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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

3xy^'-2y=(x^3)/(y^2)

3 x y^{^{'}} - 2 y = \frac{x ^{3}}{y ^{2}}

y = 3 x^{3} + c_{1} x^{2}

Solution steps

3 x y^{'} (x) - 2 y = \frac{x ^{3}}{y ^{2}}

Solve Bernoulli ODE:

y = 3 x^{3} + c_{1} x^{2}

y = 3 x^{3} + c_{1} x^{2}

x, 2 x, - x + 2

t e^{- 4 t} cos (5 t)

- x^{4}, x^{2} - 2 x

\frac{d}{d x} (\frac{1 - 3 e ^{x}}{1 - 2 cos ( x )})

\frac{\partial}{\partial y} (e^{x y} + 2 x + 3 y)

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