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Solution: Let Y(s) = L[y(t)]. Laplace transforming the ODE, we get %3D 3 s2Y - sy(0) – y (0) + 9Y = Imposing the initial conditions and rearranging, we have (s° + 9) Y = s + 3 = 8+- The solution for Y is - नंक गमरला 3 Y(s) s2 +9 s(s² +9) From the tables, = cos(3t) s2 +9 Using partial fractions, we get s (s2 +9) From the tables, it follows that s2 +9

Solution: Let Y(s) = L[y(t)]. Laplace transforming the ODE, we get %3D 3 s2Y - sy(0) – y (0) + 9Y = Imposing the initial conditions and rearranging, we have (s° + 9) Y = s + 3 = 8+- The solution for Y is - नंक गमरला 3 Y(s) s2 +9 s(s² +9) From the tables, = cos(3t) s2 +9 Using partial fractions, we get s (s2 +9) From the tables, it follows that s2 +9

Calculus: Early Transcendentals
Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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How to get this by partial fraction. be specific
Introduction to Differential Equations: Sheet 10 - MATH 201 8. Use Laplace transforms to solve the initial value problem y" + 9y = 3, y(0) = 1, y(0) = 0 Solution: Let Y(s) = L[y(t)]. Laplace transforming the ODE, we get 3 s?Y – sy(0) – y (0) + 9Y = Imposing the initial conditions and rearranging, we have (s° + 9) Y = : s+ - The solution for Y is Y(s) = s2 +9 s (s2 + 9) From the tables, L-1 = cos(3t) Using partial fractions, we get 3 s (s² + 9) From the tables, it follows that 3 s2 +9) i fT 3 1 1 L-1 s (s² + 9). 33 cos(3t) Adding these two inverse Laplace transforms, we get the solution 1 2 y(t) =+ cos(3t) 3 University of California, San Diego Page 10 of 11
Transcribed Image Text:Introduction to Differential Equations: Sheet 10 - MATH 201 8. Use Laplace transforms to solve the initial value problem y" + 9y = 3, y(0) = 1, y(0) = 0 Solution: Let Y(s) = L[y(t)]. Laplace transforming the ODE, we get 3 s?Y – sy(0) – y (0) + 9Y = Imposing the initial conditions and rearranging, we have (s° + 9) Y = : s+ - The solution for Y is Y(s) = s2 +9 s (s2 + 9) From the tables, L-1 = cos(3t) Using partial fractions, we get 3 s (s² + 9) From the tables, it follows that 3 s2 +9) i fT 3 1 1 L-1 s (s² + 9). 33 cos(3t) Adding these two inverse Laplace transforms, we get the solution 1 2 y(t) =+ cos(3t) 3 University of California, San Diego Page 10 of 11
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