What is the solution for 2sqrt(x)y^'=sqrt(1-y^2) ?

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

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2sqrt(x)y^'=sqrt(1-y^2)

2 x y^{^{'}} = 1 - y^{2}

y = sin (x + c_{1})

Solution steps

2 x y^{'} (x) = 1 - y^{2}

Solve separable ODE:

y = sin (x + c_{1})

y = sin (x + c_{1})

x^{^{'}} - 5 x = 25

\frac{d}{d θ} (r (θ)) = \frac{r ( θ ) ^{2}}{θ}, r (1) = 2

\frac{d y}{d t} + 7 y = 6, y (0) = 0

x f^{^{'}} (x) = (f (x))^{2}

y^{^{'}} = \frac{1}{2 x} y^{2} - \frac{1}{x} y - \frac{4}{x}

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