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

The solution for 3xy+y^2+(x^2+xy)y^'=0 is y=(-x^2+sqrt(x^4+c_{1)})/x ,y=-(x^2+sqrt(x^4+c_{1)})/x

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

y = \frac{- x ^{2} + x ^{4} + c _{1}}{x}, y = - \frac{x ^{2} + x ^{4} + c _{1}}{x}

Solution steps

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

Solve using substitution:

y = \frac{- x ^{2} + x ^{4} + c _{1}}{x}, y = - \frac{x ^{2} + x ^{4} + c _{1}}{x}

y = \frac{- x ^{2} + x ^{4} + c _{1}}{x}, y = - \frac{x ^{2} + x ^{4} + c _{1}}{x}

\frac{d y}{d x} = \frac{x ^{5} - 4 y}{x}

\int \frac{x}{4 - x ^{4}} d x

1 - X

t \frac{d x}{d t} - x = 2 t^{2}

\int (8 x) d x

What is the solution for 3xy+y^2+(x^2+xy)y^'=0 ?
The solution for 3xy+y^2+(x^2+xy)y^'=0 is y=(-x^2+sqrt(x^4+c_{1)})/x ,y=-(x^2+sqrt(x^4+c_{1)})/x