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}}

\int e^{4 x - 5} d x

\int \frac{x ^{2} - 8}{4 x - x ^{3}} d x

- \frac{x}{6 y}

\frac{d ^{2}}{d x ^{2}} ((x^{2} + 1)^{5})

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

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))}