Boundary Value Problem for y'' + y = A + sin(2x) with y(0) = y'(\pi/2) = 2

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Homework Help Overview

This discussion revolves around a boundary value problem (BVP) defined by the differential equation y'' + y = A + sin(2x) with boundary conditions y(0) = 2 and y'(\pi/2) = 2. The participants are exploring the implications of the parameter A, which is a real number, on the existence and uniqueness of solutions.

Discussion Character

  • Exploratory, Assumption checking

Approaches and Questions Raised

  • Participants discuss the general form of the solution, including the homogeneous and particular solutions. There is confusion regarding how to determine the values of A that lead to unique, infinitely many, or no solutions. Questions arise about the implications of A being a real number and how it affects the solution structure.

Discussion Status

Some participants express uncertainty about how to categorize the values of A in relation to the existence of solutions. There is an acknowledgment that A could potentially take on all real values, but clarity is lacking on how this translates to the types of solutions available. The discussion is ongoing, with participants seeking to clarify these points.

Contextual Notes

Participants are grappling with the definitions and implications of unique versus infinitely many solutions, as well as the conditions under which no solutions exist. There is a need for further exploration of these concepts within the context of the given BVP.

EvilKermit
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Homework Statement



a) Solve for the BVP. Where A is a real number.
b) For what values of A does there exist a unique solutions? What is the solution?
c) For what values of A do there exist infinitely many solutions?
d) For what values of A do there exist no solutions?


Homework Equations



y'' + y = A + sin(2x)
y(0) = y'([tex]\pi[/tex]/2) = 2

The Attempt at a Solution




y = yh + yp
0 = 1+ [tex]\lambda[/tex]2
yh =c1*cos(x) + c2*cos(x)

yp = A + B*sin(2x)
y = A + B*sin(2x)
y'' = -4B*sin(2x)
A + sin(2x) = A -3B*sin(2x)
A = A, B = -1/3
yp = A - 1/3*sin(2x)


y = A - 1/3*sin(2x) + c1*cos(x) + c2*sin(x)
y' = - 2/3*sin(2x) - c1*sin(x) + c2*cos(x)
2 = A + c1
2 = 2/3 + c2
c2 = 8/3

y = A - 1/3*sin(2x) + c1*cos(x) + 8/3*sin(x)

I'm confused about answering the questions. A would be equal to all real numbers, since one could solve for c1. How can I give the solution? There is a unique solution for each value of A, which I would have to write infinite solutions. And there is no value of A when there is an inifiite amount of solutions or no values.

IF A is defined, what would the answer be?
 
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EvilKermit said:

Homework Statement



a) Solve for the BVP. Where A is a real number.
b) For what values of A does there exist a unique solutions? What is the solution?
c) For what values of A do there exist infinitely many solutions?
d) For what values of A do there exist no solutions?


Homework Equations



y'' + y = A + sin(2x)
y(0) = y'([tex]\pi[/tex]/2) = 2

The Attempt at a Solution




y = yh + yp
0 = 1+ [tex]\lambda[/tex]2
yh =c1*cos(x) + c2*cos(x)

yp = A + B*sin(2x)
y = A + B*sin(2x)
y'' = -4B*sin(2x)
A + sin(2x) = A -3B*sin(2x)
A = A, B = -1/3
yp = A - 1/3*sin(2x)


y = A - 1/3*sin(2x) + c1*cos(x) + c2*sin(x)
y' = - 2/3*sin(2x) - c1*sin(x) + c2*cos(x)
2 = A + c1
So c1= 2-A

2 = 2/3 + c2
c2 = 8/3

y = A - 1/3*sin(2x) + c1*cos(x) + 8/3*sin(x)
y= A- 1/2 sin(2x)+ (2- A)cos(x)+ 8/3 sin(x)

I'm confused about answering the questions. A would be equal to all real numbers, since one could solve for c1. How can I give the solution? There is a unique solution for each value of A, which I would have to write infinite solutions. And there is no value of A when there is an inifiite amount of solutions or no values.

IF A is defined, what would the answer be?
 
So:
y= A- 1/2 sin(2x)+ (2- A)cos(x)+ 8/3 sin(x)

c) For what values of A do there exist infinitely many solutions?
d) For what values of A do there exist no solutions?

Would c and d then be no values of A have infinitely many solutions or nonexistent solution?
 
Yes.
 

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