### Applying the Inverse Laplace TransformTo solve for \( y \), we apply the inverse Laplace transform \( \mathcal{L}^{-1} \) to each term of \( \mathcal{L}\{y\} \). Doing so gives the following result. We then use linearity to rewrite the terms on the right side of the equation.\[\mathcal{L}\{y\} = \frac{1}{36} \left( \frac{1}{s} \right) - \frac{18}{35} \left( \frac{1}{s-1} \right) + \frac{823}{360} \left( \frac{1}{s-6} \right) - \frac{403}{504} \left( \frac{1}{s+6} \right)\]\[y(t) = \mathcal{L}^{-1} \left\{ \frac{1}{36} \left( \frac{1}{s} \right) - \mathcal{L}^{-1} \left\{ \frac{18}{35} \left( \frac{1}{s-1} \right) + \mathcal{L}^{-1} \left\{ \frac{823}{360} \left( \frac{1}{s-6} \right) - \mathcal{L}^{-1} \left\{ \frac{403}{504} \left( \frac{1}{s+6} \right) \right\} \right\} \right\}\]Breaking it down using linearity:\[= \frac{1}{36} \mathcal{L}^{-1} \left\{ \frac{1}{s} \right\} - \frac{18}{35} \mathcal{L}^{-1} \left\{ \frac{1}{s-1} \right\} + \frac{823}{360} \mathcal{L}^{-1} \left\{ \frac{1}{s-6} \right\} - \frac{403}{504} \mathcal{L}^{-1} \left\{ \frac{1}{s+6} \right\}\]### Laplace Transform PropertiesWe now recall the following Laplace transforms

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Next, to solve for y we apply the inverse Laplace transform - to each term of £{y}. Doing so gives the following result. We then use linearity to rewrite the terms on the right side of the equation. 1 18 1 x{y} = 3/6 (²) - 38 (5²1) + 3230 (5²6) - s 35 y(t) = y(t) = -1, = 1 18 1 -{()}. ( ² )} - 2 - ¹({ ¹8 ( ²₁ )} 36 35 1 - 2 - 1 { 1 } - 18 % - L- 36 35 Applying these gives the following result. 1 18 36 μ(1). u(t) 35 - Transform 11: If L{f(t)} = F(s) = == e². s-1 403 -6t ∙e 504 We now recall the following Laplace transforms from Appendix C. Transform 1: If £{f(t)} = F(s) ==—₁ , then f(t) = 1. S + 1 s-a "" + 823 61 360 823 360 403 1 e-1). + £¯ L X 504 L then f(t)= eat s +6 823 360 s- 1 (6 +-6)} - 2 - 1 504 ( 5 + 6) 1 11 ¹5+6] 11 S-6 403 504 L

Next, to solve for y we apply the inverse Laplace transform - to each term of £{y}. Doing so gives the following result. We then use linearity to rewrite the terms on the right side of the equation. 1 18 1 x{y} = 3/6 (²) - 38 (5²1) + 3230 (5²6) - s 35 y(t) = y(t) = -1, = 1 18 1 -{()}. ( ² )} - 2 - ¹({ ¹8 ( ²₁ )} 36 35 1 - 2 - 1 { 1 } - 18 % - L- 36 35 Applying these gives the following result. 1 18 36 μ(1). u(t) 35 - Transform 11: If L{f(t)} = F(s) = == e². s-1 403 -6t ∙e 504 We now recall the following Laplace transforms from Appendix C. Transform 1: If £{f(t)} = F(s) ==—₁ , then f(t) = 1. S + 1 s-a "" + 823 61 360 823 360 403 1 e-1). + £¯ L X 504 L then f(t)= eat s +6 823 360 s- 1 (6 +-6)} - 2 - 1 504 ( 5 + 6) 1 11 ¹5+6] 11 S-6 403 504 L

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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