What is the solution for y^'+(2xy^2+2y)/(2x^2y+2x)=0 ?

The solution for y^'+(2xy^2+2y)/(2x^2y+2x)=0 is y=(c_{1})/x

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^'+(2xy^2+2y)/(2x^2y+2x)=0

y^{^{'}} + \frac{2 x y ^{2} + 2 y}{2 x ^{2} y + 2 x} = 0

y = \frac{c _{1}}{x}

Solution steps

y^{'} (x) + \frac{2 x y ^{2} + 2 y}{2 x ^{2} y + 2 x} = 0

Solve separable ODE:

y = \frac{c _{1}}{x}

y = \frac{c _{1}}{x}

\frac{d x}{d y} = \frac{1 + x ^{2}}{1 - y ^{2}}

u^{^{'}} + n^{2} H (t) = - \frac{( - 1 ) ^{n} + 1}{π ( n ^{2} - 1 )}

\frac{d y}{d x} - 5 x = x y

y^{^{'}} - sin (x) \cdot y = 3 sin (x)

\frac{d y}{d t} = \frac{t ^{2}}{1 + y ^{2}}, y (0) = 1

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