Solve the given initial value problem using the method of Laplace transforms. 5, Ost≤7, y" + 2y' + 5y = g(t), y(0) = -2, y'(0) = 0, where g(t) = 10, 7
Solve the given initial value problem using the method of Laplace transforms. 5, Ost≤7, y" + 2y' + 5y = g(t), y(0) = -2, y'(0) = 0, where g(t) = 10, 7
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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![Solve the given initial value problem using the method of Laplace transforms.
5, 0st≤7,
y" + 2y' + 5y = g(t), y(0) = -2, y'(0) = 0, where g(t) = 10, 7<t<14,
0,
14 <t
Click here to view the table of Laplace transforms.
Click here to view the table of properties of Laplace transforms.
***
The solution has the form y(t) = p(t) + q(t)u(ta) + r(t)u(t-B), where u(t) is the unit step
function. Let x <ß. Identify the values of a and B
B =
(Simplify your answers.)](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F368b3915-4d06-4b5c-8ef4-1295bb259aad%2F50815ca5-c177-4850-a4d6-9a04e222753c%2Fdevmm3_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Solve the given initial value problem using the method of Laplace transforms.
5, 0st≤7,
y" + 2y' + 5y = g(t), y(0) = -2, y'(0) = 0, where g(t) = 10, 7<t<14,
0,
14 <t
Click here to view the table of Laplace transforms.
Click here to view the table of properties of Laplace transforms.
***
The solution has the form y(t) = p(t) + q(t)u(ta) + r(t)u(t-B), where u(t) is the unit step
function. Let x <ß. Identify the values of a and B
B =
(Simplify your answers.)
![f(t)
1
eat
t, n=1,2,...
sin bt
cos bt
eatin, n=1,2,...
eat sin bt
eat cos bt
u(t-a)
Ilab(t)
£{f+g} = £{f} + £{g}
F(s)= {f}(s)
1
£{eatf(t)} (s) = {f}(s-a)
.s>0
S
1
s-a'
n!
87+1.830
dn
b
s² + b²
S
s²+b²
S>0
s>0
s>0
n!
(s-a)n+1
b
(s-a)² + b²
s-a
(s-a)2+b2
as
S
eas-e-bs
S
{cf} c{f} for any constant c
=
s>a
s>a
s>0
Properties of Laplace Transforms
.s>a
.s>0
£{f}(s) = sc{f}(s)-f(0)
{f''(s) = s² {f}(s)-sf(0) -f'(0)
{f(n)} (s) = s"{f}(s)-sn-1f(0)-s-2'(0)--(n-1)(0)
c{tf(t)} (s) = (-1)- (c{f}(s))
ds
£¯¹ {F₁ +F₂} = £¯ ¹ {F₁} + £¯¹ {F2}
£¹(CF} = c£ ¹{F}
e{f(t-a)u(t-a))(s) = eas F(s)
¹{e-as F(s)} (t) = f(t-a)u(t-a)
{g(t)u(t-a)(s) = eas {g(t + a)(s)](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F368b3915-4d06-4b5c-8ef4-1295bb259aad%2F50815ca5-c177-4850-a4d6-9a04e222753c%2Fa2pv7lv.jpeg&w=3840&q=75)
Transcribed Image Text:f(t)
1
eat
t, n=1,2,...
sin bt
cos bt
eatin, n=1,2,...
eat sin bt
eat cos bt
u(t-a)
Ilab(t)
£{f+g} = £{f} + £{g}
F(s)= {f}(s)
1
£{eatf(t)} (s) = {f}(s-a)
.s>0
S
1
s-a'
n!
87+1.830
dn
b
s² + b²
S
s²+b²
S>0
s>0
s>0
n!
(s-a)n+1
b
(s-a)² + b²
s-a
(s-a)2+b2
as
S
eas-e-bs
S
{cf} c{f} for any constant c
=
s>a
s>a
s>0
Properties of Laplace Transforms
.s>a
.s>0
£{f}(s) = sc{f}(s)-f(0)
{f''(s) = s² {f}(s)-sf(0) -f'(0)
{f(n)} (s) = s"{f}(s)-sn-1f(0)-s-2'(0)--(n-1)(0)
c{tf(t)} (s) = (-1)- (c{f}(s))
ds
£¯¹ {F₁ +F₂} = £¯ ¹ {F₁} + £¯¹ {F2}
£¹(CF} = c£ ¹{F}
e{f(t-a)u(t-a))(s) = eas F(s)
¹{e-as F(s)} (t) = f(t-a)u(t-a)
{g(t)u(t-a)(s) = eas {g(t + a)(s)
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