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Use the Laplace transform to solve the given initial value problem. For technical reasons, write u for the Heaviside function that turns on at c,not u (t). y"-2y'-24y=8₂(t) Y(0) = 2, y'(0) = 32 -(2s) + (2s +36) (s-6) (s+4) X (s-6) (s+4) Y(s) = y(t) =

Use the Laplace transform to solve the given initial value problem. For technical reasons, write u for the Heaviside function that turns on at c,not u (t). y"-2y'-24y=8₂(t) Y(0) = 2, y'(0) = 32 -(2s) + (2s +36) (s-6) (s+4) X (s-6) (s+4) Y(s) = y(t) =

Algebra & Trigonometry with Analytic Geometry
Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN:9781133382119
Author:Swokowski
Publisher:Swokowski
Chapter6: The Trigonometric Functions
Section6.6: Additional Trigonometric Graphs
Problem 77E
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Laplace transform problem
### Solving Initial Value Problems Using the Laplace Transform #### Problem Statement Use the Laplace transform to solve the given initial value problem. For technical reasons, write \( u_C \) for the Heaviside function that turns on at \( C \), and not \( u_C(t) \). \[ y'' - 2y' - 24y = \delta_2(t) \] \[ y(0) = 2, \quad y'(0) = 32 \] #### Laplace Transform Solution First, we take the Laplace transform of the given differential equation. The Laplace transform of the given equation results in: \[ Y(s) = \frac{e^{-2s}}{(s-6)(s+4)} + \frac{(2s + 36)}{(s-6)(s+4)} \] (Note: The equation above contains an error marked in red.) To solve \( y(t) \), we need to find the inverse Laplace transform of \( Y(s) \). #### Final Answer Expressed solution for \( y(t) \): \[ y(t) = \] #### Graphs/Diagrams In this instance, no specific graphs or diagrams are provided. Instead, the primary focus is on solving the differential equation using the Laplace transform method. The provided equation \( Y(s) \) is the transformed function in the Laplace domain. To obtain \( y(t) \), you must perform the inverse Laplace transform considering the properties and initial conditions given. For additional support, refer to the detailed steps in taking the inverse Laplace transform and solving for \( y(t) \), as this involves partial fraction decomposition and applying inverse transform techniques.
Transcribed Image Text:### Solving Initial Value Problems Using the Laplace Transform #### Problem Statement Use the Laplace transform to solve the given initial value problem. For technical reasons, write \( u_C \) for the Heaviside function that turns on at \( C \), and not \( u_C(t) \). \[ y'' - 2y' - 24y = \delta_2(t) \] \[ y(0) = 2, \quad y'(0) = 32 \] #### Laplace Transform Solution First, we take the Laplace transform of the given differential equation. The Laplace transform of the given equation results in: \[ Y(s) = \frac{e^{-2s}}{(s-6)(s+4)} + \frac{(2s + 36)}{(s-6)(s+4)} \] (Note: The equation above contains an error marked in red.) To solve \( y(t) \), we need to find the inverse Laplace transform of \( Y(s) \). #### Final Answer Expressed solution for \( y(t) \): \[ y(t) = \] #### Graphs/Diagrams In this instance, no specific graphs or diagrams are provided. Instead, the primary focus is on solving the differential equation using the Laplace transform method. The provided equation \( Y(s) \) is the transformed function in the Laplace domain. To obtain \( y(t) \), you must perform the inverse Laplace transform considering the properties and initial conditions given. For additional support, refer to the detailed steps in taking the inverse Laplace transform and solving for \( y(t) \), as this involves partial fraction decomposition and applying inverse transform techniques.
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