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

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

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

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

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

Solution steps

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

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