Urgent questions

e^(xy) = x + y

xy=ln(x+y)

x(dy/dx)+y=(1+dy/dx)/(x+y)

(x+y)[x(dy/dx)+y]=(1+dy/dx)

[x(x+y)-1](dy/dx)=1-y(x+y)

dy/dx=(1-xy-y^2)/(x^2+xy-1)

2 d/dx (tan^(-1) x)=1/(1+x^2)

Taking u=tan^(-1)x

(x^2 + 1)u' = 1

Differentiating both sides of this expression by (n+1) times and using Leibniz's rule.

(x^2 + 1) d^(n+1)/dx^(n+1) u' + (n+1)[d/dx (x^2+1) d^(n)/dx^(n) u' ] + [(n+1)n/2][d^2/dx^2 (x^2+1) d^(n-1)/dx^(n-1) u' ] = 0

(x^2 + 1) y'' + (n+1)[2xy' ] + [(n+1)n/2][2y ] = 0

(x^2 + 1)y'' + 2(n + 1)x y' + n(n + 1)y = 0 for n = 1,2,3

2010-04-20 19:29:28 補充：
六呎將軍 and me show two methods in Q1. Good.

1) exy = x + y

Taking differentiation to both sides:

d(exy)/dx = 1 + dy/dx

exy d(xy)/dx = 1 + dy/dx

exy (y + x dy/dx) = 1 + dy/dx

yexy + xexy dy/dx = 1 + dy/dx

(xexy - 1) dy/dx = 1 - yexy

dy/dx = (1 - yexy)/(xexy - 1)

2) Consider u = tan-1 x, then y = u(n):

Also u' = 1/(1 + x2)

(1 + x2)u' = 1

Taking (n + 1)st derivative to both sides:

(1 + x2)u(n+2) + (n + 1)(2x)u(n+1) + [n(n + 1)/2](2)u(n) = 0

(1 + x2)u(n+2) + 2(n + 1)xu(n+1) + n(n + 1)u(n) = 0

(1 + x2)y" + 2(n + 1)xy' + n(n + 1)y = 0