What is the solution for y^'=(2x)/(1+2y),y(2)=-0.5 ?

The solution for y^'=(2x)/(1+2y),y(2)=-0.5 is y=(-1+2sqrt(x^2-4))/2 ,y=-(2sqrt(x^2-4)+1)/2

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y^'=(2x)/(1+2y),y(2)=-0.5

y^{^{'}} = \frac{2 x}{1 + 2 y}, y (2) = - 0.5

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

Solution steps

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

Solve separable ODE:

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

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

What is the solution for y^'=(2x)/(1+2y),y(2)=-0.5 ?
The solution for y^'=(2x)/(1+2y),y(2)=-0.5 is y=(-1+2sqrt(x^2-4))/2 ,y=-(2sqrt(x^2-4)+1)/2