en

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^'+y^2-y=0

Solution

y^{^{'}} + y^{2} - y = 0

Solution

y = - \frac{e ^{t + c_{1}}}{1 - e ^{t + c_{1}}}

Solution steps

y^{'} (t) + y^{2} - y = 0

Solve separable ODE:

y = - \frac{e ^{t + c_{1}}}{1 - e ^{t + c_{1}}}

y = - \frac{e ^{t + c_{1}}}{1 - e ^{t + c_{1}}}

Popular Examples

(6y^2+4xy^3)dx+(6xy+4x^2y^2)dy=0

(6 y^{2} + 4 x y^{3}) d x + (6 x y + 4 x^{2} y^{2}) d y = 0

\frac{d y}{d x} = 600 y

(dx)/(dt)+e^tx=e^t

\frac{d x}{d t} + e^{t} x = e^{t}

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

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

(dy)/(dx)=2sqrt(y+1)cos(x),y(pi)=0

\frac{d y}{d x} = 2 y + 1 cos (x), y (π) = 0

Frequently Asked Questions (FAQ)

What is the solution for y^'+y^2-y=0 ?
The solution for y^'+y^2-y=0 is y=-(e^{t+c_{1}})/(1-e^{t+c_{1)}}