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

The solution for y^'-4/x y=(y^4)/(x^8) is y=(\sqrt[3]{5}x^4)/(\sqrt[3]{-3x^5+c_{1)}}

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^'-4/x y=(y^4)/(x^8)

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

Solution steps

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

Solve Bernoulli ODE:

\int_{0}^{y + z} \frac{1}{( z + y + x ) ^{3}} d x

y^{^{'}} = \frac{1}{x} y - (\frac{x ^{2}}{3}) y^{4}

x + 2 y = 2, y - x = 1, 2 x + y = 7

\frac{d}{d x} (\frac{π}{x})

5 cos (π t)

What is the solution for y^'-4/x y=(y^4)/(x^8) ?
The solution for y^'-4/x y=(y^4)/(x^8) is y=(\sqrt[3]{5}x^4)/(\sqrt[3]{-3x^5+c_{1)}}