What is the solution for 2sqrt(xy^')=sqrt(4-y^2) ?

The solution for 2sqrt(xy^')=sqrt(4-y^2) is y=(2e^{c_{1}})/(x-e^{c_{1)}}+(2x)/(x-e^{c_{1)}}

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2sqrt(xy^')=sqrt(4-y^2)

2 x y^{^{'}} = 4 - y^{2}

y = \frac{2 e ^{c_{1}}}{x - e ^{c_{1}}} + \frac{2 x}{x - e ^{c_{1}}}

Solution steps

2 x y^{'} (x) = 4 - y^{2}

Solve separable ODE:

y = \frac{2 e ^{c_{1}}}{x - e ^{c_{1}}} + \frac{2 x}{x - e ^{c_{1}}}

y = \frac{2 e ^{c_{1}}}{x - e ^{c_{1}}} + \frac{2 x}{x - e ^{c_{1}}}

\frac{d y}{d x} + 7 x y = 8 x

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

\frac{d P}{d t} = 0.02 P (1 - \frac{P}{250000})

y^{^{'}} = - 0.00003 (y - 288.03) (y - 961.97)

\frac{d y}{d x} = 4 x (y^{2} + 1)

What is the solution for 2sqrt(xy^')=sqrt(4-y^2) ?
The solution for 2sqrt(xy^')=sqrt(4-y^2) is y=(2e^{c_{1}})/(x-e^{c_{1)}}+(2x)/(x-e^{c_{1)}}