What is the solution for 0=2x+1-2yy^' ?

The solution for 0=2x+1-2yy^' is y=sqrt(x^2+x+c_{1)},y=-sqrt(x^2+x+c_{1)}

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

0=2x+1-2yy^'

0 = 2 x + 1 - 2 y y^{^{'}}

y = x^{2} + x + c_{1}, y = - x^{2} + x + c_{1}

Solution steps

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

Solve separable ODE:

y = x^{2} + x + c_{1}, y = - x^{2} + x + c_{1}

y = x^{2} + x + c_{1}, y = - x^{2} + x + c_{1}

\int 2 [1 - cos^{2} (2 x)] d x

\int cos (x) (2 + sin (x))^{6} d x

\frac{d y}{d t} + t y^{2} = 0

\frac{4 x}{e ^{x}}

\frac{d}{d x} (\frac{1 + 1}{x})

What is the solution for 0=2x+1-2yy^' ?
The solution for 0=2x+1-2yy^' is y=sqrt(x^2+x+c_{1)},y=-sqrt(x^2+x+c_{1)}