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To use the Laplace transform to solve the given initial value problem, we first take the transform of each dy + 4y = e5t. The strategy is that the new equation can be solved for dt member of the differential equation L{y} algebraically. Once solved, transforming back to an equation for y gives the solution we need to the original differential equation. dy + L{4y} = L{e$tz dt We now recall the following, where a and b are constants. |Ap sL{y} - y(0) dt By Theorem 7.2.2.: • Using linearity of L: L{ay} = aL{y} Liebt} = 1 • By Theorem 7.1.1.: S- b Applying these gives the following result. SL{y} – y(0) + LY} S -

To use the Laplace transform to solve the given initial value problem, we first take the transform of each dy + 4y = e5t. The strategy is that the new equation can be solved for dt member of the differential equation L{y} algebraically. Once solved, transforming back to an equation for y gives the solution we need to the original differential equation. dy + L{4y} = L{e$tz dt We now recall the following, where a and b are constants. |Ap sL{y} - y(0) dt By Theorem 7.2.2.: • Using linearity of L: L{ay} = aL{y} Liebt} = 1 • By Theorem 7.1.1.: S- b Applying these gives the following result. SL{y} – y(0) + LY} S -

Advanced Engineering Mathematics
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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To use the Laplace transform to solve the given initial value problem, we first take the transform of each member of the differential equation dy - + 4y = e. The strategy is that the new equation can be solved for dt L{y} algebraically. Once solved, transforming back to an equation for y gives the solution we need to the original differential equation. dy + L{4y} = L{eSt dt We now recall the following, where a and b are constants. dy • By Theorem 7.2.2.: SL{y} – y(0) dt • Using linearity of L: L{ay} = aL{y} 1. • By Theorem 7.1.1.: S- b Applying these gives the following result. SLy} - y(0) + Liy} S -
Transcribed Image Text:To use the Laplace transform to solve the given initial value problem, we first take the transform of each member of the differential equation dy - + 4y = e. The strategy is that the new equation can be solved for dt L{y} algebraically. Once solved, transforming back to an equation for y gives the solution we need to the original differential equation. dy + L{4y} = L{eSt dt We now recall the following, where a and b are constants. dy • By Theorem 7.2.2.: SL{y} – y(0) dt • Using linearity of L: L{ay} = aL{y} 1. • By Theorem 7.1.1.: S- b Applying these gives the following result. SLy} - y(0) + Liy} S -
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