What is the solution for 2xyy^'=4x^2+3y^2 ?

The solution for 2xyy^'=4x^2+3y^2 is y=xsqrt(-4+c_{1)x},y=-xsqrt(-4+c_{1)x}

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2xyy^'=4x^2+3y^2

2 x y y^{^{'}} = 4 x^{2} + 3 y^{2}

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

Solution steps

2 x y y^{'} (x) = 4 x^{2} + 3 y^{2}

Solve Bernoulli ODE:

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

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

\int_{0}^{1} \int_{- 1}^{∣ x ∣^{- 1}} (x + y)^{2} d y d x

\int 2 x e^{x^{2} + 3} d x

x \to 8 lim (\frac{x - 8}{x ^{\frac{1}{3}} - 2})

\frac{d}{d x} (4 x^{3})

\frac{d}{d x} (x + x^{- 1})

What is the solution for 2xyy^'=4x^2+3y^2 ?
The solution for 2xyy^'=4x^2+3y^2 is y=xsqrt(-4+c_{1)x},y=-xsqrt(-4+c_{1)x}