Datrys y=A/1+e^-kx

$$y = \frac { A } { 1 + e ^ { - k x } }$$

Datrys ar gyfer A

$$A=\frac{y}{e^{kx}}+y$$

Camau ar gyfer Datrys Hafaliad Llinol

Cyfnewidiwch yr ochrau fel bod yr holl dermau newidiol ar yr ochr chwith.

$$\frac{A}{1+e^{\left(-k\right)x}}=y$$

Aildrefnu'r termau.

$$\frac{A}{e^{-kx}+1}=y$$

Mae'r hafaliad yn y ffurf safonol.

$$\frac{1}{\frac{1}{e^{kx}}+1}A=y$$

Rhannu’r ddwy ochr â \left(e^{-kx}+1\right)^{-1}.

$$\frac{\frac{1}{\frac{1}{e^{kx}}+1}A\left(\frac{1}{e^{kx}}+1\right)}{1}=\frac{y\left(\frac{1}{e^{kx}}+1\right)}{1}$$

Mae rhannu â \left(e^{-kx}+1\right)^{-1} yn dad-wneud lluosi â \left(e^{-kx}+1\right)^{-1}.

$$A=\frac{y\left(\frac{1}{e^{kx}}+1\right)}{1}$$

Rhannwch y â \left(e^{-kx}+1\right)^{-1}.

$$A=\frac{y}{e^{kx}}+y$$

Datrys ar gyfer k

$$\left\{\begin{matrix}k=-\frac{\ln(\frac{A}{y}-1)}{x}\text{, }&\left(x\neq 0\text{ and }y>A\text{ and }y<0\right)\text{ or }\left(x\neq 0\text{ and }y<A\text{ and }y>0\right)\\k\in \mathrm{R}\text{, }&\left(y=0\text{ and }A=0\right)\text{ or }\left(A=2y\text{ and }x=0\text{ and }y\neq 0\right)\end{matrix}\right.$$