en

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

Solution

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

Solution

y = ln (- \frac{c _{1}}{1 + x} + 1)

Solution steps

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

Solve separable ODE:

y = ln (- \frac{c _{1}}{1 + x} + 1)

y = ln (- \frac{c _{1}}{1 + x} + 1)

Popular Examples

(-3(2x-u)+6x+2)dx-(u-1)(2dx-du)=0

(- 3 (2 x - u) + 6 x + 2) d x - (u - 1) (2 d x - d u) = 0

(e^x+1)^2e^{-y}dx+(e^x+1)^3e^{-x}dy=0

(e^{x} + 1)^{2} e^{- y} d x + (e^{x} + 1)^{3} e^{- x} d y = 0

\frac{d y}{d t} + 2 y = 4

y^{^{'}} = y^{2} e^{- x}

y^'=yt^2-1.1y,y(0)=1

y^{^{'}} = y t^{2} - 1.1 y, y (0) = 1

Frequently Asked Questions (FAQ)

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