10. This problem indicates why we can impose only n initial conditions on a solution of an nth-order linear differential equation. (a) Given the equation y" + py' + qy = 0, explain why the value of y"(a) is determined by the values of y(a) and y'(a). (b) Prove that the equation y"- 2y'-5y 0 has a solution satisfying the conditions y(0) = 1, y'(0) = 0, and y"(0) = C if and only if C= 5. %3D %3D (a) Let a be any number. Substitute x = a in y"+ py'+ qy = 0, thus (1) Rewrite y"(a) in terms of y(a) and y'(a). Thus, y"(a) = (2) %3D Thus the value of y"(a) is determined if the values of (3) are known. (b) Given the equation y"- 2y'- 5y = 0 and initial conditions y(0) = 1, y'(0) = 0, and y"(0) = C, what should be done to show that the differential equation has one solution if and only if C= 5? %3D %3D O A. Substitute y = 5 into the equation y" - 2y'- 5y = 0. O B. Substitute y = C into the equation y"- 2y' - 5y = 0. %3D %3D C. Substitute the initial conditions into the equation y"- 2y'- 5y 0. | D. Substitute C= 5 into the equation y"- 2y'- 5y = C. Applying the substitution above results in the equation (4) Why is the proof complete? O A. The proof is complete because solving this equation yields y = %3D B. The proof is complete because solving this equation yields C= C. The proof is complete because simplifying both sides of this equation yields a contradiction. (1) O -y"(a)+ p(a)y'(a) – q(a)y(a) = 0. O y" (a) + p(a)y'(a) + q(a)y(a) = 0. O y"(a) - p(a)y'(a) – q(a)y(a) = 0. O y"(a)- p(a)y'(a) + q(a)y(a) = 0. (2) O -p(a)y'(a) – q(a)y(a). O- p(a)y'(a) + q(a)y(a). p(a)y' (a) – q(a)y(a). O p(a)y' (a) + q(a)y (a). O y'(a), q(a) and y (a) O p(a), y'(a), q(a) and y(a) O y'(a) and y (a) (3) %3D %3D p(a), and q(a) (4) O y"-2y'- 5y = 5 O C- 2(0) - 5(1) = 0. O C- 2(C)- 5(C) = 0. O 5-2(5) - 5(5) = 0
10. This problem indicates why we can impose only n initial conditions on a solution of an nth-order linear differential equation. (a) Given the equation y" + py' + qy = 0, explain why the value of y"(a) is determined by the values of y(a) and y'(a). (b) Prove that the equation y"- 2y'-5y 0 has a solution satisfying the conditions y(0) = 1, y'(0) = 0, and y"(0) = C if and only if C= 5. %3D %3D (a) Let a be any number. Substitute x = a in y"+ py'+ qy = 0, thus (1) Rewrite y"(a) in terms of y(a) and y'(a). Thus, y"(a) = (2) %3D Thus the value of y"(a) is determined if the values of (3) are known. (b) Given the equation y"- 2y'- 5y = 0 and initial conditions y(0) = 1, y'(0) = 0, and y"(0) = C, what should be done to show that the differential equation has one solution if and only if C= 5? %3D %3D O A. Substitute y = 5 into the equation y" - 2y'- 5y = 0. O B. Substitute y = C into the equation y"- 2y' - 5y = 0. %3D %3D C. Substitute the initial conditions into the equation y"- 2y'- 5y 0. | D. Substitute C= 5 into the equation y"- 2y'- 5y = C. Applying the substitution above results in the equation (4) Why is the proof complete? O A. The proof is complete because solving this equation yields y = %3D B. The proof is complete because solving this equation yields C= C. The proof is complete because simplifying both sides of this equation yields a contradiction. (1) O -y"(a)+ p(a)y'(a) – q(a)y(a) = 0. O y" (a) + p(a)y'(a) + q(a)y(a) = 0. O y"(a) - p(a)y'(a) – q(a)y(a) = 0. O y"(a)- p(a)y'(a) + q(a)y(a) = 0. (2) O -p(a)y'(a) – q(a)y(a). O- p(a)y'(a) + q(a)y(a). p(a)y' (a) – q(a)y(a). O p(a)y' (a) + q(a)y (a). O y'(a), q(a) and y (a) O p(a), y'(a), q(a) and y(a) O y'(a) and y (a) (3) %3D %3D p(a), and q(a) (4) O y"-2y'- 5y = 5 O C- 2(0) - 5(1) = 0. O C- 2(C)- 5(C) = 0. O 5-2(5) - 5(5) = 0
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
Related questions
Question
please show steps
![10. This problem indicates why we can impose only n initial conditions on a solution of an nth-order linear differential
equation.
(a) Given the equation y" + py' + qy = 0, explain why the value of y"(a) is determined by the values of y(a) and y'(a).
(b) Prove that the equation y"- 2y'-5y 0 has a solution satisfying the conditions y(0) = 1, y'(0) = 0, and y"(0) = C if
and only if C= 5.
%3D
%3D
(a) Let a be any number. Substitute x = a in y"+ py'+ qy = 0, thus (1)
Rewrite y"(a) in terms of y(a) and y'(a). Thus, y"(a) = (2)
%3D
Thus the value of y"(a) is determined if the values of (3)
are known.
(b) Given the equation y"- 2y'- 5y = 0 and initial conditions y(0) = 1, y'(0) = 0, and y"(0) = C, what should be done to
show that the differential equation has one solution if and only if C= 5?
%3D
%3D
O A. Substitute y = 5 into the equation y" - 2y'- 5y = 0.
O B. Substitute y = C into the equation y"- 2y' - 5y = 0.
%3D
%3D
C. Substitute the initial conditions into the equation y"- 2y'- 5y 0.
|
D. Substitute C= 5 into the equation y"- 2y'- 5y = C.
Applying the substitution above results in the equation (4)
Why is the proof complete?
O A. The proof is complete because solving this equation yields y =
%3D
B. The proof is complete because solving this equation yields C=
C. The proof is complete because simplifying both sides of this equation yields a contradiction.
(1) O -y"(a)+ p(a)y'(a) – q(a)y(a) = 0.
O y" (a) + p(a)y'(a) + q(a)y(a) = 0.
O y"(a) - p(a)y'(a) – q(a)y(a) = 0.
O y"(a)- p(a)y'(a) + q(a)y(a) = 0.
(2) O -p(a)y'(a) – q(a)y(a).
O- p(a)y'(a) + q(a)y(a).
p(a)y' (a) – q(a)y(a).
O p(a)y' (a) + q(a)y (a).
O y'(a), q(a) and y (a)
O p(a), y'(a), q(a) and y(a)
O y'(a) and y (a)
(3)
%3D
%3D
p(a), and q(a)
(4) O y"-2y'- 5y = 5
O C- 2(0) - 5(1) = 0.
O C- 2(C)- 5(C) = 0.
O 5-2(5) - 5(5) = 0](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F9e7b22d7-3558-4370-8e59-8331abbb5ed3%2F168b05cb-97f4-4009-a31d-e7362abe7739%2F9mvwmu.jpeg&w=3840&q=75)
Transcribed Image Text:10. This problem indicates why we can impose only n initial conditions on a solution of an nth-order linear differential
equation.
(a) Given the equation y" + py' + qy = 0, explain why the value of y"(a) is determined by the values of y(a) and y'(a).
(b) Prove that the equation y"- 2y'-5y 0 has a solution satisfying the conditions y(0) = 1, y'(0) = 0, and y"(0) = C if
and only if C= 5.
%3D
%3D
(a) Let a be any number. Substitute x = a in y"+ py'+ qy = 0, thus (1)
Rewrite y"(a) in terms of y(a) and y'(a). Thus, y"(a) = (2)
%3D
Thus the value of y"(a) is determined if the values of (3)
are known.
(b) Given the equation y"- 2y'- 5y = 0 and initial conditions y(0) = 1, y'(0) = 0, and y"(0) = C, what should be done to
show that the differential equation has one solution if and only if C= 5?
%3D
%3D
O A. Substitute y = 5 into the equation y" - 2y'- 5y = 0.
O B. Substitute y = C into the equation y"- 2y' - 5y = 0.
%3D
%3D
C. Substitute the initial conditions into the equation y"- 2y'- 5y 0.
|
D. Substitute C= 5 into the equation y"- 2y'- 5y = C.
Applying the substitution above results in the equation (4)
Why is the proof complete?
O A. The proof is complete because solving this equation yields y =
%3D
B. The proof is complete because solving this equation yields C=
C. The proof is complete because simplifying both sides of this equation yields a contradiction.
(1) O -y"(a)+ p(a)y'(a) – q(a)y(a) = 0.
O y" (a) + p(a)y'(a) + q(a)y(a) = 0.
O y"(a) - p(a)y'(a) – q(a)y(a) = 0.
O y"(a)- p(a)y'(a) + q(a)y(a) = 0.
(2) O -p(a)y'(a) – q(a)y(a).
O- p(a)y'(a) + q(a)y(a).
p(a)y' (a) – q(a)y(a).
O p(a)y' (a) + q(a)y (a).
O y'(a), q(a) and y (a)
O p(a), y'(a), q(a) and y(a)
O y'(a) and y (a)
(3)
%3D
%3D
p(a), and q(a)
(4) O y"-2y'- 5y = 5
O C- 2(0) - 5(1) = 0.
O C- 2(C)- 5(C) = 0.
O 5-2(5) - 5(5) = 0
Expert Solution
![](/static/compass_v2/shared-icons/check-mark.png)
This question has been solved!
Explore an expertly crafted, step-by-step solution for a thorough understanding of key concepts.
This is a popular solution!
Trending now
This is a popular solution!
Step by step
Solved in 2 steps with 2 images
![Blurred answer](/static/compass_v2/solution-images/blurred-answer.jpg)
Recommended textbooks for you
![Advanced Engineering Mathematics](https://www.bartleby.com/isbn_cover_images/9780470458365/9780470458365_smallCoverImage.gif)
Advanced Engineering Mathematics
Advanced Math
ISBN:
9780470458365
Author:
Erwin Kreyszig
Publisher:
Wiley, John & Sons, Incorporated
![Numerical Methods for Engineers](https://www.bartleby.com/isbn_cover_images/9780073397924/9780073397924_smallCoverImage.gif)
Numerical Methods for Engineers
Advanced Math
ISBN:
9780073397924
Author:
Steven C. Chapra Dr., Raymond P. Canale
Publisher:
McGraw-Hill Education
![Introductory Mathematics for Engineering Applicat…](https://www.bartleby.com/isbn_cover_images/9781118141809/9781118141809_smallCoverImage.gif)
Introductory Mathematics for Engineering Applicat…
Advanced Math
ISBN:
9781118141809
Author:
Nathan Klingbeil
Publisher:
WILEY
![Advanced Engineering Mathematics](https://www.bartleby.com/isbn_cover_images/9780470458365/9780470458365_smallCoverImage.gif)
Advanced Engineering Mathematics
Advanced Math
ISBN:
9780470458365
Author:
Erwin Kreyszig
Publisher:
Wiley, John & Sons, Incorporated
![Numerical Methods for Engineers](https://www.bartleby.com/isbn_cover_images/9780073397924/9780073397924_smallCoverImage.gif)
Numerical Methods for Engineers
Advanced Math
ISBN:
9780073397924
Author:
Steven C. Chapra Dr., Raymond P. Canale
Publisher:
McGraw-Hill Education
![Introductory Mathematics for Engineering Applicat…](https://www.bartleby.com/isbn_cover_images/9781118141809/9781118141809_smallCoverImage.gif)
Introductory Mathematics for Engineering Applicat…
Advanced Math
ISBN:
9781118141809
Author:
Nathan Klingbeil
Publisher:
WILEY
![Mathematics For Machine Technology](https://www.bartleby.com/isbn_cover_images/9781337798310/9781337798310_smallCoverImage.jpg)
Mathematics For Machine Technology
Advanced Math
ISBN:
9781337798310
Author:
Peterson, John.
Publisher:
Cengage Learning,
![Basic Technical Mathematics](https://www.bartleby.com/isbn_cover_images/9780134437705/9780134437705_smallCoverImage.gif)
![Topology](https://www.bartleby.com/isbn_cover_images/9780134689517/9780134689517_smallCoverImage.gif)