What is the solution for (x^4-y^2)dx(x^3-2xy)dy=0 ?

The solution for (x^4-y^2)dx(x^3-2xy)dy=0 is y=c_{1}

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

(x^4-y^2)dx(x^3-2xy)dy=0

(x^{4} - y^{2}) d x (x^{3} - 2 x y) d y = 0

y = c_{1}

Solution steps

(x^{4} - y^{2}) d x (x^{3} - 2 x y) d y = 0

Let

y

be the dependent variable. Divide by

d x

(x^{4} - y^{2}) (x^{3} - 2 x y) \frac{d y}{d x} = 0

Substitute

\frac{d y}{d x}

with

y^{'} (x)

(x^{4} - y^{2}) (x^{3} - 2 x y) y^{^{'}} (x) = 0

Isolate

y^{'} (x) : y^{'} (x) = 0

f^{^{'}} (x) = g (x)

then

f (x) = \int g (x) d x

y = \int 0 d x

\int 0 d x = c_{1}

y = c_{1}

\int - 56 sin (7 z - 5) d z

\frac{d}{d x} (3 2 x^{2})

\int \frac{1}{x ^{5} \cdot x ^{2} - 1} d x

\frac{\partial}{\partial x} (4 x y)

3 t^{2} + 2 t - 2

What is the solution for (x^4-y^2)dx(x^3-2xy)dy=0 ?
The solution for (x^4-y^2)dx(x^3-2xy)dy=0 is y=c_{1}