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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

(1+e^x)yy^'=e^x

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

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

Solution steps

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

Solve separable ODE:

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

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

\frac{d}{d x} (36 e^{- 6 x})

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

\int (9 - x^{2})^{- \frac{3}{2}} d x

- 2 e^{t}

\int \frac{e ^{x} ( 1 + x ln ( x ) )}{x} d x

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