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

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

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x^2+y^2=xyy^'

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

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

Solution steps

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

Solve Bernoulli ODE:

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

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

e^{- 2 s} (\frac{1}{4 s})

\int x (4 - x^{2})^{\frac{1}{2}} d x

\frac{\partial}{\partial y} (x y e^{x y})

\int \frac{x - x}{3} d x

x \to 0 lim (\frac{( x )}{x})

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