x(x-1)d²y/dx²+(3x-1)dy/dx+y=0
Answers
Answer: We try with a solution of the form
y(x)=∑n=0∞anxs+n,
where a0≠0 .
The first derivative is
dydx=∑n=0∞an(s+n)xs+n−1,
and the second one is
d2ydx2=∑n=0∞an(s+n)(s+n+1)xs+n−2.
Now we plug these derivative into the ODE and get
4∑n=0∞an(s+n)(s+n−1)xs+n−1+2∑n=0∞an(s+n)xs+n−1+∑n=0∞anxs+n=0.
For y to be a solution of this equation, every coefficient must be zero, therefore,
coefficient of xs−1 : a0s[4(s−1)+2]=0 , and as a0≠0 , we have 2s(2s−1)=0 .
We have two solutions, s=0 and s=1/2 , and they will generate the two lineary independent solutions of the ODE.
For every coefficient of x(n+s) , n≥0 we have
an+1(s+n+1)[4(s+n)+2]+an=0
or
an+1=−12(2s+2n+1)(s+n+1)an.
For s=0 we have
an+1=−12(n+1)(2n+1)an.
Therefore, we have
an=(−1)n2nn!(2n−1)!!,
and then
y=a0∑n=0∞(−1)n2nn!(2n−1)!!xn.
Similarly, for s=1/2 ,
an+1=−12(n+1)(2n+3)an
and then
an=−(−1)n(2n+1)!a0,
hence
y=a0∑n=0∞(−1)n(2n+1)!xn+1/2=a0sinx−−√.