✔ 最佳答案
Note: The second term should have a y'(x) present instead of y(x).
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(i) Rewrite this in standard form:
y'' + [(p - (p+2)x) / (x(1 - x))] y' + [-p/(x(1 - x))] y = 0.
Since x appears in the denominator of (either) the coefficient to y' or y,
x = 0 is a singular point.
Moreover, since x * [(p - (p+2)x) / (x(1 - x))] = (p - (p+2)x) / (1 - x)
and x^2 * [-p/(x(1 - x))] = -px/(1 - x) are both analytic (have taylor series expansions)
at x = 0, we conclude that x = 0 is a regular singular point.
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Now, we compute the DE's indicial equation.
Assume that y = Σ(n=0 to ∞) a(n) x^(n+r) for some fixed r:
x(1 - x) y'' + [p - (p + 2)x] y' - py
= x(1 - x) Σ(n=0 to ∞) (n+r)(n+r-1) a(n) x^(n+r-2)
+ [p - (p + 2)x] Σ(n=0 to ∞) (n+r) a(n) x^(n+r-1) - p Σ(n=0 to ∞) a(n) x^(n+r)
= Σ(n=0 to ∞) (n+r)(n+r-1) a(n) x^(n+r-1) + Σ(n=0 to ∞) -(n+r)(n+r-1) a(n) x^(n+r)
+ Σ(n=0 to ∞) p(n+r) a(n) x^(n+r-1) + Σ(n=0 to ∞) -(p+2)(n+r) a(n) x^(n+r)
+ Σ(n=0 to ∞) -p a(n) x^(n+r)
= Σ(n= -1 to ∞) (n+r+1)(n+r) a(n+1) x^(n+r) + Σ(n=0 to ∞) -(n+r)(n+r-1) a(n) x^(n+r)
+ Σ(n= -1 to ∞) p(n+r+1) a(n+1) x^(n+r) + Σ(n=0 to ∞) -(p+2)(n+r) a(n) x^(n+r)
+ Σ(n=0 to ∞) -p a(n) x^(n+r), by re-indexing the first and third terms
= [r(r - 1) + pr] a(0) x^(r-1)
+ Σ(n= 0 to ∞) {[(n+r+1)(n+r) + p(n+r+1)] a(n+1) - [(n+r)(n+r-1) + (p+2)(n+r) + p] a(n)} x^(n+r)
The indicial equation comes from the first term: r(r - 1) + pr = 0
Solving for r yields r = 0 or r = 1 - p.
We are guaranteed two linearly independent series solution of the assumed form above if the difference of the roots is not an integer.
In this case, this means that p is not an integer.
(ii) For r = p - 1, we have a solution of the form
y = x^(p-1) * Σ(n=0 to ∞) a(n) x^n.
As x --> 0, Σ(n=0 to ∞) a(n) x^n --> a(0). However, x^(p-1) becomes infinite, because p > 1.
Thus, this solution does not have a finite limit as x --> 0.
Since this is one of two linearly independent solutions to the DE, this means that the general solution also does not have a finite limit as x --> 0.
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I hope this helps!
