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

The solution for 6y^'-2y=y^4 is y=\sqrt[3]{(2e^{c_{1}+t})/(1-e^{t+c_{1)}}}

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6y^'-2y=y^4

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

Solution steps

6 y^{'} (t) - 2 y = y^{4}

Solve separable ODE:

y (225 - x^{2})^{\frac{1}{2}} d y = (225 + y^{2})^{\frac{1}{2}} d x

x^{^{'}} = 1 + x^{2}

y^{^{'}} + (cos (x)) y = cos (x)

(x^{2} + 1) (tan (y)) \frac{d y}{d x} = x

x (2 y + 1) d x - (x^{2} + 1) d y = 0

What is the solution for 6y^'-2y=y^4 ?
The solution for 6y^'-2y=y^4 is y=\sqrt[3]{(2e^{c_{1}+t})/(1-e^{t+c_{1)}}}