What is the solution for y^3-(6x+2)+3xy^2y^'=0 ?

The solution for y^3-(6x+2)+3xy^2y^'=0 is y=\sqrt[3]{2+3x+(c_{1})/x}

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

y^3-(6x+2)+3xy^2y^'=0

y^{3} - (6 x + 2) + 3 x y^{2} y^{^{'}} = 0

y = 3 2 + 3 x + \frac{c _{1}}{x}

Solution steps

y^{3} - (6 x + 2) + 3 x y^{2} y^{'} (x) = 0

Solve Bernoulli ODE:

y = 3 2 + 3 x + \frac{c _{1}}{x}

y = 3 2 + 3 x + \frac{c _{1}}{x}

t an g e n t (x^{3} - 4 x)^{8}

\frac{\partial}{\partial x} (x^{2} + y^{2} - 1)

d er i v a t i v e f (x) = \frac{x ^{3}}{( x ^{2} + 4 ) ^{2}}

y^{^{'}} - \frac{3 y}{x} = x

d er i v a t i v e sin (x^{2}) + \frac{( 1 - x )}{1 + x}

What is the solution for y^3-(6x+2)+3xy^2y^'=0 ?
The solution for y^3-(6x+2)+3xy^2y^'=0 is y=\sqrt[3]{2+3x+(c_{1})/x}