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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

x^2+2y^2=xyy^'

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

y = x - 1 + c_{1} x^{2}, y = - x - 1 + c_{1} x^{2}

Solution steps

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

Solve Bernoulli ODE:

y = x - 1 + c_{1} x^{2}, y = - x - 1 + c_{1} x^{2}

y = x - 1 + c_{1} x^{2}, y = - x - 1 + c_{1} x^{2}

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

y = e^{7 x} + 7 x e^{7 x}

f (x) = 6 e^{- 0.6 x}, at x = 4

\int (y - 1) d y

\int - 0.5 y d y

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