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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^'=((x^2+y^2))/((2xy))

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

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

Solution steps

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

Solve Bernoulli ODE:

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

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

f (x) = ∣ 2 x - 1 ∣

y^{^{'}} + 5 y = 6 e^{x}

\frac{\partial}{\partial y} (2 y + z)

x \to 2 lim (\frac{2 x + 3}{x - 2})

\int \frac{1}{y - 3} d y

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