# Ordinary Differential Equations

#### Combinatorics

1. The problem statement, all variables and given/known data
Find all eigenvalues and eigenfunctions:
$-y'' (x) = \lambda y(x) , x \in (a,b)$
$y(a)= y(b) =0$.

2. Relevant equations
$sin x = \frac{e^{ix} + e^{-ix} }{2i}$.

3. The attempt at a solution
So actually the only problem I have is to find the eigenfunctions (which should be something like $sin \{ \frac{n \pi (x-a) }{(b-a)} \}$ ) .
I received the eigenvalues are: $\lambda_n = \frac{n^2 \pi^2 }{(b-a)^2}$.
But how can I solve the two equations system I receive when applying these eigenvalues to the solution: $y= c_1 e^{i \sqrt{\lambda_n} x} + c_2 e^{-i \sqrt{\lambda_n} x}$.

I hope someone will be able to help me solve this two equations-system.

Thanks in advance!
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

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#### LCKurtz

Science Advisor
Homework Helper
Gold Member
1. The problem statement, all variables and given/known data
Find all eigenvalues and eigenfunctions:
$-y'' (x) = \lambda y(x) , x \in (a,b)$
$y(a)= y(b) =0$.

2. Relevant equations
$sin x = \frac{e^{ix} + e^{-ix} }{2i}$.

3. The attempt at a solution
So actually the only problem I have is to find the eigenfunctions (which should be something like $sin \{ \frac{n \pi (x-a) }{(b-a)} \}$ ) .
I received the eigenvalues are: $\lambda_n = \frac{n^2 \pi^2 }{(b-a)^2}$.
But how can I solve the two equations system I receive when applying these eigenvalues to the solution: $y= c_1 e^{i \sqrt{\lambda_n} x} + c_2 e^{-i \sqrt{\lambda_n} x}$.

I hope someone will be able to help me solve this two equations-system.

Thanks in advance!
Write the problem as $y'' +\lambda y = 0$ and consider the case where $\lambda = \mu^2 > 0$ so $y''+\mu^2y=0$. This gives you the solution pair $\{e^{i\mu x},e^{-i\mu x}\}$. But by using appropriate linear combinations of those you could use the linearly independent pair $\{\sin(\mu x),\cos(\mu x)\}$ instead. Even better, you can use any appropriate linear combination of those to get something easier to work with. Try using this pair: $\{\cos(\mu(x-a)),\sin(\mu(x-a))\}$. Take a general linear combination of those two and apply your boundary conditions. It will all drop out. At the end, remember $\lambda = \mu^2.$

#### Combinatorics

Is there any way of solving the two equations without having to do the process you mentioned?

Thanks !

#### HallsofIvy

Science Advisor
One can, by direct substitution, show that $y(x)= Acos(\lambda (x- a)+ Bsin(\lambda (x- a))$ is the general solution to the differential equation $y''+ \lambda^2 y= 0$, for any number a. Of course, then $y(0)= A= 0$.
Now $y(b)= B sin(\lambda(b- a))= 0$. You know, I presume that sin(x) is 0 if and only if x is a multiple of $\pi$ so you use that to determine possible values for the eigenvalue, $\lambda$.

#### Combinatorics

Great ! Thanks ! But how can I prove that this is the general solution ? (i.e.- it obviously solves our problem...but why is it the general solution? why is it the only form of our eigenfunctions? )

Thanks!

#### LCKurtz

Science Advisor
Homework Helper
Gold Member
Is there any way of solving the two equations without having to do the process you mentioned?

Thanks !
I suppose so, but I don't know why you would want to. Certainly you would use $y = A\cos(\mu x) + B\sin(\mu x)$ instead of the exponential version. I will show you how to get the solution from there, assuming you didn't think of the $(x-a)$ version. Applying the boundary conditions you get:$$A\cos(\mu a) + B\sin(\mu a)=0$$ $$A\cos(\mu b) + B\sin(\mu b)=0$$For this to have a non-trivial solution for $A$ and $B$ the determinant of coefficients must be 0:$$\left |\begin{array}{cc} \cos(\mu a) & \sin(\mu a)\\ \cos(\mu b) & \sin(\mu b) \end{array}\right | =\cos(\mu a)\sin(\mu b)-\cos(\mu b)\sin(\mu a)= 0$$ $$\sin(\mu(b-a))=0$$
This gives $\mu(b-a) = n\pi$ or $\mu_n = \frac{n\pi}{b-a}$

Under this condition the system of equations is dependent, so just using the first one we can write $A\cos(\mu a) = -B\sin(\mu a)$ so $A=-\frac{B\sin(\mu a)}{\cos(\mu a)}$. Substituting that for the $A$ in the general solution for $y$ gives $$y_n = -B\frac{\sin(\mu_n a)}{\cos(\mu_n a)}\cos(\mu_n x) + B\sin(\mu_n x) =(\frac{B}{\cos(\mu_n a)})(-\sin(\mu_n a)\cos(\mu_n x)+\sin(\mu_n x)\cos(\mu_n a)) =(\frac{B}{\cos(\mu_n a)})\sin(\mu_n(x-a))$$Since any constant times a solution is a solution, we can ignore the constant $\frac{B}{\cos(\mu_n a)}$ and take$$y_n = \sin(\mu_n(x-a))$$So what I showed you in my first post is something worth learning. It saves a lot of work.

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Thanks a lot !!!

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