cybla
Jul31-11, 09:00 PM
For the Frobenius Method for the exceptional case r1=r2... is the equation for the second solution
y_{2}= y_{1} ln (x) + x^{r_{1}+1}\sum_{n=0}^{\infty}b_{n}x^{n}
or
y_{2}= y_{1} ln (x) + x^{r_{1}}\sum_{n=1}^{\infty}b_{n}x^{n}
In a way both of them give the same exact answer however one begins with b_{0}x (the first one that begins at n=0) ....and the other begins with b_{1}x (the second one that begins at n=1)
Does it matter which one i use? Is one simpler than the other?
y_{2}= y_{1} ln (x) + x^{r_{1}+1}\sum_{n=0}^{\infty}b_{n}x^{n}
or
y_{2}= y_{1} ln (x) + x^{r_{1}}\sum_{n=1}^{\infty}b_{n}x^{n}
In a way both of them give the same exact answer however one begins with b_{0}x (the first one that begins at n=0) ....and the other begins with b_{1}x (the second one that begins at n=1)
Does it matter which one i use? Is one simpler than the other?