oh wow, the question actually asks for c1 to c4 and the radius of converge. which I assume doesn't mean I need to put the solutions into series form.. my apologies.
The radius of convergence is |x-x0|<R but how do you find R?
So all of my odd terms will be expressed in terms of c1 and all the even terms c0
which means I need two solutions? I'm not very familiar with power series in general, I struggled with them in first year calc too
$$ c_5=\frac{8c_1 - 14}{20} $$
which I think I'm supposed to see some sort of condensed version involving n! of some sort?
and
$$ c_6=\frac{13c_4 + 2c_2}{30} $$
but I have two equations with C2 and one with both c2 and c4?
oh wow! so many mistakes! Thank you
So for the recurrence relation, I want it in terms of cn+2
$$ c_{n+2} = \frac{[n(n-1)+1]c_n + (n-2)c_{n-2}}{(n+2)(n+1)} = 0 $$
and then I solve for n=3 giving me
c5= $$ \frac{7c_3 c_1}{20} $$ ?
definitely an algebra error present,
Y(s)(s2+6)+2=1/s2-e-s/s2-e-s/s
then bring the 2 over
Y(s)(s2+6)=1/s2-e-s/s2-e-s/s-2
divided out the (s2+6) term
Y(s)=1/(s2(s2+6)) -e-s/(s2(s2+6)) - e-s/(s(s2+6)) - 2/(s2+6)
Homework Statement
Let y(x)=\sumckxk (k=0 to ∞) be a power series solution of
(x2-1)y''+x3y'+y=2x, y(0)=1, y'(0)=0
Note that x=0 is an ordinary point.
Homework Equations
y(x)=\sumckxk (k=0 to ∞)
y'(x)=\sum(kckxk-1) (k=1 to ∞)
y''(x)=\sum(k(k-1))ckxk-2 (k=2 to ∞)
The Attempt at a Solution...
Homework Statement
Let y(x)=\sumckxk (k=0 to ∞) be a power series solution of
(x2-1)y''+x3y'+y=2x, y(0)=1, y'(0)=0
Note that x=0 is an ordinary point.
Homework Equations
y(x)=\sumckxk (k=0 to ∞)
y'(x)=\sum(kckxk-1) (k=1 to ∞)
y''(x)=\sum(k(k-1))ckxk-2 (k=2 to ∞)
The Attempt at a Solution...