- #1

- 5

- 0

I have considered the three possibilities, distinct real roots, repeated roots and conjugate complex roots and cannot find any solutions.

Are there any other possibilities to consider?

Thanks

- Thread starter gerald87
- Start date

- #1

- 5

- 0

I have considered the three possibilities, distinct real roots, repeated roots and conjugate complex roots and cannot find any solutions.

Are there any other possibilities to consider?

Thanks

- #2

- 240

- 2

Why don't you write the equation here?

- #3

- 5

- 0

Sorry, I should have been more clear in my question.

I am not talking about a specific Euler equation, but the general Euler equation,

X^2Y'' + AXY' +BY = 0.

I understand that there are three possible solutions for Y;

Y=c1x^m1 + c2x^m2 for distinct roots

Y=[c1 + c2ln(x)]x^m for repeated roots

or Y=[c1cos(bln(x))+c2sin(bln(x))]x^a for complex conjugate roots

When I put the boundary conditions Y(1) = 0, Y(2) = 0 into each of the possible solutions of Y, the only solution I can get is Y=0.

Are there any other possible solutions to an euler-cauchy problem than the three I mentioned above that should be considered?

Thanks

- #4

- 5

- 0

I have a question relating to the Euler equation,

X^2Y'' + AXY' +BY = 0.

I understand that there are three possible solutions for Y;

Y=c1x^m1 + c2x^m2 for distinct roots

Y=[c1 + c2ln(x)]x^m for repeated roots

or Y=[c1cos(bln(x))+c2sin(bln(x))]x^a for complex conjugate roots

I am to determine whether or not there are solutions (real or imaginary), other than Y=0 to an euler Equation which satisfies the boundary conditions Y(1) = 0, Y(2) = 0.

When I put the boundary conditions Y(1) = 0, Y(2) = 0 into each of the possible solutions of Y, the only solution I can get is Y=0.

Are there any other possible solutions to an euler-cauchy problem than the three I mentioned above that should be considered?

I have noticed in every boundary value euler cauchy problem has the boundary conditions with Y and the derivative of Y, rather than Y in both. Could this mean there are no solutions?

I hope this makes sense, I am a little confused by this problem.

Thanks

- #5

- 9,555

- 766

What you have here is a boundary value problem (boundary conditions at two different points) as opposed to an initial value problem where the two conditions are at the same point. In a homogeneous boundary value problem like you have here, Y identically zero always works. And if you would have found a nontrivial solution satisfying the boundary conditions, then any constant times it would have also worked.

- #6

- 5

- 0

What would happen if A and B in the initial equation were not real numbers?

Would it still be considered an euler equation? or would it have to be solved in a different way?

Thanks again.

- #7

vela

Staff Emeritus

Science Advisor

Homework Helper

Education Advisor

- 14,756

- 1,354

- #8

HallsofIvy

Science Advisor

Homework Helper

- 41,833

- 956

- #9

vela

Staff Emeritus

Science Advisor

Homework Helper

Education Advisor

- 14,756

- 1,354

- #10

HallsofIvy

Science Advisor

Homework Helper

- 41,833

- 956

Ah, thanks.

- #11

HallsofIvy

Science Advisor

Homework Helper

- 41,833

- 956

gerald87, please do NOT post the same question repeatedly.

- #12

- 5

- 0

I think I have got it under control now.

And sorry about the double posts.

- Replies
- 1

- Views
- 1K

- Last Post

- Replies
- 4

- Views
- 4K

- Replies
- 2

- Views
- 1K

- Last Post

- Replies
- 20

- Views
- 1K

- Replies
- 2

- Views
- 3K

- Last Post

- Replies
- 13

- Views
- 2K

- Replies
- 6

- Views
- 2K

- Replies
- 8

- Views
- 2K

- Replies
- 10

- Views
- 4K

- Replies
- 1

- Views
- 2K