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LaplaceTransDifferentialEq

# LaplaceTransDifferentialEq - Application of the Laplace...

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Application of the Laplace Transform to solve Linear Differential Equations The Laplace transform can be applied to solve initial value problem that contains homogeneous and non-homogeneous linear differential equations. Example : 1) Solve the I.V.P. d 2 y dt 2 + y = t y(0) = 1, y'(0) = ! 2 " # \$ % \$ Take the Laplace transform of the differential equation L {y’’(t)} + L {y(t)} = L {t} If we call Y(s) = L {y(t)}, Then L {y’’(t)} = s 2 Y(s) – s y(0) – y’(0) L {t} = 1/s 2 then s 2 Y(s) – s + 2 + Y(s) = 1/s 2 Solving for Y(s) Y(s)(s 2 + 1) = 1/s 2 + s – 2 Y(s) = 1 s 2 (s 2 + 1) + s ! 2 s 2 + 1 Now, we apply the inverse Laplace Transform y(t) = L -1 {Y(s)} = L -1 { 1 s 2 (s 2 + 1) } + L -1 { s ! 2 s 2 + 1 } 1 s 2 (s 2 + 1) = A s + B s 2 + Cs + D s 2 + 1 = (As + B) s 2 + 1 ( ) + Cs 3 + Ds 2 s 2 (s 2 + 1) (A + C) s 3 + (B + D)s 2 + As + B = 1, B = 1, A = 0, D = -1, C = 0 y(t) = L -1 { 1 s 2 } – L -1 { 1 s 2 + 1 } + L -1 { s s 2 + 1 } – 2 L -1 { 1 s 2 + 1 } = t + cost – 3 sin t. 2) Solve the I.V.P. d 2 y dt 2 ! 3 dy dt + 2y = 4e 2t y(0) = ! 3, y'(0) = 5 " # \$ % \$

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Take the Laplace transform of the differential equation L {y’’(t)} - 3 L {y’(t)} + 2 L {y(t)} =4 L {e 2t } If we call Y(s) = L {y(t)}, Then L {y’’(t)} = s 2 Y(s) – sy(0) – y’(0) L {y’(t)} = sY(s) – y(0) L {e 2t } = 1/(s-2) then s 2 Y(s) + 3s – 5 – 3sY(s) - 9 + 2Y(s) = 4/(s-2) Solving for Y(s) Y(s)(s 2 – 3s + 2) = 4/(s-2) - 3s + 14 Y(s) = 4 s ! 2 ( ) (s ! 1) s ! 2 ( ) + 14 ! 3s s ! 2 ( ) (s ! 1) = 4 + 14 ! 3s ( ) s ! 1 ( ) (s ! 2) 2 (s ! 1) = ! 3s 2 + 20s ! 24 (s ! 2) 2 (s ! 1) Now, we apply the inverse Laplace Transform y(t) = L -1 {Y(s)} = L -1 { ! 3s 2 + 20s ! 24 s ! 2 ( ) 2 (s ! 1) } ! 3s 2 + 20s ! 24 s ! 2 ( ) 2 s ! 1 ( ) = A s ! 2 + B s ! 2 ( ) 2 + C s ! 1 = A(s ! 2)(s ! 1) + B(s ! 1) + C(s ! 2) 2 s ! 2 ( ) 2 s ! 1 ( ) then (A + C) s 2 + (-3A + B – 4C)s + (2A – B + 4C) = -3s 2 + 20s - 24 A + C = -3 , -3A + B – 4C = 20, 2A – B + 4C = -24, A = 4, C = -7, B = 4 y(t) = 4 L -1 { 1 s ! 2 } + 4 L -1 { 1 s ! 2 ( ) 2 } – 7 L -1 { 1 s ! 1 } = 4e 2t + 4te 2t – 7e t . 3) Solve the I.V.P. d 3 y dt 3 ! 3 d 2 y dt 2 + 3 dy dt ! y = t 2 e t y(0) = 1, y'(0) = 0,y''(0) = 2 " # \$ % \$ Take the Laplace transform of the differential equation L { y’’’(t)} - 3 L {y’’(t)} + 3 L {y’(t)} - L {y(t)} = L {t 2 e t } If we call Y(s) = L {y(t)}, Then L {y’’’(t)} = s 3 Y(s) – s 2 y(0) – sy’(0) – y’’(0) L {y’’(t)} = s 2 Y(s) – sy(0) – y’(0) L {y’(t)} = sY(s) – y(0) L {t 2 e t } = 2/(s-1) 3 since L {t n } = n!/s n+1 then s 3 Y(s) – s 2 – 2 - 3s 2 Y(s) + 3s – 3sY(s) - 3 - Y(s) = 2/(s-1) 3
Solving for Y(s) Y(s)(s 3 - 3s 2 + 3s - 1) = 2 s ! 1 ( ) 3 + (s 2

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LaplaceTransDifferentialEq - Application of the Laplace...

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