Mathematics 38 Exam II Differential Equations March 10, 2008 1. (5 points) Determine whether the system x′ = −ty − z + t x z y′ = − − + 1 t t ′ z = x − ty is linear. If it is linear a. determine whether it is homogeneous, Solution: Yes. a. no. ′ 0 x y = −1/t 1 z b. determine its order, and c. write it in matrix form. b. 3. c. −t 0 −t t x −1 −1/t y + 1 0 z 0 0 or ~x′ = −1/t 1 −t 0 −t t −1 −1/t ~x + 1 . 0 0 2. (5 points) Given the o.d.e. (D2 − 1)x = t (N) a. find the equivalent system (SN ), b. the general solution of (N) is x(t) = c1 e−t + c2 et − t. You do not need to verify this. Use the general solution of (N) to obtain each component of the general solution of (SN ), c . write (SN ) in matrix form, d. write the general solution of (SN ) in the form ~x = c1~h1 (t) + · · · + cn~hn (t) + p~(t). ′ x1 = x2 x1 (t) = c1 e−t + c2 et − t Solution: a. . b. . ′ −t t x = x + t x (t) = −c e + c e − 1 1 2 1 2 2 −t t 0 e 0 1 e −t ′ . d. ~x = c1 ~x + + c2 c. ~x = . + t −e−t 1 0 et −1 3. (5 points) Determine whether the spring modeled by (mD2 + bD + k)x = 0 with m = 1 gram, k = 4 dynes/cm and b = 4 gram/s is undamped, underdamped, critically damped, or overdamped. Solution: Critically damped: 1 · D2 + 4 · D + 4 = (D + 2)2 . 4t + 1 t < 2 4. (5 points) Rewrite f (t) = 9 2 ≤ t < 3 in unit step function notation. 2 t t≥3 Solution: f (t) = 4t + 1 + u2 (t)(9 − (4t + 1)) + u3 (t)(t2 − 9). 1 5. (10 points) a. Make a simplified guess for a particular solution of (D2 − 2D + 1)x = tet . You do not need to determine the coefficients! Solution: k1 t2 et + k2 t3 et . √ b. Find the general solution of (D2 − 2D + 1)x = et t for t > 0. Solution: c′1 (t)et + c′2 (t)et t =0 √ c′1 (t)et + c′2 (t)et (t + 1) = tet simplifies to c′1 (t) + c′2 (t)t = 0 √ c′2 (t) = t, which gives c′1 (t) = −t3/2 and c′2 (t) = t1/2 and the general solution c1 et + c2 tet + 4 5/2 t t e. 15 6. (10 points) a. Compute L [e5t+3 ] using the definition. No credit by any other method b. State for which values of s this Laplace transform is defined. Z ∞ Z h e3 e3 −st 5t+3 3 lim [e−(s−5)h − 1] = Solution: e e dt = lim e e−(s−5)t dt = − h→∞ s − 5 h→∞ s−5 0 0 for s > 5. 7. (10 points) Find 1 ∗ cos 5t. Z t Solution: 1 · cos 5u du = (1/5)[sin 5u]t0 = (1/5) sin 5t. 0 8. (10 points) Find the Laplace transform of f (t) = te5t sin 3t. Solution: L [t sin 3t] = − 6s 6(s − 5) 3 d = 2 , so L [te5t sin 3t] = . 2 2 ds s + 9 (s + 9) ((s − 5)2 + 9)2 9. (20 points) Find the inverse Laplace transform of s+3 a. 2 . s + 4s + 5 Solution: L −1 [ b. s2 (s + 2) + 1 s+1 s+3 ] = L −1 [ ] = e−2t L −1 [ 2 ] = e−2t (cos t + sin t). 2 + 4s + 5 (s + 2) + 1 s +1 s . (s2 + 5)2 Solution: L −1 [ √ √ √ s 1 1 t s −1 √ √ sin sin ] = L [ · ] = cos 5t∗ 5t = 5t. (s2 + 5)2 (s2 + 5) (s2 + 5) 5 2 5 2 10. (20 points) Solve using the Laplace transform. No credit by any other method. a. x′′ + 4x′ + 4x = t2 e−2t , x(0) = x′ (0) = 0. 2 1 4 −2t 2 , so L [x] = and x = t e . Solution: (s + 2)2 L [x] = L [t2 e−2t ] = 3 5 (s + 2) (s + 2) 12 2 t t<2 b. (D − 1)x = , x(0) = 1. 2 t +1 t≥2 Solution: (s − 1)L [x] − 1 = L [t2 ] + L [u2 (t)] = L [x] = 2 2 e−2s −2s , so + e L [1] = + s3 s3 s 2 e−2s 1 + 3 + . s − 1 s (s − 1) s(s − 1) Now do partial fractions decompositions: setting 1 A B C E = + 2+ 3+ − 1) s s s s−1 s3 (s gives As2 (s−1)+Bs(s−1)+C(s−1)+Es3 = 1 or A+E = 0, B −A = 0, C −B = 0, −C = 1, giving C = −1, B = −1, A = −1, and E = 1. Similarly, set A B 1 = + , s(s − 1) s s−1 giving A(s − 1) + Bs = 1, or A = −1, B = 1. So, 2 e−2s 1 + + x(t) = L s − 1 s3 (s − 1) s(s − 1) 1 1 1 1 −1 −1 t −1 −1 − 2− 3+ + + u2 (t)L (t − 2) = e + 2L s s s s−1 s s−1 t2 t t = e + 2 −1 − t − + e + u2 (t) −1 + et−2 2 = 3et − 2 − 2t − t2 − u2 (t) 1 − et−2 . −1 END OF EXAMINATION ©2010 Trustees of Tufts College. All rights reserved.