I can't seem to get the right answer for this.
(x-1)y'+2y=0
I arrive at c_{n}=(2+n)c_{0}, and the answer being y(x)=c_{0}SUM((2+n)x^{n}), but the book says it is c_{0}/(1-x)^{2 }
this is very simple use separation for solving differential eqn:
dy/(2y)=-dx/(x-1)
integrate both sides
(1/2)ln(y)=-ln(x-1)+c0
cross multiplying 2 u get
ln(y)=(-2)ln(x-1)+2c0
now using log properyies and taking 2c0=ln(m0)
ln(y)=ln((x-1)^-2)+ln(m0)
use log (ab)=log a+ log b
u get
ln(y)=ln((x-1)^(-2)*m0)
comparing u get
y=m0*(1/(x-1)^(-2))
That's not what I get. I get c_{n}= (n+ 1)c_{0}.
Then $\displaystyle y= c_0\sum_{n=0}^\infty (n+1)x^n$. Note that "$\displaystyle (n+1)x^n$" is the derivative of $\displaystyle x^{n+1}$. Since such a power series is uniformly convergent inside its interval of convergence, which is -1< x< 1, we can differentiate "term by term". That is, this function is the derivative of $\displaystyle c_0\sum_{n= 0}^\infty x^{n+1}= c_0x\sum_{n=0}^\infty x^n$.
That last sum, $\displaystyle \sum_{n=0}^\infty x^n$ is a geometric series with sum $\displaystyle \frac{1}{1- x}$. So the orginal function is the derivative of $\displaystyle \frac{c_0x}{1- x}$. By the "quotient rule", that derivative is $\displaystyle \frac{c_0(1- x)+ c_0x}{(1- x)^2}= \frac{c_0}{(1- x)^2}$.
If I may expand on this a small bit, since this was a bit new to me.
What they are doing is letting
$y(x)=\displaystyle{\sum_{n=0}^\infty}c_n x^n$ then
$y'(x)=\displaystyle{\sum_{n=0}^\infty}n c_n x^{n-1}$
then if you expand out the original diff eq $(x-1)y' + 2y=0$ you get after much simplification
$\left(\displaystyle{\sum_{n=0}^\infty}\left[(n+2)c_n-(n+1)c_{n+1}\right]x^n\right)=0$
equating this to zero term by term one obtains Dr. Ivey's result of
$c_n=(n+1)c_0$