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

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}