Re: [微積] 請問要如何求y'呢?
※ 引述《rabbit31 (Dreams)》之銘言：
: 題目=>證明y=1/1+ce^-x 是y'=y-y^2的解
: 一直微不出y'=y-y^2
: orz跪求高手教教我~拜託困擾我好久~感激不盡
sol 1: plug into the eqn.
-1 -2
y = (1+Cexp(-x)) => y' = (-1)*(1+Cexp(-x)) * (Cexp(-x))' by Chain rule
-2
= (1+Cexp(-x)) * Cexp(-x)
2 -1 -2
y - y = (1+Cexp(-x)) - (1+Cexp(-x))
-2 -2
= (1+Cexp(-x)) *(1+Cexp(-x)-1) = (1+Cexp(-x)) * Cexp(-x)
sol 2: solve the eqn.
It's a Bernoulli equation.
-1 -2
Let u = y => u' = -1*y y'
Rewrite the equation
u' + u = 1
It's a inhomogeneous first order linear equation.
An obvious particular solution is u = 1.
The homogeneous solution to the equation is Cexp(-x).
Therefore, the general solution is u = 1 + Cexp(-x).
-1 -1
So, y = u = (1+Cexp(-x)) .
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11/15 15:07, , 1F
11/15 15:07, 1F
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