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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Question
Kindly provide a detailed and thorough explanation about this example figure. It is only one example in two pictures.
![Determine the general solution to (D² + 4)(f) = sin(2x) satisfying the initial condition:
f (0) = 0,
D(f(0)) = 0
Solution:
The related homogeneous ODE is (D² + 4)(f) = 0 with characteristic polynomial r2 + 4 = 0. The roots
are +2i (multiplicity 1). Thus, the general solution to the homogeneous solution is
uc(x) = c2 cos(2x) + c1 sin(2x)
Finding an annihilator for sin 2x, A(D) = (D - 2i)(D + 2i) = D² + 4. Thus, a solution to the original
equation also satisfies
(D² + 4)(D² + 4)f) = (D² + 4)(sin(2x))
(D² + 4)²(f) = 0
The characteristic polynomial is (r² + 4)² with root +2i (multiplicity 2). The corresponding solutions are
e0(x) cos(2x) = cos(2x); e0(x) sin(2x) = sin(2x)
xe0(x) cos(2x) = x cos(2x); xeº(x) sin(2x) = x sin(2x)
±2i](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F32831efc-2eba-44b3-ba15-799844cfb699%2F4f3db276-8d76-4422-9ce2-12d09dcf5c7c%2Fw4pcwlj_processed.png&w=3840&q=75)
Transcribed Image Text:Determine the general solution to (D² + 4)(f) = sin(2x) satisfying the initial condition:
f (0) = 0,
D(f(0)) = 0
Solution:
The related homogeneous ODE is (D² + 4)(f) = 0 with characteristic polynomial r2 + 4 = 0. The roots
are +2i (multiplicity 1). Thus, the general solution to the homogeneous solution is
uc(x) = c2 cos(2x) + c1 sin(2x)
Finding an annihilator for sin 2x, A(D) = (D - 2i)(D + 2i) = D² + 4. Thus, a solution to the original
equation also satisfies
(D² + 4)(D² + 4)f) = (D² + 4)(sin(2x))
(D² + 4)²(f) = 0
The characteristic polynomial is (r² + 4)² with root +2i (multiplicity 2). The corresponding solutions are
e0(x) cos(2x) = cos(2x); e0(x) sin(2x) = sin(2x)
xe0(x) cos(2x) = x cos(2x); xeº(x) sin(2x) = x sin(2x)
±2i
![Hence, our particular solution is of the form
u,(x) = c4x cos(2x) + c3x sin(2x) + C2 cos(2x) + c, sin(2x)
which can be further simplified into
up(x) = c4x cos(2x)+ c3x sin(2x)
Determining the values of the constants,
(D² + 4)(f) = sin(2x)
(D² + 4)(c4x cos(2x) + c3x sin(2x)) = sin(2x)
D²(c4x cos(2x) + c3x sin(2x)) + 4(c4x cos(2x) + c3x sin(2x)) = sin(2x)
D(-2c,x sin(2x) + C4 cos(2x) + 2c3x cos(2x) + cz sin(2x)) + 4c,x cos(2x) + 4c3x sin(2x) = sin(2x)
-4c4 sin(2x) + 4c3 cos(2x) = sin(2x)
We see that 4c3 = 0 and -4c, = 1. Hence,
%3D
C4 =
Therefore, u, (x) = -x cos(2x) and the complete solution is of the form
1
-x cos(2x) + c2 cos(2x) + c, sin(2x)
We use the initial conditions to determine c2 and c . Using the first initial condition,
f(0) = c2 cos(0) + c; sin(0) = c2 = 0
Using the second initial condition and substituting the value of c2,
1
Df
D(-x cos(2x) + c2 cos(2x) + c; sin(2x))
%3D
= D
: cos(2x) + c, sin(2x)
1
=x sin(2x) –cos(2x) + 2c, cos(2x)
1
1
D(f(0)) = -cos(0) + 2c, cos(0) = -i+ 2c, = 0
Hence, c, = Therefore, the complete solution to the problem is
1
1
f(x) = -x cos(2x) + sin(2x)](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F32831efc-2eba-44b3-ba15-799844cfb699%2F4f3db276-8d76-4422-9ce2-12d09dcf5c7c%2Fl1e6paa_processed.png&w=3840&q=75)
Transcribed Image Text:Hence, our particular solution is of the form
u,(x) = c4x cos(2x) + c3x sin(2x) + C2 cos(2x) + c, sin(2x)
which can be further simplified into
up(x) = c4x cos(2x)+ c3x sin(2x)
Determining the values of the constants,
(D² + 4)(f) = sin(2x)
(D² + 4)(c4x cos(2x) + c3x sin(2x)) = sin(2x)
D²(c4x cos(2x) + c3x sin(2x)) + 4(c4x cos(2x) + c3x sin(2x)) = sin(2x)
D(-2c,x sin(2x) + C4 cos(2x) + 2c3x cos(2x) + cz sin(2x)) + 4c,x cos(2x) + 4c3x sin(2x) = sin(2x)
-4c4 sin(2x) + 4c3 cos(2x) = sin(2x)
We see that 4c3 = 0 and -4c, = 1. Hence,
%3D
C4 =
Therefore, u, (x) = -x cos(2x) and the complete solution is of the form
1
-x cos(2x) + c2 cos(2x) + c, sin(2x)
We use the initial conditions to determine c2 and c . Using the first initial condition,
f(0) = c2 cos(0) + c; sin(0) = c2 = 0
Using the second initial condition and substituting the value of c2,
1
Df
D(-x cos(2x) + c2 cos(2x) + c; sin(2x))
%3D
= D
: cos(2x) + c, sin(2x)
1
=x sin(2x) –cos(2x) + 2c, cos(2x)
1
1
D(f(0)) = -cos(0) + 2c, cos(0) = -i+ 2c, = 0
Hence, c, = Therefore, the complete solution to the problem is
1
1
f(x) = -x cos(2x) + sin(2x)
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