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}

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(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}

y^{^{'}} = \frac{y ^{3} - 3}{( 3 x y ^{2} )}

f (x) = \frac{5 x}{e ^{x}}

\frac{1}{3 x ^{2} + 2 x}

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

\int_{0}^{1} x^{- 1} d x

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}