en

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^'=e^{(t+y)},y(0)=0

Solution

y^{^{'}} = e^{(t + y)}, y (0) = 0

Solution

y = - ln (- e^{t} + 2)

Solution steps

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

Solve separable ODE:

y = - ln (- e^{t} + 2)

y = - ln (- e^{t} + 2)

Graph

Popular Examples

x^2((dy)/(dx))+4xy-y^3=0

x^{2} (\frac{d y}{d x}) + 4 x y - y^{3} = 0

limit as x approaches 2 of (3x-1)^{x^2-3x+4}

x \to 2 lim ((3 x - 1)^{x^{2} - 3 x + 4})

(\partial)/(\partial y)(ln(x/y))

\frac{\partial}{\partial y} (ln (\frac{x}{y}))

(\partial)/(\partial y)(xe^y+ye^z+ze^x)

\frac{\partial}{\partial y} (x e^{y} + y e^{z} + z e^{x})

limit as x approaches 0 of 2/x-2/x

x \to 0 lim (\frac{2}{x} - \frac{2}{x})

Frequently Asked Questions (FAQ)

What is the solution for y^'=e^{(t+y)},y(0)=0 ?
The solution for y^'=e^{(t+y)},y(0)=0 is y=-ln(-e^t+2)