- #1
1Kris
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Hi, I'm stuck on this question from a calculus book;
Show that y'' + ((1+2n)/x)y' + y = 0 is satisfied by x-nJn(x)
Is it correct that when I differentiate that, I get these:
y= x-nJn(x)
y'=-x-nJn+1(x)
y''=nx-n-1Jn+1(x) -
x-n(dJn+1(x)/dx)?
Equation in question becomes:
x-n(2(n/x)Jn+1 - Jn - ((1+2n)/x)Jn+1 + Jn)
= x-n(-x-1Jn+1)
but this isn't 0.
Sorry if I'm repeating myself here but I could just do with some kind of a pointer.
Thanks, Kris
Show that y'' + ((1+2n)/x)y' + y = 0 is satisfied by x-nJn(x)
Is it correct that when I differentiate that, I get these:
y= x-nJn(x)
y'=-x-nJn+1(x)
y''=nx-n-1Jn+1(x) -
x-n(dJn+1(x)/dx)?
The Attempt at a Solution
Equation in question becomes:
x-n(2(n/x)Jn+1 - Jn - ((1+2n)/x)Jn+1 + Jn)
= x-n(-x-1Jn+1)
but this isn't 0.
Sorry if I'm repeating myself here but I could just do with some kind of a pointer.
Thanks, Kris