SOLVE BY POWER SERIES 4xy\'\'+2y\'-y=0 DIFFERENTIAL EQUATIONS!!! Solution Since

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Since x = 0 is a regular singular point, assume that y = S(n = 0 to 8) a(n) x^(n+r). Substituting this into the DE yields 4x S(n = 0 to 8) (n+r)(n+r-1) a(n) x^(n+r-2) + 2 S(n = 0 to 8) (n+r) a(n) x^(n+r-1) - S(n = 0 to 8) a(n) x^(n+r) = 0. Simplifying: S(n = 0 to 8) 4(n+r)(n+r-1) a(n) x^(n+r-1) + S(n = 0 to 8) 2(n+r) a(n) x^(n+r-1) - S(n = 0 to 8) a(n) x^(n+r) = 0. S(n = 0 to 8) (n+r)(4n+4r-2) a(n) x^(n+r-1) - S(n = 0 to 8) a(n) x^(n+r) = 0. Re-index the first sum: S(n = -1 to 8) (n+r+1)(4n+4r+2) a(n+1) x^(n+r) - S(n = 0 to 8) a(n) x^(n+r) = 0. ==> r(4r - 2) a(0) x^(r-1) + S(n = 0 to 8) [(n+r+1)(4n+4r+2) a(n+1) - a(n)] x^(n+r) = 0. Setting the first coefficient equal to 0 yields the indicial equation r(4r - 2) = 0 ==> r = 0 or 1/2. (i) r = 0 yields S(n = 0 to 8) [(n+1)(4n+2) a(n+1) - a(n)] x^n = 0. ==> (n+1)(4n+2) a(n+1) - a(n) = 0 ==> a(n+1) = a(n) / [(n+1)(4n+2)]. (ii) r = 1/2 yields S(n = 0 to 8) [(n + 3/2)(4n+4) a(n+1) - a(n)] x^(n + 1/2) = 0. ==> (2n+3)(2n+2) a(n+1) - a(n) = 0 ==> a(n+1) = a(n)/[(2n+2)(2n+3)]. Now, you can solve for a few coefficients (and solve the DE) as in the previous posts. I hope this helps!

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