en

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^'=e^{x-y}

Solution

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

Solution

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

Solution steps

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

Solve separable ODE:

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

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

Graph

Popular Examples

integral from-2 to 2 of 5-sqrt(4-x^2)

\int_{- 2}^{2} 5 - 4 - x^{2} d x

integral of (-1)/(x+1)

\int \frac{- 1}{x + 1} d x

tangent of f(x)=x^2-2x-3 tangent of

f (x) = x^{2} - 2 x - 3

(\partial)/(\partial x)(x^2e^{x+y}+2ycos(xy)-xy^2sin(xy))

\frac{\partial}{\partial x} (x^{2} e^{x + y} + 2 y cos (x y) - x y^{2} sin (x y))

area x^2, 2/((x^2+1))area

x^{2}, \frac{2}{( x ^{2} + 1 )}

Frequently Asked Questions (FAQ)

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