Solve
Solution Let
then and We can write the differential equation as
Substituting back into this differential equation and multiplying the x^{2} through gives
We next need to make the second term has the n^{th} power of x instead of n2. 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 2a_{2}  6a_{3}x  a_{0}
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 a_{0}
= 0
and
a_{1}
= 1 from our last equation we have 0 = 2a_{2} – a_{0} = 2a_{2} or a_{2} = 0 We also have 0 =  6a_{3} or a_{3} = 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 a_{2} = a_{3} = 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 y_{1} = x For the second linearly independent solution, we let y(0)
= 1
y’(0) =
0 this implies that a_{0}
= 1
and
a_{1}
= 0 from our last equation we have 1 = 2a_{2} – a_{0} = 2a_{2}  1_{ } or a_{2} = 1 We also have
0
=
6a_{3} or
a_{3}
= 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
