Question

(∫f(x)dx)(∫1/f(x)dx)=-1求f(x)

Answer 1

∫ ƒ(x) dx • ∫ 1/ƒ(x) dx = - 1
ƒ(x) • ∫ 1/ƒ(x) dx + 1/ƒ(x) • ∫ ƒ(x) dx = 0
ƒ²(x) • ∫ 1/ƒ(x) dx + ∫ ƒ(x) dx = 0，ƒ(x) ≠ 0
2ƒ(x)ƒ'(x) • ∫ 1/ƒ(x) dx + ƒ²(x) • 1/ƒ(x) + ƒ(x) = 0
2ƒ(x)ƒ'(x) • ∫ 1/ƒ(x) dx + ƒ(x) + ƒ(x) = 0
2ƒ(x)ƒ'(x) • ∫ 1/ƒ(x) dx = - 2ƒ(x)
ƒ'(x) • ∫ 1/ƒ(x) dx = - 1，ƒ(x) ≠ 0
∫ 1/ƒ(x) dx = - 1/ƒ'(x)
1/ƒ(x) = ƒ''(x)/ƒ'²(x)
y'² = y • y''
y • y'' - y'² = 0
令u = y'，则y'' = d(dy/dx)/dx = du/dx = du/dy • dy/dx = u • du/dy
y • u • du/dy - u² = 0
y • du/dy = u
du/u = dy/y
∫ 1/u du = ∫ 1/y dy
ln| u | = ln| y | + ln| a | = ln| ay |
u = ay
dy/dx = ay
dy/y = a • dx
∫ 1/y dy = a • ∫ dx
ln| y | = ax + b
y = e^(ax + b)，其中a和b为暂定常数
代入原式后发现∫ ƒ(x) dx • ∫ 1/ƒ(x) dx = - 1/a² = - 1
于是a = ± 1
ƒ(x) = e^(x + b) 或 ƒ(x) = e^(- x + b)，其中b为任意常数