What is the solution for y^'=(x^2+3y^2)/(2xy) ?

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

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y^'=(x^2+3y^2)/(2xy)

y^{^{'}} = \frac{x ^{2} + 3 y ^{2}}{2 x y}

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

Solution steps

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

Solve Bernoulli ODE:

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

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

t^{3} - 27 t

\int \frac{x ^{2} + 16 x - 16}{x ^{3} - 16 x} d x

\int \frac{x ^{3}}{8 - x ^{4}} d x

\frac{d}{d x} (arcsinh (x))

\frac{d y}{d x} y^{3} = 7 x^{3} + 2 x

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