Solve the given second order differential equation

[chwala]

Homework Statement

My own question (set by me):

$4x^2 \dfrac{d^2u}{dx^2} +12x\dfrac{du}{dx} -4u=0$

Relevant Equations

Cauchy-Euler equations

Going through ode's,
My lines,

Let solution be of the form, $u =x^s$
...
Characteristic equation is;

$4(s(s-1)+12s-4=0$

$4s^2+8s-4=0$

$s_1=\sqrt 2 -1$ and $s_2 = -(\sqrt 2+1)$

thus the two linearly indepedent solutions are,

$u_1(x)= x^{\sqrt 2 -1}$ and $u_2 = \dfrac{1}{x^{\sqrt 2+1}}$

i think that's fine but of course any insight is welcome.... i find such problems to be convenient ie as long as you know the steps you're good to go.

Supposing i had,

$4x^4 \dfrac{d^2u}{dx^2} +12x^2\dfrac{du}{dx} -4u=0$ would one make use of another variable? say let $v=x^2?$

What about,

$4x^5 \dfrac{d^2u}{dx^2} +12x^2\dfrac{du}{dx} -4u=0$


Cheers

 

[Orodruin]

amending latex

This is quite common in your posts. Please consider using the preview functionality.

 

[chwala]

This is quite common in your posts. Please consider using the preview functionality.

I am unable to see the equations in preview mode; i am aware of the functionality ...

 

[pasmith]

$4x^4\dfrac{d^2u}{dx^2} +12x^2\dfrac{du}{dx} -4u=0$ we will make use of anothe variable? say let $v=x^2?$

If you expect $v^{\alpha}$ as a solution, then you can look directly for $x^{2\alpha}$ as a solution. This ODE does not admit solutions of that form.

Also you can divide that entire ODE by 4.

 

[erobz]

Out of curiosity, what motivates looking for a solution like $u = x^s$ ( or any other form for that matter)? I've never understood. Is it just because the coefficients seem to be powers of $x$? Saying ok, I believe you, and applying the machinery is different from understanding the "why".

 

[pasmith]

Out of curiosity, what motivates looking for a solution like $u = x^s$ ( or any other form for that matter)? I've never understood. Is it just because the coefficients seem to be powers of $x$? Saying ok, I believe you, and applying the machinery is different from understanding the "why".

This follows from $$x^n\frac{d^n}{dx^n}x^m = m(m-1) \dots (m-n+1) x^m$$ so we get $$\sum_{n=0}^N a_n x^n \frac{d^n}{dx^n}(x^m) = P(m)x^m$$ for some polynomial $m$.

 

[chwala]

Out of curiosity, what motivates looking for a solution like $u = x^s$ ( or any other form for that matter)? I've never understood. Is it just because the coefficients seem to be powers of $x$? Saying ok, I believe you, and applying the machinery is different from understanding the "why".

I tend to agree ...in most of such problems, the series solutions are dependant on the $x$ coefficients of the given ode. Look at the legendre differential equation (frobenius method) for eg. which i used to study a while back...the steps are similar ... cancelling out the x"s with only a slight deviation on this part of transforming $n$ to $n+2$,

$\sum_{n=0}^\infty n(n-1)a_n x^{n-2} = \sum_{n=-2}^\infty (n+2)(n+1)a_{n+2} x^n$

 

[Orodruin]

This follows from $$x^n\frac{d^n}{dx^n}x^m = m(m-1) \dots (m-n+1) x^m$$ so we get $$\sum_{n=0}^N a_n x^n \frac{d^n}{dx^n}(x^m) = P(m)x^m$$ for some polynomial $m$.

Just to add on: An ODE of this form is known as a Cauchy-Euler equation. With the ansatz $y = x^s$ it boils down to finding the roots of the polynomial $P(s)$.

Common occurrences in physics include the radial part of the homogeneous and spherically symmetric (or cylindrically symmetric) Laplace equation.

 

[DaveE]

I am unable to see the equations in preview mode; i am aware of the functionality ...

You usually will need to reload the page for the LaTeX formatting to render, unless there's already LaTeX in the thread. Try it.

 

[erobz]

Just to add on: An ODE of this form is known as a Cauchy-Euler equation. With the ansatz $y = x^s$ it boils down to finding the roots of the polynomial $P(s)$.

Common occurrences in physics include the radial part of the homogeneous and spherically symmetric (or cylindrically symmetric) Laplace equation.

So , I guess my question In general for any ODE(not just this one) is what other forms of a solution could one find? Is it completely arbitrary? The Wiki says "We assume a trial solution $y = x^m$". What happens if I assumed something else, does the ability to compute the solution then break down somewhere in the process, etc...What happens? Mind you I have only ever had an Introductory course in differential equations 15 years ago. It was the "engineers version" ( open note exams, with astonishingly similar pre-test practice sheets- I don't think anyone failed if they at least showed up on exam day). My actual mathematical knowledge is quite low.

 

[Orodruin]

What happens if I assumed something else, does the ability to compute the solution then break down somewhere in the process, etc...What happens?

It just won’t work because the linearly independent solutions will not have another form. If you have an ODE of degree $n$, you know that there are $n$ independent solutions. Making an ansatz is therefore not a bad thing if it lets you find those solutions. If it doesn’t, then it is a bad ansatz.

With different ODEs there are different reasons to make different guesses. In the case of a Cauchy-Euler equation, the reason is covered in post #6. The ODE goes from solving an $n$th order differential equation to finding the roots of a polynomial of degree $n$ - which has $n$ roots representing the $n$ independent solutions to the ODE.

 

[erobz]

It just won’t work because the linearly independent solutions will not have another form. If you have an ODE of degree $n$, you know that there are $n$ independent solutions. Making an ansatz is therefore not a bad thing if it lets you find those solutions. If it doesn’t, then it is a bad ansatz.

Is that just true with linear ODE's or all ODE's?

 

[Orodruin]

Is that just true with linear ODE's or all ODE's?

Which part?

 

[erobz]

Which part?

The part about having to have $n$ linear independent solutions for a degree $n$ ODE.

 

[Orodruin]

The part about having to have $n$ linear independent solutions for a degree $n$ ODE.

Linearly independent, no. You will not be able to construct new solutions by taking linear combinations of the solutions you found because the general ODE is, well, non-linear.

However, you will always have $n$ integration constants to fix.