What is the solution for x*y^'=sqrt(x^2-2xy-y^2)+y ?

The solution for x*y^'=sqrt(x^2-2xy-y^2)+y is y=sqrt(2)xsin(ln(x)+c_{1})-x

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

x*y^'=sqrt(x^2-2xy-y^2)+y

x \cdot y^{^{'}} = x^{2} - 2 x y - y^{2} + y

y = 2 x sin (ln (x) + c_{1}) - x

Solution steps

x \cdot y^{'} (x) = x^{2} - 2 x y - y^{2} + y

Solve using substitution:

y = 2 x sin (ln (x) + c_{1}) - x

y = 2 x sin (ln (x) + c_{1}) - x

\int 3 x 2 x^{2} + 1 d x

\frac{\partial}{\partial y} (4 z e^{x yz})

x \to - 9 - lim (\frac{- 6 x}{x + 9})

\frac{d}{d x} (\frac{x + 3}{x ^{2} + 3})

x = 28 - 7 y^{2}, x = 7 y^{2} - 28

What is the solution for x*y^'=sqrt(x^2-2xy-y^2)+y ?
The solution for x*y^'=sqrt(x^2-2xy-y^2)+y is y=sqrt(2)xsin(ln(x)+c_{1})-x