The exponents of the equation is then,



The first solution is given by






I can't seem to find the second solution,


The recurrence formula is given by

$$c_n=\frac{c_{n-1}}{1-(n-1)^2},n\neq 2$$

We then have linearly independent solution

For $n>2$





I notice that the two are the same. How shall I find the linearly independent $y_2$ with log in it. Any help would be appreciated.


First, you might note that your $c_n$ are given for $n > 2$ by $c_n = \frac{2(-1)^nc_2}{n!(n-2)!}$ (which we can get from your closed-form solution for the first sequence by simply shifting $n$ by $2$). This should allow you to construct your $y_2$. However, there is an easier way to get the solution by making this equation look like a more familiar ODE. Make the substitution $x = \frac{s^2}{4}$. Then, \begin{align*} \frac{\mathrm{d}y}{\mathrm{d}x} &= \frac{2}{s}\frac{\mathrm{d}y}{\mathrm{d}s} \\ \frac{\mathrm{d}^2y}{\mathrm{d}x^2} &= \frac{4}{s^2}\frac{\mathrm{d}^2y}{\mathrm{d}s^2}-\frac{4}{s^3}\frac{\mathrm{d}y}{\mathrm{d}s} \end{align*} This allows us to rewrite our equation as $$\frac{1}{4}\left(s^2\frac{\mathrm{d}^2y}{\mathrm{d}s^2}+s\frac{\mathrm{d}y}{\mathrm{d}s}+(s^2-4)y\right) = 0$$ Notice that this is Bessel's equation and will therefore have solutions $$y = c_1J_2(s)+c_2Y_2(s) = c_1J_2(2\sqrt{x})+c_2Y_2(2\sqrt{x})$$ The Bessel functions will have the same series representations as those produced by the Frobenius method you have used.

  • $\begingroup$ The answer for second solution should looks like $$C_2x^{-1}(-\frac{1}{2}-\frac{x}{2}+\frac{29x^2}{144}+...)+\frac{1}{4}y_1(x)ln|x|$$. I still cannot reach this form using Frobenius method. Mind if you help me? $\endgroup$ – Crazy Jul 29 '17 at 4:57

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