en

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

Solution

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

Solution

y = ln (x + ln (x) + c_{1})

Solution steps

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

Solve separable ODE:

y = ln (x + ln (x) + c_{1})

y = ln (x + ln (x) + c_{1})

Graph

Popular Examples

y^'+y^2cos(x)=0,y(0)=x(x)

y^{^{'}} + y^{2} cos (x) = 0, y (0) = x (x)

\frac{d y}{d t} = 2 y - 5

\frac{d v}{d t} = - 4 v

x^'=x^2+1/49 ,x(0)=-7

x^{^{'}} = x^{2} + \frac{1}{49}, x (0) = - 7

(xsqrt(1-y^2))dx+(ysqrt(1-x^2))dy=0

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

Frequently Asked Questions (FAQ)

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