en

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^'+e^yx=0

Solution

y^{^{'}} + e^{y} x = 0

Solution

y = - ln (\frac{x ^{2}}{2} - c_{1})

Solution steps

y^{'} (x) + e^{y} x = 0

Solve separable ODE:

y = - ln (\frac{x ^{2}}{2} - c_{1})

y = - ln (\frac{x ^{2}}{2} - c_{1})

Popular Examples

(dx)/(dt)=20-0.6*x

\frac{d x}{d t} = 20 - 0.6 \cdot x

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

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

(dy)/(dx)=e^{2x-y},y(0)=0

\frac{d y}{d x} = e^{2 x - y}, y (0) = 0

dy=x(2ydx-xdy),y(2)=3

d y = x (2 y d x - x d y), y (2) = 3

(dy)/(dx)=y(4x^2+5)

\frac{d y}{d x} = y (4 x^{2} + 5)

Frequently Asked Questions (FAQ)

What is the solution for y^'+e^yx=0 ?
The solution for y^'+e^yx=0 is y=-ln((x^2)/2-c_{1})