What is the solution for y^'=(2y^4-2)/(4xy^3) ?

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^'=(2y^4-2)/(4xy^3)

y^{^{'}} = \frac{2 y ^{4} - 2}{4 x y ^{3}}

Solution steps

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

Solve separable ODE:

n = 3 \sum \infty 1 + \frac{1}{n ^{2}}

x \cdot e^{x}

\frac{d}{d x} (x^{2} + \frac{1}{x})

(∣ x ∣)^{^{'}}

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

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