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

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

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(dy)/(dx)=-y^2,y(1)= 1/2

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

y = \frac{1}{x + 1}

Solution steps

\frac{d y}{d x} = - y^{2}

Solve separable ODE:

y = \frac{1}{x + 1}

y = \frac{1}{x + 1}

\frac{d y}{d x} = \frac{1}{y}

\frac{d}{d x} (sin (x) - cos (y) - 2) = 0

y^{^{'}} - 0.1 x y = x

\frac{d y}{d x} = \frac{15 x ^{2} - 6 x + 2}{( y + 2 )}

\frac{d y}{d x} = \frac{1 + x}{x y ^{6}}, y (1) = 5

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