. cos2 x − (− sin2 x) cos x sin x − sin x cos x Z u2 = 1dx = x + C. Thus, a particular solution of the nonhomogeneous equation is yp = u1 y1 + u2 y2 = cos x ln | cos x| + x sin x and the general solution is y = c1 cos x + c2 sin x + cos x ln | cos x| + x sin x. 3 4. (a) Define what it means for a set of functions {y1 , y2 , y3 } to be a fundamental set of solutions of a 3rd–order homogeneous differential equation on an interval I. (10 points) Solution: {y1 , y2 , y3 } is a set of linearly independent solutions to differential equation over I. (b) Determine whether the set of functions {x4 , −x4 } is linearly independent on the open interval (0, 2). (10 points) Solution: The functions are linearly dependent since 1 · x4 + 1 · (−x4 ) = 0 for every x in (0, 2). (Alternatively, the two functions are scalar multiples of each other.) 4 5. State a 1st-order homogeneous differential equation for which the functions {x4 , −x4 } are solutions on the interval (0, 2). Find the general solution of your differential equation. (5 points) Solution: Observe that letting y = x4 implies that y 0 = 4x3 . Since 4x4 − 4x4 = 0, y = x4 is a solution (on the interval (0, 2)) to the differential equation xy 0 − 4y = 0. Similarly, y = −x4 is also a solution of this differential equation since xy 0 − 4y = x(−4x3 ) − 4x4 = −4x4 + 4x4 = 0 for all x ∈ (0, 2). Observe that this is a 1st–order homogeneous Cauchy–Euler equation. The general solution (a one–parameter family) is given by y = cx4 . 5 6. State a 2nd-order homogeneous differential equation for which the functions {x4 , −x4 } are solutions on the interval (0, 2). Find the general solution of your differential equation. (5 points) Solution: Letting y = x4 gives y 0 = 4x3 and y 00 = 12x2 . Since 12x4 − 12x4 = 0, y = x4 is a solution (on the interval (0, 2)) to x2 y 00 − 12y = 0. Since this is a 2nd–order Cauchy–Euler equation, we solve the auxillary equation m(m − 1) − 12 = 0. Thus, m2 − m − 12 = (m + 3)(m − 4) = 0 has solutions m = −3, 4. Therefore, the general solution is y = c1 x−3 + c2 x4 . Note: There are infinitely many correct answers to this problem. For example, y = ±x4 are also solutions to the Cauchy–Euler equation x2 y 00 − xy 0 − 8y = 0 which has auxillary equation 0 = m(m − 1) − m − 8 = m2 − 2m − 8 = (m + 2)(m − 4) and general solution y = c1 x−2 + c2 x4 . 6 7. Solve the homogeneous Cauchy-Euler equation 2x2 y 00 + 2xy 0 − 8y = 0 (10 points) Solution: This equation has the same solution set as x2 y 00 + xy 0 − 4y = 0. The auxillary equation is 0 = m(m − 1) + m − 4 = m2 − 4 = (m − 2)(m + 2). The general solution is y = c1 x−2 + c2 x2 . 7 8. (Bonus!) Explain why one should not attempt to use the method of undetermined coefficients to solve y 00 + y = ln x. (+2 points) Solution: The method of undetermined coefficients does not apply in this case because the function ln x is not a finite sum of finite products of polynomials, exponentials emx , and trigonmetric functions sin βx and cos βx.. 8