# Solve the differential equation $x^2\frac{d^2y}{dx^2}+x\frac{dy}{dx}+(x-1)y=0$

$$x^2\frac{d^2y}{dx^2}+x\frac{dy}{dx}+(x-1)y=0$$

$$\sum_{n=0}^{\infty}(n+r)(n+r-1)c_nx^{n+r}+\sum_{n=0}^{\infty}(n+r)c_nx^{n+r}+\sum_{n=1}^{\infty}c_{n-1}x^{n+r}-\sum_{n=0}^{\infty}c_nx^{n+r}$$

The exponents of the equation is then,

$$r^2-1=0$$

$$r_1=1,r_2=-1$$

The first solution is given by

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

$$c_1=-\frac{c_0}{3}$$

$$c_2=-\frac{c_1}{8}$$

$$c_3=-\frac{c_2}{15}$$

$$y_1(x)=1+2\sum_{n=1}^{\infty}\frac{(-1)^n(x^n)}{(n!)(n+2)!}$$

I can't seem to find the second solution,

$$r=r_2=-1$$

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 $n>2$

$$c_3=-\frac{c_2}{3}$$

$$c_4=-\frac{c_3}{8}$$

$$c_5=-\frac{c_4}{15}$$

$$c_6=-\frac{c_5}{24}$$

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.
• 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? – Crazy Jul 29 '17 at 4:57