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

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

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

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

Solution steps

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

Solve Bernoulli ODE:

f (x) = \frac{6 x + 16}{x + 3}

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

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4 e^{4 x} cos (5 x)

y^{^{''}} + 9 y = 2 sin (2 t), y (0) = 0, y^{^{'}} (0) = - 1

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