Multiplying on x2
x2y"(x)+xy'(x)+x2y(x)=0
y(x)=∑n=0∝anxr+n , for a not equal 0, x>0
y′(x)=∑n=0∝an(r+n)xr+n−1
y"(x)=∑n=0∝an(r+n)(r+n−1)xr+n−2
y′(x)=∑n=0∝an(r+n)(r+n−1)xr+n+∑n=0∝an(r+n)xr+n+∑n=0∝anxr+n+2
a0r2xr+a1(r+1)2xr+1+∑n=2∝(an(r+n)2an−2)xr+n=0
The indicial equation is r2= 0, and hence r1 = r2 = 0.
a1 = 0; the recurrence relation is
an=(r+n)2an−2 ,n=2,3...
We conclude a1= a3 = a5 = … = 0, and since r = 0,
a2m=(2m)2a2m−2 ,m=1,2...
a2m=22m(m!)(−1)ma0 ,m=1,2...
y1(x)=a0(1+∑n=1∝22m(m!)(−1)ma0)
as y(1)=1, a0=1 so y1(x)=1+∑n=1∝22m(m!)(−1)m
J0(x)=∑n=1∝22m(m!)(−1)m
Since indicial equation has repeated roots, coeficients in second solution can be found using an′(r)∣r=0
a0(r)r2xr+a1(r)(r+1)2xr+1+∑n=2∝(an(r)(r+n)2an−2(r))xr+n=0
a1(r)=0,a1′(0)=0
an(r)=r+nan−2(r) , n=2,3..
a2m+1′(0)=0 , m=1,2...
a2m(r)=r+2ma2m−2(r)=(r+2)2...(r+2m)2(−1)ma0
a2m(r)a2m′(r)=2(r+21+r+41...+r+2m1)
a2m′(0)=2(21+41...+2m1)
Hm=21+41...+2m1
y2(x)=J0(x)lnx1+∑m=1∝22m(m!)(−1)m+1Hmx2m,x>0
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