4.5 Initial-Value Problems with Piecewise Continuous Nonhomogeneous Terms In this section we illustrate the application of Laplace transforms to the solution of nonhomogeneous initial-value problems in which the forcing function f is piecewise continuous. Example 1. Find the solution of the initial-value problem y 0 + 2y = f (x); where f (x) = ( x 5 y(0) = 4 0≤x<3 x ≥ 3. SOLUTION The function f has a jump discontinuity at x = 3. Therefore, the first step is to express f in terms of u3 (x) = u(x − 3). You should verify that f (x) = x−x u(x−3)+5u(x−3) = x−[(x−3)+3]u(x−3)+5u(x−3) = x−(x−3)u(x−3)+2u(x−3). Now, taking the Laplace transform of the equation, we get L[y 0 + 2y] = sL[y] − y(0) + 2L[y] = 1 e−3s e−3s − + 2 s2 s2 s Applying the initial condition and solving for L[y], we find that L[y] = Y (s) = = 1 e−3s e−3s 4 + + 2 + 2 2 s (s + 2) s (s + 2) s(s + 2) s + 2 4s2 + 1 (2s + 1)e−3s + . s2 (s + 2) s2 (s + 2) By partial fraction decomposition 4s2 + 1 s2 (s + 2) (2s + 1)e−3s s2 (s + 2) = − = 17 1 11 1 1 + + 2 4 s 2s 4 s+2 3 e−3s 1 e−3s 3 e−3s − + 4 s 2 s2 4 s+2 Therefore, Y (s) = − 11 1 1 17 1 3 e−3s 1 e−3s 3 e−3s + − + + + 4 s 2 s2 4 s+2 4 s 2 s2 4 s+2 and −2x + 3 u(x − 3) + 1 (x − 3) u(x − 3) − 3 e−2(x−3) u(x − 3) y(x) = L−1 [Y (s)] = − 14 + 12 x + 17 4 e 4 2 4 ( 1 1 −2x − 4 + 2 x + 17 0≤x<3 4e = 17 −2x 3 −2(x−3) x−1+ 4 e − 4e x≥3 151 The graph of y is y x 3 Example 2. Find the solution of the initial-value problem y 00 + 4y = f (x); where f (x) = ( 1 −2 y(0) = 1, y 0(0) = 0 0 ≤ x < π/2 x ≥ π/2. SOLUTION The function f has a jump discontinuity at x = π/2. Therefore, we express f in terms of u(x − π/2): f (x) = 1 − 3u(x − π/2). Applying the Laplace transform to the equation and using the initial conditions, we get L[y 00 + 4y] = s2 L[y] − sy(0) − y 0 (0) + 4L[y] = (s2 + 4)L[y] − s = 1 3e−πs/2 − s s 1 3e−πs/2 − . s s Solving for L[y] = Y (s), we have Y (s) = 1 1 s − 3e−πs/2 2 + 2 . + 4) s(s + 4) s + 4 s(s2 By partial fraction decomposition, 1 11 1 s = − . + 4) 4 s 4 s2 + 4 s(s2 Therefore, Y (s) = 11 1 s 3 e−πs/2 3 −πs/2 s s − − + e + 2 2 2 4 s 4 s +4 4 s 4 s +4 s +4 and y(x) = L−1 [Y (s)] = = 1 4 − 14 cos 2x − 34 u(x − π/2) + 34 cos 2(x − π/2)u(x − π/2) + cos 2x   14 + 34 cos 2x 0 ≤ x < π/2  − 12 x ≥ π/2. 152 The graph of y is y 1 x Π 2 -0.5 -1 Exercises 4.5 Solve the initial-value problem. 1. y 0 − 2y = f (x); y(0) = 2 where f (x) = 2. y 0 + 3y = f (x); y(0) = 1 4. y 00 + 4y = f (x); 1 0≤x<1 2 x≥1 . where f (x) = 3. y 00 + y = f (x); ( ( sin x 0≤x<π 0 x≥π . y(0) = 0, y 0(0) = 1 where ( 1 0≤x<1 f (x) = . 0 x≥1 y(0) = 1, y 0 (0) = 0 ( where 0 0≤x<2 1 x≥2 f (x) = 5. y 00 + 2y 0 + y = f (x); 6. y 00 + 4y = f (x); . y(0) = 0, y 0(0) = 0 ( 1 f (x) = x−1 where 0≤x<2 . x≥2 y(0) = 0, y 0 (0) = 0 where ( sin x 0 ≤ x < 2π f (x) = . 0 x ≥ 2π 153 7. y 00 − 4y 0 + 3y = f (x); 8. y 00 − 3y 0 + 2y = f (x); 9. y 00 + 2y 0 + y = f (x); 10. y 00 − 4y 0 + 4y = f (x); y(0) = 0, y 0 (0) = 0 ( −1 f (x) = 1 where 0≤x<1 . x≥1 y(0) = 0, y 0 (0) = 0 ( 0 f (x) = 2x − 4 where y(0) = 3, y 0(0) = −1 ( ex f (x) = ex − 1 where 0≤x<2 . x≥2 0≤x<1 . x≥1 y(0) = 1, y 0 (0) = −1 where ( e2x 0≤x<2 f (x) = . −e2x x≥2 154