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

[EvilKermit]

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'(\pi/2) = 2

The Attempt at a Solution

y = y_h + y_p
0 = 1+ \lambda²
y_h =c₁*cos(x) + c₂*cos(x)

y_p = 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
y_p = 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?

 

[HallsofIvy]

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'(\pi/2) = 2

The Attempt at a Solution

y = y_h + y_p
0 = 1+ \lambda²
y_h =c₁*cos(x) + c₂*cos(x)

y_p = 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
y_p = 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?

 

[EvilKermit]

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?

 

[HallsofIvy]

Yes.