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

The solution for 2y^'=y^3*cos(x) is y=sqrt(-1/(sin(x)+c_{1))},y=-sqrt(-1/(sin(x)+c_{1))}

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2y^'=y^3*cos(x)

2 y^{^{'}} = y^{3} \cdot cos (x)

y = - \frac{1}{sin ( x ) + c _{1}}, y = - - \frac{1}{sin ( x ) + c _{1}}

Solution steps

2 y^{'} (x) = y^{3} \cdot cos (x)

Solve separable ODE:

y = - \frac{1}{sin ( x ) + c _{1}}, y = - - \frac{1}{sin ( x ) + c _{1}}

y = - \frac{1}{sin ( x ) + c _{1}}, y = - - \frac{1}{sin ( x ) + c _{1}}

What is the solution for 2y^'=y^3*cos(x) ?
The solution for 2y^'=y^3*cos(x) is y=sqrt(-1/(sin(x)+c_{1))},y=-sqrt(-1/(sin(x)+c_{1))}