# Problem

• Feb 26th 2009, 07:15 PM
MathTiger
Problem
Let http://msr02.math.mcgill.ca/webwork2...6cc830a571.png be a positive constant. Consider the equation

with boundary conditions http://msr02.math.mcgill.ca/webwork2...dbbe9016e1.png and http://msr02.math.mcgill.ca/webwork2...bd2a708161.png.

For certain discrete values of http://msr02.math.mcgill.ca/webwork2...6cc830a571.png, this equation can have non-zero solutions. Find the three smallest values of http://msr02.math.mcgill.ca/webwork2...6cc830a571.png for which this is the case.

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I've been trying to do this, but I get nowhere. Maybe I'm doing something wrong... Let me know.

So the characteristic equation is k^2 + 8k + a = 0.

The roots are:

k1 = -4 + sqrt(16-a)
k2 = -4 - sqrt(16-a)

General solution:

y = Ae^[(-4+sqrt(16-a))x] + Be^[(-4-sqrt(16-a))x]

From the given conditions:

0 = A+B --> B = -A

0 = Ae^[-8+2sqrt(16-a)] + Be^[-8-2sqrt(16-a)]

Substituting -A for B and taking e^-8 out:

0 = (Ae^(-8))[e^(2sqrt(16-a)) - e^(-2sqrt(16-a)]

So I thought that for A and B to not equal 0, I need to make the term in the second parantheses 0 and solve for a. However I only get one answer: a=0 which seems to be wrong.

Maybe I'm completely missing the point here. I would appreciate it if anyone could help me.
• Feb 26th 2009, 08:18 PM
Soroban
Hello, MathTiger!

I too much be missing something . . . I found only one solution.

Quote:

Let $\displaystyle a$ be a positive constant.

Consider the equation: .$\displaystyle \frac{d^2y}{dx^2} + 8\,\frac{dy}{dx} + ay \:=\:0$

with boundary conditions: .$\displaystyle y(0) = 0,\:y(2) = 0$

For certain discrete values of $\displaystyle a$, this equation can have non-zero solutions.
Find the three smallest values of $\displaystyle a$ for which this is the case.

Your preliminary work is correct . . .

Quote:

So the characteristic equation is: .$\displaystyle k^2 + 8k + a \:= \:0$

The roots are: .$\displaystyle \begin{array}{ccc}k_1 &=& -4 +\sqrt{16-a} \\ k_2 &=& -4 - \sqrt{16-a} \end{array}$

General solution: .$\displaystyle y \;= \;Ae^{(-4+\sqrt{16-a})x} + Be^{(-4-\sqrt{16-a})x}$

From the given conditions:

. . $\displaystyle 0 \:=\: A+B\quad\Rightarrow\quad B \:=\: -A$

. . $\displaystyle 0 \:= \:Ae^{-8+2\sqrt{16-a}} + Be^{-8-2\sqrt{16-a}}$

Substituting $\displaystyle -A$ for $\displaystyle B$ and taking out $\displaystyle e^{-8}\!:\;\;Ae^{-8}\,\bigg[e^{2\sqrt{16-a}} - e^{-2\sqrt{16-a}}\bigg] \:=\:0$

So I thought that for $\displaystyle A, B \neq 0$,
. . I need to make the second group equal to 0 and solve for $\displaystyle a.$ . Right!

We have: .$\displaystyle e^{2\sqrt{16-a}} - e^{-2\sqrt{16-a}} \;=\;0$

Multiply by $\displaystyle e^{2\sqrt{16-a}}\!:\;\;e^{4\sqrt{16-a}} - 1 \:=\:0 \quad\Rightarrow\quad e^{4\sqrt{16-a}} \:=\:1\:=\:e^0$

Then: .$\displaystyle 4\sqrt{16-a} \:=\:0 \quad\Rightarrow\quad \sqrt{16-a} \:=\:0 \quad\Rightarrow\quad 16-a \:=\:0 \quad\Rightarrow\quad \boxed{a \:=\:16}$

• Feb 26th 2009, 08:26 PM
MathTiger
Correction
Oops, I did mean I only get a=16, not a=0. But the online submission system says that a=16 is neither of these three lowest values of a.
• Feb 26th 2009, 11:06 PM
CaptainBlack
Quote:

Originally Posted by MathTiger
Let http://msr02.math.mcgill.ca/webwork2...6cc830a571.png be a positive constant. Consider the equation

with boundary conditions http://msr02.math.mcgill.ca/webwork2...dbbe9016e1.png and http://msr02.math.mcgill.ca/webwork2...bd2a708161.png.

For certain discrete values of http://msr02.math.mcgill.ca/webwork2...6cc830a571.png, this equation can have non-zero solutions. Find the three smallest values of http://msr02.math.mcgill.ca/webwork2...6cc830a571.png for which this is the case.

-----

I've been trying to do this, but I get nowhere. Maybe I'm doing something wrong... Let me know.

So the characteristic equation is k^2 + 8k + a = 0.

The roots are:

k1 = -4 + sqrt(16-a)
k2 = -4 - sqrt(16-a)

General solution:

y = Ae^[(-4+sqrt(16-a))x] + Be^[(-4-sqrt(16-a))x]

From the given conditions:

0 = A+B --> B = -A

0 = Ae^[-8+2sqrt(16-a)] + Be^[-8-2sqrt(16-a)]

Substituting -A for B and taking e^-8 out:

0 = (Ae^(-8))[e^(2sqrt(16-a)) - e^(-2sqrt(16-a)]

So I thought that for A and B to not equal 0, I need to make the term in the second parantheses 0 and solve for a. However I only get one answer: a=0 which seems to be wrong.

Maybe I'm completely missing the point here. I would appreciate it if anyone could help me.

let $\displaystyle \lambda_1$ and $\displaystyle \lambda_2$ be the roots of the charateristic equation.

Then if $\displaystyle A\ne 0$ we have:

$\displaystyle e^{2\lambda_1}=e^{2\lambda_2}$

which implies that:

$\displaystyle 2\lambda_1 - 2\lambda_2=\text{\bf{i}}\ 2\ n\ \pi$

for some $\displaystyle n \in \mathbb{Z}.$

CB
• Feb 27th 2009, 12:23 AM
MathTiger
Quote:

Originally Posted by CaptainBlack
let $\displaystyle \lambda_1$ and $\displaystyle \lambda_2$ be the roots of the charateristic equation.

Then if $\displaystyle A\ne 0$ we have:

$\displaystyle e^{2\lambda_1}=e^{2\lambda_2}$

which implies that:

$\displaystyle 2\lambda_1 - 2\lambda_2=\text{\bf{i}}\ 2\ n\ \pi$

for some $\displaystyle n \in \mathbb{Z}.$

CB

Sorry... I'm not sure how you get that equation... And even so, how do I choose n to get the lowest values of a?
• Feb 27th 2009, 04:17 AM
CaptainBlack
Quote:

Originally Posted by MathTiger
Sorry... I'm not sure how you get that equation... And even so, how do I choose n to get the lowest values of a?

The equation holds because for any integer $\displaystyle n$ :

$\displaystyle e^{a+2 \text{\bf{i}}\ n\ \pi}=e^ae^{2 \text{\bf{i}}\ n\ \pi}=e^a$

So if two powers of $\displaystyle e$ are equal then the powers differ by an integer multiple of $\displaystyle 2 \text{\bf{i}}\ \pi$

CB