What is the solution for y^'-4/x y=(y^3)/(x^5),y(1)=1 ?

The solution for y^'-4/x y=(y^3)/(x^5),y(1)=1 is y=(sqrt(2)x^4)/(sqrt(-x^4+3))

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

y^'-4/x y=(y^3)/(x^5),y(1)=1

y^{^{'}} - \frac{4}{x} y = \frac{y ^{3}}{x ^{5}}, y (1) = 1

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

Solution steps

y^{'} (x) - \frac{4}{x} y = \frac{y ^{3}}{x ^{5}}

Solve Bernoulli ODE:

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

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

What is the solution for y^'-4/x y=(y^3)/(x^5),y(1)=1 ?
The solution for y^'-4/x y=(y^3)/(x^5),y(1)=1 is y=(sqrt(2)x^4)/(sqrt(-x^4+3))