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December 19, 2014

December 19, 2014

Posted by **Jen** on Saturday, April 21, 2012 at 5:18pm.

(1-x)y'' + y = 0; x0 = 0

- math -
**MathMate**, Saturday, April 21, 2012 at 9:16pmI'll give it a try.

(1-x)y"+y=0 at x0=0

A. Check for singular points.

Write as y"+(1/1-x)y=0

and find Lim x->0 (1/1-x) = 1

Therefore x0=0 is not a singular point.

B. Assume y=∑an x^n for n=0 to ∞

then

(1-x)y"=∑n=0 to ∞ (n)(n-1)an x^(n-2)

=∑n=2 to ∞ (n)(n-1)an x^(n-2) -∑n=1 to ∞ (n)(n-1)an x^(n-1)

Now switch indices

=∑n=0 to ∞ (n+2)(n+1)an+2 x^(n) -∑n=0 to ∞ (n+1)(n)an+1 x^(n)

Add to y gives

y(x)

∑n=0 to ∞ (n+2)(n+1)an+2 x^(n) -∑n=0 to ∞ (n+1)(n)an+1 x^(n)+∑ n=0 to ∞ an x^n =0

This gives

an+1 = [(n)(n+1)an+1 - an]/[(n+2)(n+1)]

The coefficients an are dependent on values of a0 and a1, which are arbitrary constants depending on the given initial conditions.

Check my arithmetic.

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