System of 4 equations in 4 unknowns

[kingwinner]

Homework Statement

0 = A+B+D
0 = A-B+C
0 = A exp(k) + B exp(-k) + C sin(k) + D cos(k)
0 = A exp(k) - B exp(-k) + C cos(k) - D sin(k)

Solve for A, B, C, D in the above system. (k is a positive real number)

Homework Equations

N/A

The Attempt at a Solution

1st equation=> D = -A-B
2nd equation=> C = B-A
Put these into the 3rd and 4th equation, we get:
0 = A exp(k) + B exp(-k) + (B-A) sin(k) + (-A-B) cos(k)
0 = A exp(k) - B exp(-k) + (B-A) cos(k) - (-A-B) sin(k)

How should I continue??

Just wondering: In a system of 4 equations in 4 unknowns, is it POSSIBLE to have infinitely many solutions? or must the solution be unique?

Any help is greatly appreciated!

 

[w3390]

Now you have two equations and two unknowns, so solve one equation for A or B and plug that into the other equation.

 

[kingwinner]

Now you have two equations and two unknowns, so solve one equation for A or B and plug that into the other equation.

But I don't think we can divide by that stuff involoving sin, cos, and exp because it may be zero.

Also, I am expecting to have infinitely many solutions because this actually comes from an eigenvalue problem: ODE X'''' = λ X with boundary conditions X(0)=X'(0)=X(1)=X'(1)=0
Here λ=k^4, with k>0
where k is solution cos(k) cosh(k) = 1.
To find the eigenfunctions, I got the above system of 4 equations in 4 unknowns. Since any nonzero multiple of an eigenfunction is again an eigenfunciton, I am expecting the solution of the system to have one arbitrary constant.

Can someone please help??

 

[flatmaster]

If you know linear algebra, do it in a matrix. You should be able to get all unknowns A,B,C, and D. k will remain an arbitrary constant.

 

[kingwinner]

But I think k here is a FIXED (given) positive real number??

 

[HallsofIvy]

Yes, that's what flatmaster just said: " k will remain an arbitrary constant."