Ordinary differential equations. Series method.

[LagrangeEuler]

Question:
Why equations
x(1-x)\frac{d^2y}{dx^2}+[\gamma-(\alpha+\beta+1)x]\frac{dy}{dx}-\alpha \beta y(x)=0
should be solved by choosing
$y(x)=\sum^{\infty}_{m=0}a_mx^{m+k}$
and not
$y(x)=\sum^{\infty}_{m=0}a_mx^{m}$?
How to know when we need to choose one of the forms.
Also when I sum over $m$, then $\sum^{\infty}_{m=0}a_mx^{m+k}=y(x,k)$. Right?

 

[MarcusAgrippa]

Question:
Why equations
x(1-x)\frac{d^2y}{dx^2}+[\gamma-(\alpha+\beta+1)x]\frac{dy}{dx}-\alpha \beta y(x)=0
should be solved by choosing
$y(x)=\sum^{\infty}_{m=0}a_mx^{m+k}$
and not
$y(x)=\sum^{\infty}_{m=0}a_mx^{m}$?
How to know when we need to choose one of the forms.
Also when I sum over $m$, then $\sum^{\infty}_{m=0}a_mx^{m+k}=y(x,k)$. Right?

Ultimately, you are looking for polynomial solutions. A polynomial need not begin at x^0. So you allow it to begin at x^k.

You could begin by substituting into the equation the series
$y(x)=\sum^{\infty}_{m=0}a_mx^{m}$
But then, if the solution polynomial does begin with the x^k term, your first set of coefficients
$a_0, a_1, ..., a^{k-1}$
would all work out to be zero. It is thus less work to allow for this eventuality by starting your series at
$x^k$
You can the solve, among other things, for the value of k.

Technically, the notation you suggest,
$\sum^{\infty}_{m=0}a_mx^{m+k}=y(x,k)$
is correct, but redundant.

 

[LagrangeEuler]

Ok. Fair enough. I will try to solve tomorrow this with both methods. Could you just tell me how do you know when the first few will be zero or how you know that here when you see equation? Or you always use $\sum^{\infty}_{m=0}a_mx^{m+k}$?

 

[MarcusAgrippa]

Ok. Fair enough. I will try to solve tomorrow this with both methods. Could you just tell me how do you know when the first few will be zero or how you know that here when you see equation? Or you always use $\sum^{\infty}_{m=0}a_mx^{m+k}$?

Always use $\sum^{\infty}_{m=0}a_mx^{m+k}$

 

[vela]

Question:
Why equations
x(1-x)\frac{d^2y}{dx^2}+[\gamma-(\alpha+\beta+1)x]\frac{dy}{dx}-\alpha \beta y(x)=0
should be solved by choosing
$y(x)=\sum^{\infty}_{m=0}a_mx^{m+k}$
and not
$y(x)=\sum^{\infty}_{m=0}a_mx^{m}$?
How to know when we need to choose one of the forms.
Also when I sum over $m$, then $\sum^{\infty}_{m=0}a_mx^{m+k}=y(x,k)$. Right?

See http://mathworld.wolfram.com/FrobeniusMethod.html

x=0 is a singular point of the differential equation. That's why you have to use the first form and not a plain old Taylor series.

 

[MidgetDwarf]

The method you are inquiring about is the Method of Frobenius. You have to use the Method of Frobenius whenever there is a singular point in the differential equation.
A singular point can be broken up into 2 categories. They are: ordinary singular point and irregular singular point (not sure if this is the correct name). I suggest reading the section action in your DE book. In other words, no taylor series expansion exist at a singular point. So we change where the Taylor Series is centered, however it gets really messy, so we employ the Method of Frobenius.

There is also a method of getting the values of K, before you even start massaging the problem.