What is the solution for 6y^'-2y=x(y^4),y(0)=-2 ?

The solution for 6y^'-2y=x(y^4),y(0)=-2 is y=(2\sqrt[3]{e^x})/(\sqrt[3]{-4e^xx+4e^x-5)}

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6y^'-2y=x(y^4),y(0)=-2

6 y^{^{'}} - 2 y = x (y^{4}), y (0) = - 2

y = \frac{2 3 e ^{x}}{3 - 4 e ^{x} x + 4 e ^{x} - 5}

Solution steps

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

Solve Bernoulli ODE:

y = \frac{2 3 e ^{x}}{3 - 4 e ^{x} x + 4 e ^{x} - 5}

y = \frac{2 3 e ^{x}}{3 - 4 e ^{x} x + 4 e ^{x} - 5}

y^{^{'}} - y^{(\frac{5}{3})} = 0

\frac{x ^{2}}{x ^{4} + 16}

\frac{\partial}{\partial x} (y^{8} + 2 x^{6} + 2 x y)

f (x) = 2 x^{3} - 5 x^{2} + 3 x - 7

3 + 5 cos (3 t)

What is the solution for 6y^'-2y=x(y^4),y(0)=-2 ?
The solution for 6y^'-2y=x(y^4),y(0)=-2 is y=(2\sqrt[3]{e^x})/(\sqrt[3]{-4e^xx+4e^x-5)}