# sol4 - ACM 95b/100b Solutions for Problem Set 4 Sean Mauch...

This preview shows pages 1–4. Sign up to view the full content.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: ACM 95b/100b Solutions for Problem Set 4 Sean Mauch [email protected] February 8, 2002 1 Problem 1 Problem. Use the shifting theorems to assist in solving the following initial value problems. 1. 4 y 00- 4 y + 37 y = 0 , y (0) = 3 , y (0) = 3 2 2. y 00 + y = r ( t ) , y (0) = 0 , y (0) = 0 , r ( t ) = ( t, < t < 1 , t > 1 Solution. 1. We consider the problem 4 y 00- 4 y + 37 y = 0 , y (0) = 3 , y (0) = 3 2 . We take the Laplace transform of the differential equation and solve for ˆ y . 4( s 2 ˆ y- sy (0)- y (0))- 4( s ˆ y- y (0)) + 37ˆ y = 0 (4 s 2- 4 s + 37)ˆ y = 12 s- 6 ˆ y = 6 2 s- 1 4 s 2- 4 s + 37 ˆ y = 3 s- 1 / 2 s 2- s + 37 / 4 ˆ y = 3 s- 1 / 2 ( s- 1 / 2) 2 + 9 We take the inverse Laplace transform to obtain the solution. y = 3e t/ 2 cos(3 t ) Alternate Method. We could also solve this problem without the Laplace transform. We substitute y = e λt into the differential equation. 4 λ 2- 4 λ + 37 = 0 λ = 1 2 ± ı 3 1 The general solution of the differential equation is y = c 1 e t/ 2 cos(3 t ) + c 2 e t/ 2 sin(3 t ) . We substitute this solution into the initial conditions to determine the constants. y (0) = 3 c 1 = 3 y (0) = 3 2 3 2 + 3 c 2 = 3 2 c 2 = 0 y = 3e t/ 2 cos(3 t ) 2. We consider the problem y 00 + y = r ( t ) , y (0) = 0 , y (0) = 0 , r ( t ) = ( t, < t < 1 , t > 1 We take the Laplace transform of the differential equation. s 2 ˆ y- sy (0)- y (0) + ˆ y = Z 1 t e- st d t ( s 2 + 1 ) ˆ y = 1 s 2- 1 s + 1 s 2 e- s ˆ y = 1 s 2 ( s 2 + 1)- 1 s ( s 2 + 1) + 1 s 2 ( s 2 + 1) e- s ˆ y = 1 s 2- 1 s 2 + 1 + s s 2 + 1- 1 s + 1 s 2 + 1- 1 s 2 e- s We take the inverse Laplace transform to obtain the solution. y = t- sin t + (cos( t- 1)- 1 + sin( t- 1)- ( t- 1)) H ( t- 1) y = t- sin t + (cos( t- 1) + sin( t- 1)- t ) H ( t- 1) Alternate Method. We could also solve this problem without the Laplace transform. First we solve the problem on the interval [0 ... 1]. The homogeneous solution is y h = c 1 cos t + c 2 sin t. We determine a particular solution y p = t by inspection. The general solution is y = c 1 cos t + c 2 sin t + t. We substitute this solution into the initial conditions to determine the constants. y (0) = 0 c 1 = 0 y (0) = 0 c 2 + 1 = 0 y = t- sin t, for t ∈ [0 ... 1] 2 Now we solve problem on the interval [1 ... ∞ ). The general solution is y = c 1 cos t + c 2 sin t. We determine initial conditions at t = 1 from the solution on the interval [0 ... 1]. y (1) = 1- sin(1) , y (1) = 1- cos(1) We substitute the solution into the initial conditions to determine the constants. y (1) = 1- sin(1) , y (1) = 1- cos(1) c 1 cos(1) + c 2 sin(1) = 1- sin(1) ,- c 1 sin(1) + c 2 cos(1) = 1- cos(1) c 1 = cos(1)- sin(1) , c 2 = cos(1) + sin(1)- 1 y = (cos(1)- sin(1))cos t + (cos(1) + sin(1)- 1)sin t Now we have the solution of the problem....
View Full Document

{[ snackBarMessage ]}

### Page1 / 13

sol4 - ACM 95b/100b Solutions for Problem Set 4 Sean Mauch...

This preview shows document pages 1 - 4. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online