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

The solution for y^'=((3x^2+y^2))/(xy) is y=xsqrt(6ln(x)+c_{1)},y=-xsqrt(6ln(x)+c_{1)}

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

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

y = x 6 ln (x) + c_{1}, y = - x 6 ln (x) + c_{1}

Solution steps

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

Solve Bernoulli ODE:

y = x 6 ln (x) + c_{1}, y = - x 6 ln (x) + c_{1}

y = x 6 ln (x) + c_{1}, y = - x 6 ln (x) + c_{1}

\int co x d x

\int_{1}^{16} \frac{x}{x} d x

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

- \frac{4}{( 5 ) ^{3}}

x \to - 1 lim (x^{2} - x + 7)

What is the solution for y^'=((3x^2+y^2))/(xy) ?
The solution for y^'=((3x^2+y^2))/(xy) is y=xsqrt(6ln(x)+c_{1)},y=-xsqrt(6ln(x)+c_{1)}