Solutions to Math 3D Homework 8: Laplace Transforms and Convolusions, Assignments of Mathematics

Download Solutions to Math 3D Homework 8: Laplace Transforms and Convolusions and more Assignments Mathematics in PDF only on Docsity! Math 3D - Homework 8 Solutions Paul Macklin March 14, 2005 Page 250, #6 (10 points): Solve the given initial-value problem. y′′ + y′ + y = 2δ(t− 1)− δ(t− 2); y(0) = 1, y′(0) = 0 Solution: First, we take the Laplace transform of both sides: L(y′′ + y′ + y) = (s2Y − sy(0)− y′(0)) + (sY − y(0)) + Y = (s2 + s + 1)Y − s− 1 = 2L(δ(t− 1))− L(δ(t− 2)) = 2e−s − e−2s. Thus, Y (s) = s + 1 s2 + s + 1 + 2e−s 1 s2 + s + 1 − e−2s 1 s2 + s + 1 = s + 12 s2 + s + 1 + 1 2 1 s2 + s + 1 + 2e−s 1 s2 + s + 1 − e−2s 1 s2 + s + 1 . Now, 1 s2 + s + 1 = 1( s + 12 )2 + 34 = 1 ( s + 12 )2 + (√ 3 2 )2 = 2√ 3 √ 3 2( s + 12 )2 + (√ 3 2 )2 , and so L−1 ( 1 s2 + s + 1 ) = 2√ 3 e− 1 2 t sin (√ 3 2 t ) . So, L−1 ( e−cs 1 s2 + s + 1 ) = 2√ 3 Hc(t)e− 1 2 (t−c) sin (√ 3 2 (t− c) ) . Similarly, L−1 ( s + 12 s2 + s + 1 ) = e− 1 2 t cos (√ 3 2 t ) . Putting all this together, we have y(t) = L−1 ( s + 12 s2 + s + 1 + 1 2 1 s2 + s + 1 + 2e−s 1 s2 + s + 1 − e−2s 1 s2 + s + 1 ) = e− 1 2 t ( cos (√ 3 2 t ) + 1√ 3 sin (√ 3 2 t )) + 4√ 3 H1(t)e− 1 2 (t−1) sin (√ 3 2 (t− 1) ) − 2√ 3 H2(t)e− 1 2 (t−2) sin (√ 3 2 (t− 2) ) . 1