What is the solution for y^'=((y^4-1))/(xy^3) ?

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

y^'=((y^4-1))/(xy^3)

y^{^{'}} = \frac{( y ^{4} - 1 )}{x y ^{3}}

y = 4 c_{1} x^{4} + 1, y = - 4 c_{1} x^{4} + 1

Solution steps

y^{'} (x) = \frac{( y ^{4} - 1 )}{x y ^{3}}

Solve separable ODE:

y = 4 c_{1} x^{4} + 1, y = - 4 c_{1} x^{4} + 1

y = 4 c_{1} x^{4} + 1, y = - 4 c_{1} x^{4} + 1

l a pl a ce t r an s f or m - 8 sin (0.2 t)

d er i v a t i v e 2 x - cot^{2} (x)

\frac{d}{d x} (\frac{sin ( x y )}{x ^{2} + y ^{2}})

\int 2^{3} sin (x^{2}) d x

\frac{d}{d x} (\frac{2}{4 x ^{2} + 1})

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