What is the solution for 2xyy^'=x^2-y^2 ?

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

2xyy^'=x^2-y^2

2 x y y^{^{'}} = x^{2} - y^{2}

y = \frac{x ^{2}}{3} + \frac{c _{1}}{x}, y = - \frac{x ^{2}}{3} + \frac{c _{1}}{x}

Solution steps

2 x y y^{'} (x) = x^{2} - y^{2}

Solve Bernoulli ODE:

y = \frac{x ^{2}}{3} + \frac{c _{1}}{x}, y = - \frac{x ^{2}}{3} + \frac{c _{1}}{x}

y = \frac{x ^{2}}{3} + \frac{c _{1}}{x}, y = - \frac{x ^{2}}{3} + \frac{c _{1}}{x}

x \to 0 lim (csc (6 x))

y^{^{'}} = 0.5 y - 450

\int_{1}^{2} (1 + 2 y) d y

x \to 0 lim (3 x^{2})

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

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