What is the solution for 2yy^'+2=y^2+2x ?

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

2yy^'+2=y^2+2x

2 y y^{^{'}} + 2 = y^{2} + 2 x

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

Solution steps

2 y y^{'} (x) + 2 = y^{2} + 2 x

Solve Bernoulli ODE:

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

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

\frac{d}{d x} (\frac{x ^{2}}{4})

\frac{sin ( x )}{x}

\int \frac{12}{x ^{4}} d x

arctan (\frac{x}{2})

\frac{d y}{d x} - 50 y = - y^{2}

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