Solve the differential equation using power series.

$\displaystyle y^{'} = {\frac{y}{x}} + 1$
cannot be solved for $y$ as a power series $x$. Solve this equation for $y$ as a power series in powers for $x-1$.: Introduce $t=x-1$ as a new independent variable and solve the resulting equation for y as a power series in $t$.)

So far I have proved that it cannot be solved for y as power series in terms of x because you get negative exponents. I did solve for x, which equal t+1.

So, the equation looks like $\displaystyle y^{'} = {\frac{y}{t+1}} + 1$.

Can I multiply both sides by $(t+1)$ so the equation looks like $\displaystyle (t+1)y^{'} = y + (t+1)$?

My solution so far:

$y^{'} = \displaystyle \sum^{\infty}_{n=1}{n}{c_n}{(t+1)^{n-1}}$
$y = \displaystyle \sum^{\infty}_{n=0}{c_n}{(t+1)^{n}}$

$\displaystyle (t+1)y^{'} = y + (t+1)$

$(t+1)\displaystyle \sum^{\infty}_{n=1}{n}{c_n}{(t+1)^{n-1}} = \displaystyle \sum^{\infty}_{n=0}{c_n}{(t+1)^{n}} + (t+1)$

$\displaystyle \sum^{\infty}_{n=1}{n}{c_n}{(t+1)^{n}} - \displaystyle \sum^{\infty}_{n=0}{c_n}{(t+1)^{n}} =(t+1)$

$\displaystyle \sum^{\infty}_{n=0}{n}{c_n}{(t+1)^{n}} - \displaystyle \sum^{\infty}_{n=0}{c_n}{(t+1)^{n}} =(t+1)$

$\displaystyle \sum^{\infty}_{n=0}[nc_n-c_n]*(t+1)^n = (t+1)$

$\displaystyle nc_n-c_n = (t+1)$

$\displaystyle c_n(n-1) = (t+1)$

$\displaystyle c_n = \frac {(t+1)}{n-1}$

Substituing x back in we get:

$\displaystyle c_n = \frac {(x)}{n-1}$

Did I do this correctly?

$\displaystyle \sum^{\infty}_{n=0}[nc_n-c_n]*(t+1)^n = (t+1)$

$\displaystyle nc_n-c_n = (t+1)$

$\displaystyle c_n(n-1) = (t+1)$

$\displaystyle c_n = \frac {(t+1)}{n-1}$

Substituing x back in we get:

$\displaystyle c_n = \frac {(x)}{n-1}$

Did I do this correctly?

From this line is is not.

$$\sum^{\infty}_{n=0}[nc_n-c_n]*(t+1)^n = (t+1) \implies \forall n\neq 1\ \ \ nc_n-c_n = 0$$

At the end, $c_n$ should not depend on $x$!

Note that you can solve the equation directly with the substitution $$y(x) = r(x)x$$