Math 54. Selected Solutions for Week 10

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1 Math 54. Selected Solutions for Week 10 Section 4.1 (Page 399) 9. Find a synchronous solution of the form A cos Ωt+B sin Ωt to the given forced oscillator equation using the method of Example 4 to solve for A and B : + 2y + 4y = 6 cos 2t + 8 sin 2t, Ω = 2 We have y = A cos 2t + B sin 2t ; y = 2A sin 2t + 2B cos 2t ; = 4A cos 2t 4B sin 2t. Plugging these into the equation then gives + 2y + 4y = 4A cos 2t 4B sin 2t + 2( 2A sin 2t + 2B cos 2t) + 4(A cos 2t + B sin 2t) = ( 4A + 4B + 4A) cos 2t + ( 4B 4A + 4B) sin 2t = 4B cos 2t 4A sin 2t, so (equating coefficients) 4B = 6 and 4A = 8. Therefore B = 3/2 and A = 2, so the solution is y = 2 cos 2t + 3 sin 2t. 2 Section 4.2 (Page 407) 32. Use Definition 1 to determine whether the functions y 1 0, y 2 = e t are linearly dependent on the interval (0, 1). y 2. These are linearly dependent, because y 1 = 0 y 2, so it is a constant multiple of 1

2 2 35. Linear Dependence of Three Functions. Three functions y 1 (t), y 2 (t), y 3 (t) are said to be linearly dependent on an interval I if, on I, at least one of these functions is a linear combination of the remaining two [e.g., if y 1 (t) = c 1 y 2 (t)+c 2 y 3 (t) ]. Equivalently (compare problem 33), y 1, y 2, and y 3 are linearly dependent on I if there exist constants C 1, C 2, and C 3, not all zero, such that C 1 y 1 (t) + C 2 y 2 (t) + C 3 y 3 (t) = 0 for all t I. Otherwise, we say that these functions are linearly independent on I. For each of the following, determine whether the given three functions are linearly dependent or linearly independent on (, ) : (a). y 1 (t) = 1, y 2 (t) = t, y 3 (t) = t 2. (b). y 1 (t) = 3, y 2 (t) = 5 sin 2 t, y 3 (t) = cos 2 t. (c). y 1 (t) = e t, y 2 (t) = te t, y 3 (t) = t 2 e t. (d). y 1 (t) = e t, y 2 (t) = e t, y 3 (t) = cosh t. a. Linearly independent, because C 1 y 1 (t) + C 2 y 2 (t) + C 3 y 3 (t) = C 3 t 2 + C 2 t + C 1. The only polynomial that is zero for all values of t is the zero polynomial, and this forces C 1 = C 2 = C 3 = 0. b. Linearly dependent, because y 2 (t) + 5y 3 (t) = 5 sin 2 t + t cos 2 t = 5 = 5 3 y 1(t). c. Linearly independent, because C 1 y 1 (t) + C 2 y 2 (t) + C 3 y 3 (t) = C 1 e t + C 2 te t + C 3 t 2 e t = (C 3 t 2 + C 2 t + C 1 )e t. Since e t is never zero, the only way for this to be zero for all values of t is for the factor C 3 t 2 + C 2 t + C 1 to be zero for all t, and this forces C 1 = C 2 = C 3 = 0. d. Linearly dependent, because by definition y 3 = cosh t = et + e t = y 1(t) y 2(t).

3 3 Section 4.3 (Page 415) 30. Using the representation for e (α+iβ)t in (6), verify the differentiation formula (7). d dt e(α+iβ)t = d ( e αt (cos βt + i sin βt) ) dt = αe αt (cos βt + i sin βt) + e αt ( β sin βt + i cos βt) = αe (α+iβ)t + iβe t (cos βt + i sin βt) = (α + iβ)e (α+iβ)t. 37. The auxiliary equations for the following differential equations have repeated complex roots. Adapt the repeated root procedure of Section 4.2 to find their general solution: (a). y iv y = 0 (b). y iv y + 16y = 0. [Hint: The auxiliary equation is (r 2 + 2r + 4) 2 = 0.] a. The characteristic polynomial is r 4 + 2r = (r 2 + 1) 2, which has double roots ±i (two double roots, for a total of four). Therefore a general solution is y(t) = c 1 cos t + c 2 t cos t + c 3 sin t + c 4 t sin t. (In other words, you have solutions cos t and sin t coming from one of the roots, and the fact that it s a double root leads you to multiply them by t to get another pair of solutions.) b. As noted in the hint, the characteristic polynomial has double complex roots 1 ± 3i, so a general solution is y(t) = c 1 e t cos 3t + c 2 te t cos 3t + c 3 e t sin 3t + c 4 te t sin 3t. Section 4.4 (Page 424) 5. Decide whether or not the method of undetermined coefficients can be applied to find 2ω (x) 3ω(x) = 4x sin 2 x + 4x cos 2 x. Yes. Since sin 2 x + cos 2 x = 1, the right-hand side simplifies to 4x, so the method can be applied in this case (with trial solution c 1 x + c 2 ).

4 4 6. Decide whether or not the method of undetermined coefficients can be applied to find (θ) + 3y (θ) y(θ) = sec θ. No. There s no way to get sec θ into the form P (θ)e rθ, P (θ)e αθ cos βθ, or P (θ)e αθ sin βθ for a polynomial P (θ). 8. Decide whether or not the method of undetermined coefficients can be applied to find 8z (x) 2z(x) = 3x 100 e 4x cos 25x. Yes, since the function on the right-hand side is of the form Ct m e αt cos βt. Section 4.5 (Page 429) 36. Determine the form of a particular solution for the differential equation. Do not solve. 4y + 4y = t 2 e 2t e 2t. The characteristic polynomial of the associated homogeneous equation is r 2 4r + 4 = (r 2) 2. By the Method of Undetermined Coefficients with s = 2, the trial solution is therefore 46. Show that the boundary value problem t 2 (A 2 t 2 + A 1 t + A 0 )e 2t. + λ 2 y = sin t ; y(0) = 0, y(π) = 1 has a solution if and only if λ ±1, ±2, ±3,.... First find a general solution to the equation + λ 2 y = sin t. The characteristic polynomial is y 2 +λ 2, and it has roots ±λi. Note that if λ = 0 then the characteristic equation has a double real root, but if λ 0 then it has non-real complex conjugate roots. So we have to consider λ = 0 and λ 0 separately. If λ = 0 then it is easy to see that y p (t) = sin t is a particular solution, and that a general solution is given by y(t) = sin t + c 1 t + c 0.

5 The condition y(0) = 0 implies c 0 = 0, and y(π) = 1 then gives c 1 = 1/π. Therefore y(t) = sin t + t/π is a solution to the equation. Next assume that λ 0. We may assume that λ > 0, since changing λ to λ does not change the equation. A general solution to the associated homogeneous equation is then y(t) = c 1 sin λt + c 2 cos λt. If λ 1 then y p (t) = c 3 cos t + c 4 sin t is a trial solution for the method of undetermined coefficients. We have p + λ 2 y p = c 3 cos t c 4 sin t + λ 2 (c 3 cos t + c 4 sin t) = (λ 2 1)(c 3 cos t + c 4 sin t). Setting this equal to sin t gives c 3 = 0 and c 4 = 1/(λ 2 1). This gives y p (t) = sin t λ 2 1, and a general solution to the differential equation is y(t) = sin t λ c 1 sin λt + c 2 cos λt. The condition y(0) = 0 implies c 2 = 0, and y(π) = 1 gives c 1 = 1/ sin λπ, provided that sin λπ 0. This is true if and only if λ is not an integer, so in this case we have shown that a solution exists if and only if λ ±1, ±2, ±3,.... Finally, consider the case λ = 1. In this case a general solution to the associated homogeneous differential equation is y(t) = c 1 cos t + c 2 sin t. Since sin t occurs in this general solution, the trial solution for the method of undetermined coefficients is We have so that y p = c 3 t cos t + c 4 t sin t. y p = c 3 cos t c 3 t sin t + c 4 sin t + c 4 t cos t ; p = 2c 3 sin t c 3 t cos t + 2c 4 cos t c 4 t sin t, p + y p = 2c 3 sin t + 2c 4 cos t. Setting this equal to sin t gives c 3 = 1/2 and c 4 = 0, so y p (t) = (t/2) cos t, and a general solution to the nonhomogeneous equation is y(t) = t 2 cos t + c 1 cos t + c 2 sin t. The condition y(0) = 0 then gives c 1 = 0. However, since sin π = 0, the condition y(π) = 1 gives the equation π 2 ( 1) = 1, which cannot hold. Therefore there is no solution when λ = 1. 5