急問amathe高手!!!10分+++++

1
dy/dx = 3x^2 - 6x - 1 (given)
y = ∫3x^2 - 6x - 1 dx
y = x^3 - 3x^2 - x + C ... (1)

Substitute (0,1) into (1),
C = 1

Equation of C is y = x^3 - 3x^2 - x + 1


2.
∫(sinx-cosx)^2 dx
= ∫sin^2 x + cos^2 x - 2sinx cosx dx
= ∫1 - sin 2x dx
= x + cos2x / 2 + C

3. y = tan (1/x)
dy/dx = [sec^2 (1/x)] (-1/x^2)
x^2 (dy/dx) = - [1 + tan^2 (1/x)]
x^2 (dy/dx) + (y^2 + 1) = 0　(Since y = tan (1/x))

Differentiate both sides of the above result with respect to x,
x^2 (d^2 y/dx^2) + dy/dx (2x) + 2y (dy/dx) = 0

Dividing both sides by x^2,
(d^2 y/dx^2) + [2(x + y) / x^2] (dy/dx) = 0

2008-03-11 16:51:58 補充：
1. dy/dx = 3x^2 - 6x - 1 (已知)
y = ∫3x^2 - 6x - 1 dx
y = x^3 - 3x^2 - x + C ... (1)

把 (0,1) 代入 (1),
C = 1

C 的方程是 y = x^3 - 3x^2 - x + 1

3. 將 "Differentiate both sides of the above result with respect to x" 改為 "兩邊對 x 微分"
"Dividing both sides by x^2," 改為 "兩邊除以 x^2"