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

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

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

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

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

Solution steps

2 y^{'} (x) = y^{3} 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}}

x \to 0 lim (x^{4} sin (\frac{2}{x}))

g (u) = 7 u + 13 u

\frac{\partial}{\partial x} (x e^{3 x y})

\int \frac{cos ( 2 x )}{x} d x

\int 8 cos (θ) cos^{5} (sin (θ)) d θ

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