What is the solution for xyy^'=x^6+3y^2 ?

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

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xyy^'=x^6+3y^2

x y y^{^{'}} = x^{6} + 3 y^{2}

y = x^{3} 2 ln (x) + c_{1}, y = - x^{3} 2 ln (x) + c_{1}

Solution steps

x y y^{'} (x) = x^{6} + 3 y^{2}

Solve Bernoulli ODE:

y = x^{3} 2 ln (x) + c_{1}, y = - x^{3} 2 ln (x) + c_{1}

y = x^{3} 2 ln (x) + c_{1}, y = - x^{3} 2 ln (x) + c_{1}

\int 3 x sin (x) cos (x) d x

\frac{d}{d x} (ln (x \cdot x^{2} - 1))

f (x) = x^{9} cos (x)

t^{2} e^{- 3 t}

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

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