What is the solution for y^'-8/x y=(y^4)/(x^{18)} ?

The solution for y^'-8/x y=(y^4)/(x^{18)} is y=(\sqrt[3]{7}x^8)/(\sqrt[3]{-3x^7+c_{1)}}

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^'-8/x y=(y^4)/(x^{18)}

y^{^{'}} - \frac{8}{x} y = \frac{y ^{4}}{x ^{18}}

Solution steps

y^{'} (x) - \frac{8}{x} y = \frac{y ^{4}}{x ^{18}}

Solve Bernoulli ODE:

\int X^{2} d X

\frac{\partial}{\partial x} (2 y^{3} + 4 x)

\int \frac{2 - 8 x}{9 - 4 x ^{2}} d x

y = 2 x^{2} - 9 x + 7

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

What is the solution for y^'-8/x y=(y^4)/(x^{18)} ?
The solution for y^'-8/x y=(y^4)/(x^{18)} is y=(\sqrt[3]{7}x^8)/(\sqrt[3]{-3x^7+c_{1)}}