I am using the generalized Bessel function to obtain a solution to an ODE. The derivative of the general solution is something like
\begin{align} \frac{dT_{1}}{dx} = \frac{3}{4}Ax^{\frac{1}{4}} \Big[c_1 I_{-\frac{1}{3}}\Big(Ax^{\frac{3}{4}}\Big) + c_2 I_{\frac{1}{3}}\Big(Ax^{\frac{3}{4}}\Big)\Big] \end{align}
where A is a constant. Applying the boundary condition
\begin{align} \frac{dT_{1}}{dx}\Big|_{x=0} = 0 \end{align}
results in
\begin{equation} 0 = (0)^{\frac{1}{4}} \Big[c_1 I_{-\frac{1}{3}}\Big(0\Big) + c_2 I_{\frac{1}{3}}\Big(0\Big)\Big] \end{equation}
where
\begin{equation} I_{-\frac{1}{3}}\Big(0\Big) \longrightarrow \infty \end{equation}
\begin{equation} I_{\frac{1}{3}}\Big(0\Big) \longrightarrow 0 \end{equation}
The problem is with the first zero on the right hand side. It seems like this would make the entire right hand side zero and make the boundary condition useless in determining the constants. However, I found this same problem in a textbook and they determined that $c_1=0$, but they did not give details and said they will let the reader verify this on their own. I can see where $c_1=0$ if the first zero is ignored but can that really be done? Is the zero really a problem and if so, what can be done about it?
If it is helpful, another boundary condition is \begin{align} \frac{dT_{1}}{dx}\Big|_{x=L} = 0 \end{align}
Any help would be appreciated. Thanks
Your case might be different from the case in the textbook because there is $x^{1/4}$ multiplying the Bessel function.
The function $x^{\frac{1}{4}} I_{-\frac{1}{3}}\Big(Ax^{\frac{3}{4}}\Big)$ doesn't tends to infinity for $x$ tending to $0$. $$c_1x^{\frac{1}{4}} I_{-\frac{1}{3}}\Big(Ax^{\frac{3}{4}}\Big)=\frac{c_1}{(2A)^{1/3}\Gamma(2/3)}+O(x^2)$$ So, the condition for this term to be equal to $0$ is $c_1=0$