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





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


     a2  =  0

We also have

       0  =   - 6a3


      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


       a2  =  -1

We also have

       0  =  -6a3  


         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


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