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

The solution for y^'-4/x y=(y^4)/(x^4),y(1)=1 is y=(\sqrt[3]{3}x^4)/(\sqrt[3]{-x^9+4)}

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^4),y(1)=1

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

Solution steps

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

Solve Bernoulli ODE:

\frac{\partial}{\partial x} (- x z)

\int ln (1 - 2 x) d x

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

\int 4 x e^{- 5 x} d x

x \to a lim (\frac{1}{x})

What is the solution for y^'-4/x y=(y^4)/(x^4),y(1)=1 ?
The solution for y^'-4/x y=(y^4)/(x^4),y(1)=1 is y=(\sqrt[3]{3}x^4)/(\sqrt[3]{-x^9+4)}