What is the solution for (dy)/(dx)=(e^{-2x})/(1+e^{-2x)} ?

The solution for (dy)/(dx)=(e^{-2x})/(1+e^{-2x)} is y=-1/2 ln(1+e^{-2x})+c_{1}

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

(dy)/(dx)=(e^{-2x})/(1+e^{-2x)}

\frac{d y}{d x} = \frac{e ^{- 2 x}}{1 + e ^{- 2 x}}

y = - \frac{1}{2} ln (1 + e^{- 2 x}) + c_{1}

Solution steps

\frac{d y}{d x} = \frac{e ^{- 2 x}}{1 + e ^{- 2 x}}

Substitute

\frac{d y}{d x}

with

y^{'} (x)

y^{^{'}} (x) = \frac{e ^{- 2 x}}{1 + e ^{- 2 x}}

f^{^{'}} (x) = g (x)

then

f (x) = \int g (x) d x

y = \int \frac{e ^{- 2 x}}{1 + e ^{- 2 x}} d x

\int \frac{e ^{- 2 x}}{1 + e ^{- 2 x}} d x = - \frac{1}{2} ln (1 + e^{- 2 x}) + c_{1}

y = - \frac{1}{2} ln (1 + e^{- 2 x}) + c_{1}

10 cos (x)

\int x e^{- \frac{1}{2} x^{2}} d x

x \to - \infty lim (e^{a x})

e^{\frac{- x ^{2}}{2}}

\int \frac{1}{x ( x ^{2} - 8 )} d x

What is the solution for (dy)/(dx)=(e^{-2x})/(1+e^{-2x)} ?
The solution for (dy)/(dx)=(e^{-2x})/(1+e^{-2x)} is y=-1/2 ln(1+e^{-2x})+c_{1}