en

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^'=e^{-y},y(0)=0

Solution

y^{^{'}} = e^{- y}, y (0) = 0

Solution

y = ln (t + 1)

Solution steps

y^{'} (t) = e^{- y}

Solve separable ODE:

y = ln (t + 1)

y = ln (t + 1)

Popular Examples

(dy)/(dx)= 2/3 xsqrt(1-9y^2),y(0)=0

\frac{d y}{d x} = \frac{2}{3} x 1 - 9 y^{2}, y (0) = 0

(dv)/(dt)=sqrt(2/9)v

\frac{d v}{d t} = \frac{2}{9} v

derivative of sqrt(x)+sqrt(y)=1

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

(1+e^x)y*y^'=e^y

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

xy^'-y=0,y(0)=sqrt(x)

x y^{^{'}} - y = 0, y (0) = x

Frequently Asked Questions (FAQ)

What is the solution for y^'=e^{-y},y(0)=0 ?
The solution for y^'=e^{-y},y(0)=0 is y=ln(t+1)