By using the method of Laplace transforms, find the solution of the following initial value-problem: { J +y = u2(1)e-2(1-2) y(0) = 1
By using the method of Laplace transforms, find the solution of the following initial value-problem: { J +y = u2(1)e-2(1-2) y(0) = 1
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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![Table 6.1
Frequently Encountered Laplace Transforms.
y(t) Y)
Y(s) = Lly)
y(1) = L='1Y)
Y(s) = ly)
y(1) = ear
n!
Y (s) =
(s > 0)
Y(s) =
(s > a)
y(1) = "
y(1) = sin ar
Y(s) =
y(1) = cos at
y(1) = eat sinor
Y(s) =
y(t) = A coS ut
Y(s) =
(s – a)² + a?
(s - a)2 + w?
Zas
y(1) =I sin ar
Y(s) =
y(1) =I cos aue
Y(s) =
ru) -
-as
y(1) = ua(t)
Y(s) =
(s > 0)
y() = 8a(1)
Y(s) =e-as
Table 6.2
Rules for Laplace Transforms:
Given functions y(t) and w(t) with L[y] = Y(s) and L[w] = W(s) and constants a and a.
Rule for Laplace Transform
Rule for Inverse Laplace Transform
= sL[y]- y(0) = sY(s) - y(0)
Lly + w] = L[y] + L[w} = Y(s) + W(s)
L-IY + W) = L-(Y)+'(W) = y() + w(1)!
%3D
%3D
%3D
%3D
Llay) = aL[y] = aY(s)
(aY] = aL-(Y] = ay(1)
Llua(1)y(1 - a)] = e-a L[y] = e=afY (s)
Llee" y(1)] = Y(s - a)
-IY(s - a)) = L-Y) = y()](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fbb7f1ce8-2552-49b5-ba1d-534a58de11ae%2Fec706d1a-e266-452b-a164-73b420983d0c%2Fefrzafs_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Table 6.1
Frequently Encountered Laplace Transforms.
y(t) Y)
Y(s) = Lly)
y(1) = L='1Y)
Y(s) = ly)
y(1) = ear
n!
Y (s) =
(s > 0)
Y(s) =
(s > a)
y(1) = "
y(1) = sin ar
Y(s) =
y(1) = cos at
y(1) = eat sinor
Y(s) =
y(t) = A coS ut
Y(s) =
(s – a)² + a?
(s - a)2 + w?
Zas
y(1) =I sin ar
Y(s) =
y(1) =I cos aue
Y(s) =
ru) -
-as
y(1) = ua(t)
Y(s) =
(s > 0)
y() = 8a(1)
Y(s) =e-as
Table 6.2
Rules for Laplace Transforms:
Given functions y(t) and w(t) with L[y] = Y(s) and L[w] = W(s) and constants a and a.
Rule for Laplace Transform
Rule for Inverse Laplace Transform
= sL[y]- y(0) = sY(s) - y(0)
Lly + w] = L[y] + L[w} = Y(s) + W(s)
L-IY + W) = L-(Y)+'(W) = y() + w(1)!
%3D
%3D
%3D
%3D
Llay) = aL[y] = aY(s)
(aY] = aL-(Y] = ay(1)
Llua(1)y(1 - a)] = e-a L[y] = e=afY (s)
Llee" y(1)] = Y(s - a)
-IY(s - a)) = L-Y) = y()
Transcribed Image Text:By using the method of Laplace transforms, find the solution of the following initial
value-problem:
dy
+y = u2(1)e-2(1-2)
y(0)
dt
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