Question

I'm trying to change this sturm liouville equation into a boundary value problem. To no avail I cannot get it work. Can anyone help me out <3

Answer 1

https://gyazo.com/dedacc3095f13aec942736a8be994996

Answer 2

with \(x = e^t\)

Answer 3

.

Answer 4

what does \(.\) mean

Answer 5

What did you try?

Answer 6

a bunch of scribbles getting no where I AM LOST

Answer 7

this is fked i swear
i should ahve done operating systems

Answer 8

with x = e^t i should get to this
https://gyazo.com/2405ad6a1ffabe622aa8c8d0939c65c0

Answer 9

but i am not getting there i am going in circles

Answer 10

i am getting like a humongous bunch of eqns

Answer 11

please try this substitution:

\(\huge y=kx^s\)

Answer 12

i need this x = e^t

Answer 13

You shoulda done linear algebra bb

Answer 14

sry u don't know the chain rule I can't help you.

Answer 15

so help me with chain rule

Answer 16

I'm afraid this is beyond my current level, but if you want to learn about chain rules, you can go here:

https://www.khanacademy.org/math/differential-calculus/taking-derivatives/chain_rule/v/chain-rule-introduction

Khan Academy always provides free, detailed tutorials on these things. c:

Answer 17

getting to that thing you posted is not that hard.

forget about sturm-louiville for a bit, that's the bit that's throwing me. i don't know what that reference means in the problem.

this is a euler-caucy 2nd order linear DE, and you using the [much better, IMHO] change of independent variable substitution, \(x=e^t\), to solve ot

this means \(xy' = Y'\), and \(x^2y'' = Y'' - Y'\) which you sub straight into your DE now in terms of independent variable t.[i presume your uppercase Y signifies that we are now working with our new independent varable t so i gave followed that convention.]

so your DE in x becomes in t : \(Y'' - Y' + Y'+ \lambda Y = 0\) or \(Y'' + \lambda Y = 0\) which is what you want.

to change the boundary values, again using \(x = e^t\) we have \(y(1)\) = 0, so \( Y(0) = 0\) as \(1 = e^0\), and \(y'(e) = Y'(1) = 0\).

as for eigen stuff, i assume you are now going to create a system?

Answer 18

and if you can wait, someone like @SithsAndGiggles or @freckles is well worth waiting for.....

Answer 19

Borrowing from @IrishBoy123's comment, a change of variables \(x=e^t\), \(y(x)=Y(t)\) yields the linear ODE,
\[Y''+\lambda Y=0~~\implies~~Y=C_1\cos(\sqrt\lambda t)+C_2\sin(\sqrt\lambda t)\]which is equivalent to
\[y=C_1\cos(\sqrt\lambda \ln x)+C_2\sin(\sqrt\lambda \ln x)\]With the given boundary conditions, you get the following system:
\[\begin{cases}C_1\cos0+C_2\sin0=0\\[1ex]
-C_1\dfrac{\sqrt\lambda}{e}\sin(\sqrt \lambda)+C_2\dfrac{\sqrt\lambda}{e}\cos(\sqrt\lambda)=0\end{cases}\implies\begin{cases}C_1=0\\[1ex]C_2\cos(\sqrt\lambda)=0\end{cases}\]You want a nontrivial solution, so you require that \(C_2\neq0\). It should be easy to find the eigenvalues now.

Answer 20

Finding the eigenfunctions is easy enough once you determine \(\lambda\), so you would have \(y=\sin(\sqrt\lambda\ln x)\).