\begin{align*}
(x-1)y'' – xy' + y = 0 &\iff (x-1)\sum_{n=2}^{\infty} n(n-1)c_n x^{n-2} – x\sum_{n=1}^{\infty} nc_n x^{n-1} + \sum_{n=0}^{\infty} c_n x^n = 0 \\
&\iff \sum_{n=2}^{\infty} n(n-1)c_n x^{n-1} – \sum_{n=2}^{\infty} n(n-1)c_n x^{n-2} – \sum_{n=1}^{\infty} nc_n x^n + \sum_{n=0}^{\infty} c_n x^n = 0 \\
&\iff \sum_{k=1}^{\infty} (k+1)kc_{k+1} x^k – \sum_{k=0}^{\infty} (k+2)(k+1)c_{k+2} x^k – \sum_{k=1}^{\infty} kc_k x^k + \sum_{k=0}^{\infty} c_k x^k = 0 \\
&\iff -2c_2 + c_0 + \sum_{k=1}^{\infty} x^k \left[ k(k+1)c_{k+1} – (k+2)(k+1)c_{k+2} – kc_k + c_k \right] = 0 \\
\end{align*}
Therefore, $-2c_2 + c_0 = 0 \Rightarrow c_2 = \frac{c_0}{2!}$. And,
$$c_{k+2} = \dfrac{c_k (1-k) + k(k+1)c_{k+1}}{(k+2)(k+1)}$$
To find the solution, first we let $c_0 = 1$ and $c_1 = 0$, then $c_2 = \frac{1}{2!}$. Then, for $k=1$ we have:
$$c_3 = \dfrac{2c_2}{3\cdot 2} = \dfrac{1}{3!}$$
For $k=2$ we have:
$$c_4 = \dfrac{-2c_2 + 6c_3}{4 \cdot 3} = \dfrac{1}{4!}$$
And the pattern continues so for this solution say $y_1$ we have:
$$y_1 = c_0 + c_1 x + c_2 x^2 + c_3 x^3 + \dots = 1 + \frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^4}{4!} + \dots = e^x – x$$
Now for $y_2$ we let $c_0 = 0$ and $c_1 = 1$, so that $c_2 = 0$ too. Then, we find that $c_3 = c_4 = \dots = 0$, so that
$$y_2 = c_0 + c_1 x + c_2 x^2 + \dots = x$$
Then, the final solution should be $a_0 y_1 + a_1 y_2 = a_0 e^x + (a_1 – a_0)x$. However, from wolfram it says the solution is $y = a_0 e^x + a_1 e^{-x}$. Where did I go wrong?
Best Answer
other approach
The equation can be written as
$$(x-1)(y''-y')-(y'-y)=0$$ or $$(x-1)z'-z=0$$ with $$z=y'-y $$
the solution is $$z=\lambda (x-1)=-7 (x-1) $$
Now look for series solution of
$$y'-y=-7 (x-1) $$