Specific Solution to Differential Equation Solve \(\displaystyle{x}{\left({x}-{1}\right)}^{{2}}{g}{''}{\left({x}\right)}+{\left({x}-{1}\right)}{g}'{\left({x}\right)}-{x}{g{{\left({x}\right)}}}={\frac{{{x}-{1}}}{{{x}}}},\qquad{x}\geq{1}.\) I already

Oliver Carson

Oliver Carson

Answered question

2022-03-31

Specific Solution to Differential Equation
Solve x(x1)2g(x)+(x1)g(x)xg(x)=x1x,x1.
I already know the general solutions to this as
g(x)=c1P(x)+c2Q(x)+gs(x).
Here, P(x)=xPα2(2x1),Q(x)=xQα2(2x1),α=512
P and Q are the associated Legendre functions and gs(x) is a special solution to this equation. I also know that the special solution can be determined by an integral
gs(x)=2Q(x)P(x)x3dx2P(x)Q(x)x3dx.
Can anyone help me find a particular solution to this equation?

Answer & Explanation

Malia Booth

Malia Booth

Beginner2022-04-01Added 16 answers

Step 1
By letting gs(x)=f1(x)+1x1, you can get:
x(x1)f1 (2x1)f1+f1=1x.
If you let f1=a1x+f2, then:
x(x1)f2 (2x1)f2+f2=1+5a1x+3a1x2.
We set a1=15 to discard the first term. Then we do f2=a2x2+f3:
x(x1)f3 (2x1)f3+f3=35+11a2x2+8a2x3.
We set a2=355. And so on. In general, we find a recursive relation:
a1=15, an+1=ann(n+2)n2+5n+5.
Step 2
You can solve this recursive relation using raising factorials:
an=-151(n-1)3(n-1)(1+r1)(n-1)(1+r2)(n-1)=-(n-1)!(n+1)!10Γ(r1+1)Γ(r2+1)Γ(r1+n)Γ(r2+n),
where r1,2=125±5 are roots of n2+5n+5=0. First terms of f1(x) are:
f1(x)=15x355x2241045x3726061x41728248501x5+o(x5)

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