What is the type of DE and solving method for y'' + (x^m)y = 0?

[Saladsamurai]

Homework Statement

I have the following DE:

$$y'' + x^my = 0$$

Homework Equations

Maybe Frobenius or Power series or Bessel.

The Attempt at a Solution

I am having a hard time nailing down the type of DE this is. Clearly it is 2nd order homogeneous with variable coefficients. If m is a * positive* (which I will assume) integer, then it has no singular points. So that would mean I can cross Frobenius off the list right?

I think that is my biggest question here: We only use Frobenius when there are (regular) singular points right? Else, we can just use Power series?

EDIT: Power series leads me to:

$$\sum_0^\infty (n-1)na_nx^{n-2} +\sum_0^\infty a_nx^{n+m} = 0$$

Seems like this might be difficult to get the exponents to match.

 

[Dickfore]

Is m an integer?

 

[Saladsamurai]

Is m an integer?

Does not specify unfortunately. It's a made-up problem from professor.

 

[Dickfore]

So, if $m$ is not a non-negative integer, then the point $x = 0$ is a pole. For example, what if $m = -1$? Furthermore, even if $m > 0$, but non-integer, for example, $m = 1/2$, the point $x = 0$, although not a pole, is still a branch point in the complex plane, and, therefore, singular.

I would say, go with the Frobenius method. The equation is:

$$\sum_{n = 0}^{\infty}{(n + \sigma)(n + \sigma - 1) \, c_{n} \, x^{n + \sigma - 2}} + \sum_{n = 0}^{\infty}{c_{n} \, x^{n + \sigma + m}} = 0$$

You can always write:

$$m = [m] + \{m\}, \; 0 \le \{m\} < 1$$

Therefore, you will get:

$$\sum_{n = -2}^{\infty}{(n + \sigma + 2)(n + \sigma + 1) c_{n + 2} \, x^{n + \sigma}} + \sum_{n = [m]}^{\infty}{c_{n - [m]} \, x^{n + \sigma + \{m\}}} = 0$$

 

[Saladsamurai]

Just talked to the TA. Evidently the intended solution is Bessel. I am going to work through it now with the hint he gave me. i.e. reducible to Bessel equation.

 

[Saladsamurai]

I have no idea how to do this

I have that if an equation is of the form

$$\frac{d}{dx}\left(x^a\frac{dy}{dx}\right) + bx^cy = 0\qquad(2)$$

then I can use some formula in my book. So clearly I need to find a,b, and c. Expanding (2) we have:

$$x^ay'' + ax^{a-1}y' + bx^c = 0 = x^2y'' + x^{m+2}y \qquad(3)$$

where the right hand side of (3) was obatined by multiplying the original EQ ny x². Usually, to solve for undetermined coefficients, we compare 'like powers of x' but here those powers are the unknowns. Should I instead compare "like derivatives of y" ?

i.e.,

$$x^a = x^2$$

$$ax^{a-1} = 0$$

$$bx^{c} = x^{m+2}$$

I don't think this works ...

 

[Dickfore]

First, you should do a substitution of the argument:

$$x = a \, t^{\alpha}$$

What does the equation transform to?

 

[Saladsamurai]

First, you should do a substitution of the argument:

$$x = a \, t^{\alpha}$$

What does the equation transform to?

I'm sorry Dickfore. What am I substituting now? Am I putting $x = a \, t^{\alpha}$ inside the differential operator? And also, why did you choose that particular substitution?EDIT: Ok. Here is how I can solve this. I know it is not rigorous at all, but on the not-so-off chance one shows up on my exam tonight, I would like to just be able to solve it. If a DE takes the form

$$\frac{d}{dx}\left(x^a\frac{dy}{dx}\right) + bx^cy = 0\qquad(2)$$

we can calculate a,b,c and use in a formula. Looking at the original equation

$$y' + x^my = 0$$

and comparing it to (2) we note that $x^my = bx^cy \Rightarrow c = m \text{ and } b = 1$ and comparing what is left we have:

$$y' = \frac{d}{dx}\left(x^a\frac{dy}{dx}\right) = x^ay'' + ax^{a-1}y' \Rightarrow a = 0$$

where I used y'' = x^a to find "a" since my y' term does not exist.

Thoughts?

 

[Dickfore]

You should simply use the chain rule:

$$\frac{d y}{d x} = \frac{d t}{d x} \, \frac{d y}{d t}$$

but you need to express everything on the rhs as a function of t. So, you have to use:

$$\frac{d t}{d x} = \frac{1}{d x/d t} = \frac{1}{a \, \alpha \, t^{\alpha - 1}}$$

So, the first derivative becomes:

$$\frac{d y}{d t} = \frac{\dot{y}}{a \, \alpha \, t^{\alpha - 1}}$$

where $\dot{y} = d y/d t$.

What should you get for the second derivative with respect to x?

 

[Dick]

I'm sorry Dickfore. What am I substituting now? Am I putting $x = a \, t^{\alpha}$ inside the differential operator? And also, why did you choose that particular substitution?


EDIT: Ok. Here is how I can solve this. I know it is not rigorous at all, but on the not-so-off chance one shows up on my exam tonight, I would like to just be able to solve it. If a DE takes the form

$$\frac{d}{dx}\left(x^a\frac{dy}{dx}\right) + bx^cy = 0\qquad(2)$$

we can calculate a,b,c and use in a formula. Looking at the original equation

$$y' + x^my = 0$$

and comparing it to (2) we note that $x^my = bx^cy \Rightarrow c = m \text{ and } b = 1$ and comparing what is left we have:

$$y' = \frac{d}{dx}\left(x^a\frac{dy}{dx}\right) = x^ay'' + ax^{a-1}y' \Rightarrow a = 0$$

where I used y'' = x^a to find "a" since my y' term does not exist.

Thoughts?

If you have a formula for 2) how about using a=0, b=1, c=m?