What is the solution for xy^'+y=-2x^6y^4 ?

The solution for xy^'+y=-2x^6y^4 is y= 1/(x\sqrt[3]{2x^3+c_{1)}}

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xy^'+y=-2x^6y^4

x y^{^{'}} + y = - 2 x^{6} y^{4}

Solution steps

x y^{'} (x) + y = - 2 x^{6} y^{4}

Solve Bernoulli ODE:

\frac{\partial}{\partial x} (y^{2} sin (x))

e^{- π s} \frac{2}{s ^{3}}

y^{^{'}} - 2 \frac{y}{x} = x, y (1) = 2

y^{'} = - y^{3}

y = sin^{3} (x^{2})

What is the solution for xy^'+y=-2x^6y^4 ?
The solution for xy^'+y=-2x^6y^4 is y= 1/(x\sqrt[3]{2x^3+c_{1)}}