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         Back to Series Solution Page