Chapter 6 6.4 DEs with Discontinuous Force Functions
Discontinuous Force Functions Using Laplace Transform, as in 6.2, we solve nonhomogeneous linear second order DEs with constant coefficients. The only difference, from 6.2, is that the nonhomogeneous part g(t) is discontinuous. Methods are as in 6.2. We use the chart in page 321.
Solve the IVP y +y = f(t) = { 1 0 t < 3π 0 3π t < y(0) = 0, y (0) = 1. We write f(t) = u 0 (t) u 3π (t). So, we have y +y = u 0 (t) u 3π (t) We write Y(s) = L{y}. So, L{y }+L{y} = L{u 0 } L{u 3π } =
(s 2 Y sy(0) y (0))+Y = 1 s e 3πs s (s 2 + 1)Y = 1+ 1 s e 3πs s = s + 1 s e 3πs s Y = s + 1 s(s 2 + 1) e 3πs s(s 2 + 1) = a s + bs + d ( d s 2 + 1 e 3πs s + es + g ) s 2 + 1 = 1 s + s + 1 ( 1 s 2 + 1 e 3πs s s ) s 2 + 1 = =
So, Y = 1 s s s 2 + 1 + 1 s 2 + 1 e 3πs s So, the solution: { 1 y = L 1 {Y} = L 1 s { } e L 1 3πs s + s s 2 + 1 e 3πs } { } { } s 1 L 1 +L 1 s 2 + 1 s 2 + 1 { } s +L 1 s 2 + 1 e 3πs
Use the chart in page 321 { } s y = 1 cos t + sin t u 3π (t)+l 1 s 2 + 1 e 3πs (1) To use formula 13, write H(s) = s s 2 + 1, hence h(t) = L 1 {H(s)} = cos t From (1) and formula 13: y = 1 cos t + sin t + u 3π (t)+u 3π (t)h(t 3π) = 1 cos t + sin t + u 3π (t)+u 3π (t) cos(t 3π) = 1 cos t + sin t u 3π (t) u 3π (t) cos t
Solve the IVP y +3y + 2y = u 2 (t) y(0) = 0, y (0) = 1. We write Y(s) = L{y}. There are two steps: Compute Y(s), by application of Laplace transform L. Compute y = L 1 {Y(s)} by application of Inverse Laplace transform L.
We have L{y }+3L{y }+2L{y} = L{u 2 (t)} (s 2 Y(s) sy(0) y (0))+3(sY(s) y(0))+2y(s) = 1 s e 2s (s 2 Y(s) 1)+3sY(s)+2Y(s) = 1 s e 2s Y(s) = = 1 1 s(s 2 + 3s + 2) e 2s + s 2 + 3s + 2 1 s(s 2 + 3s + 2) e 2s + 1 s + 1 + 1 s + 2
Apply L 1 : { } y = L 1 {Y(s)} = L 1 1 s(s 2 + 3s + 2) e 2s +e t e 2t To use formula 13, write (2) H(s) = 1 s(s 2 + 3s + 2), h(t) = L 1 {H(s)} Do partial fraction: H(s) = 1 s(s + 1)(s + 2) = a s + b s + 1 + c s + 2
So, 1 1 1 3 2 1 2 0 0 a b c = 0 0 1 So, 1 0 0 0 1 0 0 0 1 H(s) = a b c =.5 1.5 a =.5, b = 1, c =.5 1 s(s + 1)(s + 2) =.5 s 1 s + 1 +.5 s + 2
So, h(t) = L 1 {H(s)} =.5 e t +.5e 2t From (2) and formula 13: y = u 2 (t)h(t 2)+e t e 2t = u 2 (t) [.5 e (t 2) +.5e 2(t 2)] + e t e 2t
Solve the IVP y +y = g(t) = { t/2 0 t < 6 3 t 6 y(0) = 0, y (0) = 1. We write Y(s) = L{y}. There are two steps: Compute Y(s), by application of Laplace transform L. Compute y = L 1 {Y(s)} by application of Inverse Laplace transform L.
Write g(t) = (1 u 6 (t)) t 2 + 3u 6(t) = t 2 u 6(t) t 6 2 By formula 13 So, Apply L: L{g(t)} = 1 2s e 6s 2 2 L{t} = 1 2s e 6s 2 2s 2 L{y }+L{y} = L{g(t)} = 1 2s e 6s 2 2s 2 s 2 Y(s) sy(0) y (0)+Y(s) = 1 2s e 6s 2 2s 2
s 2 Y(s) 1+Y(s) = 1 2s e 6s 2 2s 2 Y(s) = 1 s 2 + 1 + 1 2s 2 (s 2 + 1) e 6s 2s 2 (s 2 + 1) = 1 s 2 + 1 +.5 s.5 2 s 2 + 1 e 6s 2s 2 (s 2 + 1) =.5 s 2 + 1 +.5 s e 6s 2 2s 2 (s 2 + 1)
So, y = L 1 {Y(s)} =.5 sin t+.5t.5l 1 u 6 (t)h(t 6) (3) Where h(t) = L 1 {H(s)} and H(s) = 1 s 2 (s 2 + 1) = 1 s 1 2 s 2 + 1 seen above. So, h(t) = L 1 {H(s)} = t sin t
Finally, by (3), y =.5 sin t +.5t.5L 1 u 6 (t)((t 6) sin(t 6))
6.4 Assignments and Homework Read Example 1, 2 (They are helpful). Homework: 6.4 See Homework site