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How to determinate the linearly independence between some special functions defined by ODE? For example:

  1. ${}_1F_1(a;b;x)$ , $x^{1-b}{}_1F_1(a-b+1;2-b;x)$ when $b$ is integer

  2. ${}_2F_1(a,b;c;x)$ , $x^{1-c}{}_2F_1(a-c+1,b-c+1;2-c;x)$ when $c$ is integer

  3. HeunC$(\alpha,\beta,\gamma,\delta,\eta;x)$ , $x^{-\beta}$HeunC$(\alpha,-\beta,\gamma,\delta,\eta;x)$ when $\beta$ is integer

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    Please note that this question base on the information of http://eqworld.ipmnet.ru/en/solutions/ode/ode0210.pdf , http://eqworld.ipmnet.ru/en/solutions/ode/ode0222.pdf and http://www.maplesoft.com/support/help/Maple/view.aspx?path=HeunC , so this question should be correctly asked.2012-06-15

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The usual route to proving the linear independence of two solutions $f_1(z)$, $f_2(z)$ is to compute the Wronskian determinant,

$\begin{vmatrix}f_1(z)&f_2(z)\\f_1^\prime(z)&f_2^\prime(z)\end{vmatrix}$

If this determinant is identically zero, then your two functions are linearly dependent.

For your first pair, the Wronskian is $\dfrac{\exp\,z\sin\,\pi b}{\pi z^b}$; for integer $b$, this Wronskian is clearly zero, and thus your pair is linearly dependent. For your second, the Wronskian is $(1-c)\dfrac{(1-z)^{c-a-b-1}}{z^c}$; this zeroes out only if $c=0$, so apart from that case, your two functions are linearly independent.

Abel's identity is useful if you do not happen to have explicit expressions for your two solutions.

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    @John, you're right (though on the other hand, does a linear combination of your pair satisfy a certain second-order differential equation?)2012-09-24