math derivative 2

dy/dx = 2[x^2 + sqrt(x^2 -1)][1 + x/sqrt(x^2 -1)]
= 2(sqrt y)(sqrt(x^2 -1) + x)/sqrt(x^2 -1)
= 2(sqrt y)(sqrt y)/sqrt(x^2 -1)
= 2y/sqrt(x^2 -1)
that is
sqrt(x^2 -1) y' = 2y
Differentiate both sides, we get
sqrt(x^2 -1)y" + y'[x/sqrt(x^2 -1)] = 2y'
(x^2 -1)y" + xy' = 2sqrt(x^2 -1)y'
(x^2 -1)y" + xy' = 2sqrt(x^2 -1)[2y/sqrt(x^2 -1)]
(x^2 -1)y" + xy' = 4y
so (x^2 -1)y" + xy' - 4y = 0.




2009-01-23 14:05:06 補充：
Correction: The 1st line should be dy/dx = y' = 2[x + sqrt(x^2 -1)][1 + x/sqrt(x^2 -1)].

y = {x + √[(x^2)-1]}^2
y = x^2 + (x^2-1) + 2x*√(x^2-1)
y = 2x^2 + 2x(x^2-1)^(1/2) - 1
dy/dx = 4x + [2(x^2-1)^(1/2) + 2x(1/2)(2x)(x^2-1)^(-1/2)]
dy/dx = 4x + 2(x^2-1)^(1/2) + 2(x^2)[(x^2-1)^(-1/2)]
(d/dx)(dy/dx) = 4 + 2(2x)(1/2)(x^2-1)^(-1/2) + [2(2x)(x^2-1)^(-1/2) + 2(x^2)(-1/2)(2x)(x^2-1)^(-3/2)]
(d/dx)(dy/dx) = 4 + 2x(x^2-1)^(-1/2) + 4x(x^2-1)^(-1/2) - 2(x^3)(x^2-1)^(-3/2)
(d/dx)(dy/dx) = 4 + 6x(x^2-1)^(-1/2) - 2(x^3)(x^2-1)^(-3/2)
(x^2-1)*[(d/dx)(dy/dx)] = 4(x^2-1) + 6x(x^2-1)^(1/2) - 2(x^3)(x^2-1)^(-1/2)
x*(dydx) = 4x^2 + 2x(x^2-1)^(1/2) + 2x(x^2)(x^2-1)^(-1/2)
4y = 8x^2 + 8x(x^2-1)^(1/2) - 4
(x^2-1)*[(d/dx)(dy/dx)] + x*(dydx) - 4y = 4(x^2-1) + 6x(x^2-1)^(1/2) - 2(x^3)(x^2-1)^(-1/2) + [4x^2 + 2x(x^2-1)^(1/2) + 2x(x^2)(x^2-1)^(-1/2)] - [8x^2 + 8x(x^2-1)^(1/2) - 4]
= 4x^2 - 4 + 6x(x^2-1)^(1/2) - 2(x^3)(x^2-1)^(-1/2) + 4x^2 + 2x(x^2-1)^(1/2) + 2(x^3)(x^2-1)^(-1/2) - 8x^2 - 8x(x^2-1)^(1/2) + 4]
= [4x^2 + 4x^2 - 8x^2] + [6x(x^2-1)^(1/2) + 2x(x^2-1)^(1/2) - 8x(x^2-1)^(1/2)] - 2(x^3)(x^2-1)^(-1/2) + 2(x^3)(x^2-1)^(-1/2) - 4 + 4
= 0