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cybla
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Frobenius Method Exceptional case r1=r2
For the Frobenius Method for the exceptional case r1=r2... is the equation for the second solution
y[itex]_{2}[/itex]= y[itex]_{1}[/itex] ln (x) + x[itex]^{r_{1}+1}[/itex][itex]\sum_{n=0}^{\infty}b_{n}x^{n}[/itex]
or
y[itex]_{2}[/itex]= y[itex]_{1}[/itex] ln (x) + x[itex]^{r_{1}}[/itex][itex]\sum_{n=1}^{\infty}b_{n}x^{n}[/itex]
In a way both of them give the same exact answer however one begins with [itex]b_{0}[/itex]x (the first one that begins at n=0) ...and the other begins with [itex]b_{1}[/itex]x (the second one that begins at n=1)
Does it matter which one i use? Is one simpler than the other?
For the Frobenius Method for the exceptional case r1=r2... is the equation for the second solution
y[itex]_{2}[/itex]= y[itex]_{1}[/itex] ln (x) + x[itex]^{r_{1}+1}[/itex][itex]\sum_{n=0}^{\infty}b_{n}x^{n}[/itex]
or
y[itex]_{2}[/itex]= y[itex]_{1}[/itex] ln (x) + x[itex]^{r_{1}}[/itex][itex]\sum_{n=1}^{\infty}b_{n}x^{n}[/itex]
In a way both of them give the same exact answer however one begins with [itex]b_{0}[/itex]x (the first one that begins at n=0) ...and the other begins with [itex]b_{1}[/itex]x (the second one that begins at n=1)
Does it matter which one i use? Is one simpler than the other?
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