Re: [微積] 請問要如何求y'呢?

[harveyhs (Hango)]

※ 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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