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x^2y''-3xy'-12y=0

with initial conditions y(1)=0 and y'(1)=7

I'm supposed to solve for y in the form y=c1y1+c2y2

y1 = x^6 by inspection

Now to solve for y2

y2=y1v

v can be solved for by the equation

y1v''+(2y1'+py1)v'=0

where p is the function in front of y' when there is no function in front of y'':

x^2y''-3xy'-12y=0 divide by x^2

y''-3/xy'-12/x^2y=0

so p=-3/x

x^6v''+(2*6x^5+(-3/x)x^6)v'=0

x^6v''+(12x^5-3x^5)v'=0

x^6v''=-9x^5v'

so v'=e^(-x^2/18)

and v= -x/9(e^(-x^2/18)+c

so y2=x^6(-x/9(e^(-x^2/18)))

y2=-x^(7)/9*e^(-x^2/18)

Now y=c1y1+c2y2 and my initial conditions are y(1)=0 and y'(1)=7

c1*x^6+c2*-x^(7)/9*e^((-x^2)/18)

y'=c1y1'+c2y2'

= c1*6*x^5+

c2*((-7/9*x^6*e^((-x^2)/18)+((-x^7)/9*(-x/9)*e^((-x^2)/18)

Subbing in 1 for x for both equations I get

c1+c2*-1/9*e^(-1/18)

and

6c1+c2*(-7/9*e^(-1/18)+(-1/9*-1/9*e^(-1/18))

=

6c1+c2*(-63/81*e^(-1/18)-1/81*e^(-1/18)

=

6c1-64/81*e^(-1/18)c2

I solved for c1 in the first equation:

c1=c2*1/9*e^(-1/18)

and plugged that into the second equation

6*c2*1/9*e^(-1/18)-64/81*e^(-1/18)*c2=7

c2*e^(-1/18)*(54/81-64/81)=7

c2*e^(-1/18)*-10/81=7

c2=-81/10*7*e^(1/18)

so c1=(-81/10*7*e^(1/18))*1/9*e^(-1/18)

c1=-567/10*e^(1/18)*1/9*e^(-1/18)

=-567/90

Now I plug those back in the equation c1y1+c2y2=y and that's not the right answer. There's a good chance I messed up in there somehow, although I've double checked everything. Does anyone see what I did wrong, or better yet, an easier way to solve this problem. This way is just ridiculous.

Thanks a lot.