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
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](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F67a79aa2-f715-406c-8bee-8178252bb86d%2F5e77f987-12de-44b3-b7de-032b2ab30ec3%2F1a6559n_processed.jpeg&w=3840&q=75)
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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