Finding a constant that makes the differential equation hold

[s3a]

I get a = -49/4, 0, 49/4 by using the characteristic equation and quadratic formula and searching for values of a inside of the square root for which a is 0 or greater which isn't correct apparently. I've also wondered if the inside of the root should be less than 0 such that I can get a solution involving complex numbers using natural exponents which could lead me to the trigonometry involving reals but I don't have any concrete plans on getting this to work if this is what I am supposed to do.

Any help would be greatly appreciated!
Thanks in advance!

Edit: Added the question.

 

[HallsofIvy]

I'm sorry but since you haven't said what the differential equation is, it's impossible to understand what you are doing.

 

[s3a]

Oops! I just added the question!

 

[HallsofIvy]

Solutions of such a second order, linear, differential equation can have three different forms:
1) If the characteristic equation has a double real root, \alpha, then y= e^{\alpha x}(C_1+ C_2x).

2) If the characteristic equation has two real roots, \alpha and \beta, y= C_1e^{\alpha x}+ C_2e^{\beta x}

3) If the characteristic equation has two complex roots, \alpha+ \beta i and \alpha- \beta i, y= e^{\alpha x}(C_1 cos(\beta x)+ C_2 sin(\beta))

Now use the conditions y(0)= 0, y(7)= 0 to determine what those constants must be. What must a be in order that we are not forced to C_1= C_2= 0?

 

[Ray Vickson]

I get a = -49/4, 0, 49/4 by using the characteristic equation and quadratic formula and searching for values of a inside of the square root for which a is 0 or greater which isn't correct apparently. I've also wondered if the inside of the root should be less than 0 such that I can get a solution involving complex numbers using natural exponents which could lead me to the trigonometry involving reals but I don't have any concrete plans on getting this to work if this is what I am supposed to do.

Any help would be greatly appreciated!
Thanks in advance!

Edit: Added the question.

You need to show your work.

RGV

 

[s3a]

I attached my initial work.

Also, HallsofIvy, I don't know how to proceed but I understand the constants should not have to be 0. Do the next steps involve trigonometry?

 

[HallsofIvy]

You don't give any reason for "choosing" a as you say.

I pointed out that if the characteristic equation has a double root (49- 4a= 0) the solution must be of the form e^{\alpha x}(C_1+ C_2x). Now apply the boundary conditions: y(0)= e^0(C_1+ C_2(0))= C_1= 0 and y(7)= e^{7\alpha}(C_2(7))= 0 which gives C_2= 0. So if a= 49/4, both C_1 and C_2 must be 0 which means y is never non-zero. a= 49/4 is NOT an answer.

Now, try the other two cases.

 

[s3a]

My current reasoning is to find a value to substitute for x which involves an a such that the whole thing is equal to 0 since x = 0 and x = 7 satisfy the boundary conditions but that has also failed. I attached my latest work.