# Finding a constant that makes the differential equation hold

1. Feb 18, 2012

### 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!

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Last edited: Feb 18, 2012
2. Feb 18, 2012

### HallsofIvy

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

3. Feb 18, 2012

### s3a

Oops! I just added the question!

4. Feb 18, 2012

### HallsofIvy

Staff Emeritus
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 $C_1= C_2= 0$?

Last edited: Feb 18, 2012
5. Feb 18, 2012

### Ray Vickson

You need to show your work.

RGV

6. Feb 18, 2012

### 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?

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7. Feb 18, 2012

### HallsofIvy

Staff Emeritus
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.

8. Feb 18, 2012

### 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.

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