What is the solution for (2y^2+4x^2)dx=xydy ?

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

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(2y^2+4x^2)dx=xydy

(2 y^{2} + 4 x^{2}) d x = x y d y

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

Solution steps

(2 y^{2} + 4 x^{2}) d x = x y d y

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}

g (u) = 10 u + 7 u

x (\frac{d y}{d x}) + y = e^{x}

x \to 3 lim (- \frac{x - 4}{x - 3})

t^{2} - \frac{1}{t ^{3}}

\frac{\partial}{\partial x} (\frac{y - 3 x}{z})

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