What is the solution for (dy)/(dx)=y^2cos(x),y(0)=1 ?

The solution for (dy)/(dx)=y^2cos(x),y(0)=1 is y=-1/(sin(x)-1)

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(dy)/(dx)=y^2cos(x),y(0)=1

\frac{d y}{d x} = y^{2} cos (x), y (0) = 1

y = - \frac{1}{sin ( x ) - 1}

Solution steps

\frac{d y}{d x} = y^{2} cos (x)

Solve separable ODE:

y = - \frac{1}{sin ( x ) - 1}

y = - \frac{1}{sin ( x ) - 1}

2 y \cdot y^{^{'}} - y^{2} x - 2 x = 0

y^{^{'}} = - \frac{2 y}{2 x + 100}

\frac{( x ^{2} - 9 )}{x} \cdot \frac{d y}{d x} + y = 0

\frac{d y}{d x} = \frac{x ^{2} + 7 x + 3}{y ^{2}}

(x y^{2} - x) d x + (2 x^{2} + 4) d y = 0

What is the solution for (dy)/(dx)=y^2cos(x),y(0)=1 ?
The solution for (dy)/(dx)=y^2cos(x),y(0)=1 is y=-1/(sin(x)-1)