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

\frac{\partial}{\partial y} (\frac{1}{x + y ^{2}})

\int \frac{x - 3}{6 x - x ^{2}} d x

\frac{d}{d x} (ln (cos (x)))

\int \frac{tan ^{2} ( x )}{sec ^{5} ( x )} d x

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}