Solutions to Quiz 13 in MATH 214: Using Laplace Transform to Solve Initial Value Problems , Quizzes of Differential Equations

Download Solutions to Quiz 13 in MATH 214: Using Laplace Transform to Solve Initial Value Problems and more Quizzes Differential Equations in PDF only on Docsity! MATH 214 – QUIZ 13 – SOLUTIONS Use the Laplace transform (and the table below) to solve the initial value problem y′′ + 3y′ + 2y = u10(t), y(0) = 0, y ′(0) = 0. Solution: Taking the Laplace transform of both sides gives L{y′′ + 3y′ + 2y} = L{u10(t)} L{y′′}+ 3L{y′}+ 2L{y} = e −10s s s2 L{y} − s y(0)− y′(0) + 3(sL{y} − y(0)) + 2L{y} = e −10s s (s2 + 3s+ 2)L{y} = e −10s s so that L{y} = e−10s 1 s(s+ 2)(s+ 1) . Expanding this last term in partial fractions gives 1 s(s+ 2)(s+ 3) = A s + B s+ 2 + C s+ 1 = A(s+ 1)(s+ 2) +Bs(s+ 1) + Cs(s+ 2) (s+ 2)(s+ 1) so that A(s+ 1)(s+ 2) +Bs(s+ 1) + Cs(s+ 2) = 1. Plugging in s = 0 gives A = 1/2, plugging in s = −2 gives B = 1/2 and plugging in s = −1 gives C = −1. Therefore L−1 { 1 2 1 s + 1 2 1 (s+ 2) − 1 (s+ 1) } = 1 2 + 1 2 e−2t − e−t. Finally, y = u10(t) ( 1 2 + 1 2 e−2(t−10) − e−(t−10) ) . Use the Laplace transform (and the table below) to solve the initial value problem y′′ − 4y′ + 4y = δ(t− 5) + u10(t), y(0) = 0, y′(0) = 0. Solution: Taking the Laplace transform of both sides gives L{y′′ − 4y′ + 4y} = L{δ(t− 5)}+ L{u10(t)} L{y′′} − 4L{y′}+ 4L{y} = e−5t + e −10s s s2 L{y} − s y(0)− y′(0)− 4(sL{y} − y(0)) + 4L{y} = e−5t + e −10s s (s2 − 4s+ 4)L{y} = e−5t + e −10s s so that L{y} = e−5t 1 (s− 2)2 + e−10t 1 s(s− 2)2 . Expanding the second term in partial fractions gives 1 s(s− 2)2 = A s + B s− 2 + C (s− 2)2 = A(s− 2)2 +Bs(s− 2) + Cs s(s− 2)2 so that A(s− 2)2 +Bs(s− 2) + Cs = 1. Plugging in s = 0 gives A = 1/4, plugging in s = 2 gives C = 1/2 and taking a derivative gives the equation 2A(s− 2) +Bs+B(s− 2) +C = 0 and plugging s = 2 into this equation gives B = −1/4. Therefore L−1 { 1 4 1 s − 1 4 1 (s− 2) + 1 2 1 (s− 2)2 } = 1 4 − 1 4 e2t + 1 2 t e2t. Since also L−1 { 1 (s− 2)2 } = t e2t, we finally arrive at y = u5(t) ( (t− 5) e2(t−5) ) + u10(t) ( 1 4 − 1 4 e2(t−10) + 1 2 (t− 10) e2(t−10) ) .