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

The solution for y^'=(cos(x))/(2y),y(0)=-1 is y=-sqrt(sin(x)+1)

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y^'=(cos(x))/(2y),y(0)=-1

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

y = - sin (x) + 1

Solution steps

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

Solve separable ODE:

y = - sin (x) + 1

y = - sin (x) + 1

(x^{3} + 5 x^{3} y^{4}) d x + e^{x^{4}} y^{3} d y = 0

\frac{d y}{d x} = - \frac{6}{9} x^{5} y^{9}

y^{^{'}} = e^{x^{2}} 64 - y^{2}

y^{^{'}} = \frac{y ^{2}}{1 + x ^{2}}

y^{^{'}} = \frac{e ^{x}}{e ^{y}}, y (0) = ln (2)

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