What is the solution for y^3-(18x+6)+3xy^2y^'=0 ?

The solution for y^3-(18x+6)+3xy^2y^'=0 is y=\sqrt[3]{3(3x+2)+(c_{1})/x}

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^3-(18x+6)+3xy^2y^'=0

y^{3} - (18 x + 6) + 3 x y^{2} y^{^{'}} = 0

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

Solution steps

y^{3} - (18 x + 6) + 3 x y^{2} y^{'} (x) = 0

Solve Bernoulli ODE:

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

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

\int \frac{arctan ( x )}{x} d x

\frac{d}{d x} (6 x^{2} e^{- x})

y^{^{''}} + 4 y = 2 e^{x} + sin (2 x)

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

\frac{\partial}{\partial x} (4 - x^{2} - 4 Y (x)^{2})

What is the solution for y^3-(18x+6)+3xy^2y^'=0 ?
The solution for y^3-(18x+6)+3xy^2y^'=0 is y=\sqrt[3]{3(3x+2)+(c_{1})/x}