Solve
Solution
Let
then
and
We can write the differential equation as
Substituting back into this differential equation and multiplying the x2 through gives
We next need to make the second term has the nth power of x instead of n-2. For this term, we let
u
= n – 2,
n =
u + 2
The second term becomes
now changing this back to n and placing the term back into the differential equation gives
Since they sums do not all start at the same number, we pull out the n = 0 and n = 1 terms to get
or
We can now combine the series to get
-2a2 - 6a3x - a0
We are looking for two linearly independent solutions, so we let the first one be such that
y(0)
= 0
y’(0) =
1
this implies that
a0
= 0
and
a1
= 1
from our last equation we have
0 = -2a2 – a0 = -2a2
or
a2 = 0
We also have
0 = - 6a3
or
a3 = 0
The terms from the series must all be zero, since that is what it means for a polynomial to be zero. Hence
Notice that since
a2 = a3 = 0
All of the rest of the coefficients must be zero, since they are each a multiple of the coefficient two before them. Therefore the first linear independent solution is
y1 = x
For the second linearly independent solution, we let
y(0)
= 1
y’(0) =
0
this implies that
a0
= 1
and
a1
= 0
from our last equation we have
1 = -2a2 – a0 = -2a2 - 1
or
a2 = -1
We also have
0
=
-6a3
or
a3
= 0
we still have
so notice that the odd terms are all zero. For the even terms, we have
This one has the series representation
