What is the solution for y^'+2y^2cos(x+1)=0 ?

The solution for y^'+2y^2cos(x+1)=0 is y=-1/(-2sin(1+x)+c_{1)}

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y^'+2y^2cos(x+1)=0

y^{^{'}} + 2 y^{2} cos (x + 1) = 0

y = - \frac{1}{- 2 sin ( 1 + x ) + c _{1}}

Solution steps

y^{'} (x) + 2 y^{2} cos (x + 1) = 0

Solve separable ODE:

y = - \frac{1}{- 2 sin ( 1 + x ) + c _{1}}

y = - \frac{1}{- 2 sin ( 1 + x ) + c _{1}}

\frac{d}{d x} (f (x) \cdot f (x))

x \to 4 - lim (arcsin (\frac{x}{4}))

x \to 0 lim (\frac{x ^{2} h + 3 x h ^{2} + h ^{3}}{2 x h + 5 h ^{2}})

y = x + 2, y = \frac{1}{x + 1}, x = 2, x = 0

\frac{\partial}{\partial x} (e^{y l n (x)})

What is the solution for y^'+2y^2cos(x+1)=0 ?
The solution for y^'+2y^2cos(x+1)=0 is y=-1/(-2sin(1+x)+c_{1)}