What is the solution for x^2y^'+xy=3y^3 ?

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

x^2y^'+xy=3y^3

x^{2} y^{^{'}} + x y = 3 y^{3}

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

Solution steps

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

Solve Bernoulli ODE:

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

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

x \to 4 lim (\frac{f ( x )}{g ( x )})

x \to 0 lim (x e^{- x^{2}})

\int 54 sec^{4} (x) d x

\int \frac{3}{16} y^{2} d y

\frac{\partial}{\partial y} (- 8 e^{x} ln (y))

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