Laplace Transforms Solutions for Various Differential Equations, Assignments of Differential Equations

Download Laplace Transforms Solutions for Various Differential Equations and more Assignments Differential Equations in PDF only on Docsity! Solutions HW 11 10.1 3 (a) F (s) = 1 s− 3 (b) F (x) = 2 s+ 1 + 1 s (c) F (x) = 1 s− ln 2 + 6 s (d) F (x) = 1 s+ ln 3 8 We have cosh kx = 12 ( ekx + e−kx ) , so L(cosh kx) = 12 [ L(ekx) + L(e−kx) ] = 1 2(s− k) + 1 2(s+ k) . Similarly, L(sinh kx) = 12 [ L(ekx)− L(e−kx) ] = 1 2(s− k) − 1 2(s+ k) . 10 (a) 1 s2 + 5s+ 6 = 1 s+ 2 − 1 s+ 3 , so the inverse transform is e−2t − e−3t. (b) 1 s(s+ 1) = 1 s − 1 s+ 1 , so the inverse transform is 1− e−t. (c) 1 s(s+ 3) = 1 3 s − 1 3 s+ 3 , so the inverse transform is 13 ( 1− e−3t ) . (d) 1 s2 − 4 = 1 4 s− 2 − 1 4 s+ 2 , so the inverse transform is 14 ( e2t − e−2t ) = 1 2 sinh 2t. 1 2 (e) 1 s2 + 7s+ 12 = 1 s+ 3 − 1 s+ 4 , so the inverse transform is e−3t − e−4t. (f) 1 s2 + 6s+ 8 = 1 2(s+ 2) − 1 2(s+ 4) , so the inverse transform is 12 ( e−2t − e−4t ) . (g) 1 (s+ 1)(s+ 2)(s+ 3) = 1 2 s+ 1 − 1 s+ 2 + 1 2 s+ 3 , so the inverse transform is 12e −t − e−2t + 12e −3t. (h) s+ 4 (s+ 1)(s+ 2)(s+ 3) = 3 2 s+ 1 − 2 s+ 2 + 1 2 s+ 3 , so the inverse transform is 32e −t − 2e−2t + 12e −3t. 10.2 1 (a) The Laplace transform of the equation leads to X(s) = x0 s− k = 10 s− k , which implies x(t) = 10ekt. (b) The Laplace transform of the equation leads to X(s) = x0 − 100/s s− k = 1000− 100/s s− k , which implies x(t) = 1000ekt + 100 ( 1− ekt ) /k. (c) The Laplace transform of the equation leads to X(s) = 10x0 + 100/s 10s+ 1 = 30 + 100/s 10s+ 1 which implies x(t) = 100− 97e−t/10. 2 (a) The Laplace transform of the equation gives X(s) = s+ 1 s2 + 2 + s (s2 + 1)(s2 + 2) = s s2 + 1 + 1 s2 + 2 , which implies that x(t) = cos t+ sin √ 2 t√ 2 .