What is the solution for yy^'=(e^x)/(1+e^x),y(0)=1 ?

The solution for yy^'=(e^x)/(1+e^x),y(0)=1 is y=sqrt(2ln(1+e^x)+1-2ln(2))

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

yy^'=(e^x)/(1+e^x),y(0)=1

y y^{^{'}} = \frac{e ^{x}}{1 + e ^{x}}, y (0) = 1

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

Solution steps

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

Solve separable ODE:

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

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

\frac{d y}{d x} = (y + 1) (x + 1) cos (x^{2} + 2 x)

\frac{d y}{d x} = 3 x^{2} y^{2} + 3 x^{2}

y^{^{'}} = 3 t^{2} \cdot e^{(y - 2)}

y^{^{'}} - 3 y = 79

7 x y \frac{d y}{d x} = 2 x, y > 0

What is the solution for yy^'=(e^x)/(1+e^x),y(0)=1 ?
The solution for yy^'=(e^x)/(1+e^x),y(0)=1 is y=sqrt(2ln(1+e^x)+1-2ln(2))