What is the solution for y^2y^'=(1+y^3)cos(x) ?

The solution for y^2y^'=(1+y^3)cos(x) is y=\sqrt[3]{e^{3sin(x)+c_{1}}-1}

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y^2y^'=(1+y^3)cos(x)

y^{2} y^{^{'}} = (1 + y^{3}) cos (x)

y = 3 e^{3 s i n (x) + c_{1}} - 1

Solution steps

y^{2} y^{'} (x) = (1 + y^{3}) cos (x)

Solve separable ODE:

y = 3 e^{3 s i n (x) + c_{1}} - 1

y = 3 e^{3 s i n (x) + c_{1}} - 1

\frac{d y}{d t} = y (y - 1)

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

y^{^{'}} = \frac{x ^{4}}{y ^{2}}

\frac{d y}{d x} = x y + 6 x + 5 y + 30

\frac{d P}{d t} = 0.05 P - 0.0005 P^{2}

What is the solution for y^2y^'=(1+y^3)cos(x) ?
The solution for y^2y^'=(1+y^3)cos(x) is y=\sqrt[3]{e^{3sin(x)+c_{1}}-1}