What is the solution for y^'-(2/x y)=((y^3)/(x^3)),y(1)=1 ?

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

y^'-(2/x y)=((y^3)/(x^3)),y(1)=1

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

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

Solution steps

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

Solve Bernoulli ODE:

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

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

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