ODE Problem with boundary conditions

1. Apr 4, 2012

1. The problem statement, all variables and given/known data
Solve:

y'' - λy = 0

where y(0)=y(1)=0, y=y(t)

2. Relevant equations

3. The attempt at a solution

Hi everyone,

This is part of a PDE question, I just need to solve this particular ODE. I know how to do it in the case for y'' + λy = 0, where you get the solution:

λ = (nπ)^2,
y = C.sin(nπt), C a constant

However, I don't know how to go about doing it for this case, other than getting the trivial zero solution of course. Am I correct in thinking I should rewrite it as:

y'' + (-λ)y = 0

and then solve as for the positive case?

I would be really grateful if you could please tell me if this is correct.

2. Apr 4, 2012

HallsofIvy

Staff Emeritus
In other words, you are assuming $\lambdas< 0$, right? Okay, how about setting $\lambda= -\alpha^2$ where $\alpha$ is any non-zero real number?

What is the characteristic equation for $y''+ \alpha^2y= 0$. What characteristic roots do you get and what is the general solution?

3. Apr 4, 2012

Hi,

Assume λ<0

Let λ = -a^2, for some real, non-zero number a

Then

y'' + (a^2)y = 0

y(t) = c1.sin(at) + c2.cos(at)

y(0)=0, therefore c2 = 0

y(1)=0, therefore c1 = 0 (trivial solution)

or sin(a) = 0

a = n∏

so then λ = -(n∏)^2

So then y(t) = c1.sin(n∏t)

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Then one of the conditions I was given in the question was that

∫(y^2)dt = 2, with limits of integration 0 to 1

Integrating this out:

(c1)^2.∫[sin(n∏t)]^2.dt =2

so then (1/2)(c1)^2.∫(1-cos(n∏t))dt =2

which gives us:

(c1)^2 = 4

c1 = ±2

So then y(t) = c1.sin(n∏t), c1=±2

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I don't like that I've ended up with two possible values for c1, but this solution makes sense to me. Please could you tell me if I am correct?