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

The solution for xyy^'=2(2x^2+y^2) is y=xsqrt(-4+c_{1)x^2},y=-xsqrt(-4+c_{1)x^2}

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

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

y = x - 4 + c_{1} x^{2}, y = - x - 4 + c_{1} x^{2}

Solution steps

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

Solve Bernoulli ODE:

y = x - 4 + c_{1} x^{2}, y = - x - 4 + c_{1} x^{2}

y = x - 4 + c_{1} x^{2}, y = - x - 4 + c_{1} x^{2}

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

0.0018425 (t + 7)^{4.5}

y = x^{2} ln (2 x)

\int \frac{1 - x}{x} d x

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

What is the solution for xyy^'=2(2x^2+y^2) ?
The solution for xyy^'=2(2x^2+y^2) is y=xsqrt(-4+c_{1)x^2},y=-xsqrt(-4+c_{1)x^2}