Can the 2nd ode x*y''-c*y=0 be solved exactly?

[sufive]

As the tittle, can the 2nd ode (B.C not fixed, I need the general solution)

y''-c/x*y=0

be solved exactly? Or can it be translated into some
special mathphysical equations, such as Bessel?
Hypergeometric? etc

any comments or references are welcome.

Thank you for advance!

 

[Raskolnikov]

Would this work?

<br /> <br /> y&#039;&#039; - \frac{cy}{x} = 0, can be rewritten as
\frac{dydy}{y} = \frac{c}{x}dxdx, which upon integrating gives
ln|y|dy = c*ln|x|dx, and then we would just have to integrate one more time?

 

[Mark44]

I would go with a series solution, with y = c₀ + c₁x + c₂x² + ... + c_nxⁿ + ...

 

[gomunkul51]

I would go with a series solution, with y = c₀ + c₁x + c₂x² + ... + c_nxⁿ + ...

Don't you need to search for a solution not from the type:

<br /> y(x)=\sum_{n=0}^{\infty}a_{n}x^{n}<br />

but from the form:

<br /> y(x)=\sum_{n=0}^{\infty}a_{n}x^{n+r}<br />

<br /> a_{0}\neq 0<br />

because of the singularity at x=0 ?

I don't think it is possible to find both linearly independent series' if you search the solution in the first form, I couldn't. but may be I'm rusty in the Olde ODE's :)

 

[HallsofIvy]

Would this work?

<br /> <br /> y&#039;&#039; - \frac{cy}{x} = 0, can be rewritten as
\frac{dydy}{y} = \frac{c}{x}dxdx, which upon integrating gives
ln|y|dy = c*ln|x|dx, and then we would just have to integrate one more time?

No, that is not valid.
\frac{d^2y}{dx^2}\ne \frac{(dy)(dy)}{(dx)(dx)}

 

[LCKurtz]

The solution can be expressed in terms of Bessel functions. See:

http://eqworld.ipmnet.ru/en/solutions/ode/ode0206.pdf

but you will have to come to those by series methods.

 

[Mark44]

Don't you need to search for a solution not from the type:

<br /> y(x)=\sum_{n=0}^{\infty}a_{n}x^{n}<br />

but from the form:

<br /> y(x)=\sum_{n=0}^{\infty}a_{n}x^{n+r}<br />

<br /> a_{0}\neq 0<br />

because of the singularity at x=0 ?

The equation as originally given is xy'' - cy = 0, which is defined when x = 0. It is only by dividing by x that you introduce an singularity.

I don't think it is possible to find both linearly independent series' if you search the solution in the first form, I couldn't. but may be I'm rusty in the Olde ODE's :)

 

[sufive]

Thans to all of you. Your browse or reply helps me.

To this point, LCKurtz's message is the most nearest to
my need!

 

[LCKurtz]

The equation as originally given is xy'' - cy = 0, which is defined when x = 0. It is only by dividing by x that you introduce an singularity.

No, this DE has a regular singular point at x = 0. gomunkul51 is correct about the form the infinite series would take. See

http://banach.millersville.edu/~bob/math365/singular.pdf

for the definition of a singular point and discussion.

 

[vela]

You might also try a change of variables to transform the differential equation into the Bessel differential equation. For instance, u=sqrt(x) gets you a differential equation that sort of looks like the Bessel differential equation.

 

[Mark44]

No, this DE has a regular singular point at x = 0. gomunkul51 is correct about the form the infinite series would take. See

http://banach.millersville.edu/~bob/math365/singular.pdf

for the definition of a singular point and discussion.

OK, thanks.