Explicitly put in this behavior and use a power series expansion to solvethe full equation.

R=yℓ

∑∞

k=0

akyke−y

(^2) / 2

∑∞

k=0

akyℓ+ke−y

(^2) / 2
We’ll need to compute the derivatives.
dR
dy

=

∑∞

k=0

ak[(ℓ+k)yℓ+k−^1 −yℓ+k+1]e−y

(^2) / 2
d^2 R
dy^2

=

∑∞

k=0

ak[(ℓ+k)(ℓ+k−1)yℓ+k−^2 −(ℓ+k)yℓ+k

−(ℓ+k+ 1)yℓ+k+yℓ+k+2]e−y

(^2) / 2
d^2 R
dy^2

=

∑∞

k=0

ak[(ℓ+k)(ℓ+k−1)yℓ+k−^2

−(2ℓ+ 2k+ 1)yℓ+k+yℓ+k+2]e−y

(^2) / 2
We can now plug these into the radial equation.
d^2 R
dy^2

+

2

y

dR

dy

−y^2 R−

ℓ(ℓ+ 1)

y^2

R+

2 E

̄hω

R= 0

Each term will contain the exponentiale−y

(^2) / 2
, so we can factor that out. We can also run a single
sum over all the terms.
∑∞
k=0
ak

[

(ℓ+k)(ℓ+k−1)yℓ+k−^2 −(2ℓ+ 2k+ 1)yℓ+k+yℓ+k+2

+2(ℓ+k)yℓ+k−^2 − 2 yℓ+k−yℓ+k+2−ℓ(ℓ+ 1)yℓ+k−^2 +

2 E

̄hω

yℓ+k

]

= 0

The terms for largeywhich go likeyℓ+k+2and some of the terms for smallywhich go likeyℓ+k−^2
should cancel if we did our job right.

∑∞

k=0

ak

[

[(ℓ+k)(ℓ+k−1)−ℓ(ℓ+ 1) + 2(ℓ+k)]yℓ+k−^2

+

[

2 E

̄hω

− 2 −(2ℓ+ 2k+ 1)

]

yℓ+k

]

= 0

∑∞

k=0

ak

[

[ℓ(ℓ−1) +k(2ℓ+k−1)−ℓ(ℓ+ 1) + 2ℓ+ 2k]yℓ+k−^2

+

[

2 E

̄hω

− 2 −(2ℓ+ 2k+ 1)

]

yℓ+k

]

= 0

∑∞

k=0

ak

[

[k(2ℓ+k+ 1)]yℓ+k−^2 +

[

2 E

̄hω

−(2ℓ+ 2k+ 3)

]

yℓ+k

]

= 0