What is the solution for xyy^'=x^6+3y^2,y(e)=-e^3 ?

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

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xyy^'=x^6+3y^2,y(e)=-e^3

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

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

Solution steps

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

Solve Bernoulli ODE:

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

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

\frac{d}{d x} ((3 x^{3} + 4) (x^{4} - 2 x))

x \to 0 lim (\frac{x ^{2} ln ( x )}{x})

\frac{\partial}{\partial x} ((y - 2) (arctan (x)))

\frac{d}{d x} (\frac{csc ( x )}{2 ^{x}})

x \to 0 lim (\frac{x ^{2} - 2 x}{x})

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