What is the solution for y^'=(2x)/(1+2y),y(3)=0 ?

The solution for y^'=(2x)/(1+2y),y(3)=0 is y=(-1+sqrt(4x^2-35))/2

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y^'=(2x)/(1+2y),y(3)=0

y^{^{'}} = \frac{2 x}{1 + 2 y}, y (3) = 0

y = \frac{- 1 + 4 x ^{2} - 35}{2}

Solution steps

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

Solve separable ODE:

y = \frac{- 1 + 4 x ^{2} - 35}{2}

y = \frac{- 1 + 4 x ^{2} - 35}{2}

\frac{d y}{d x} = \frac{x y - x ^{3} y}{1 - y}

x d y - y d x = 0, y (1) = 2

\frac{d y}{d x} = \frac{y}{x}, y (5) = 20

\frac{d y}{d x} = - \frac{3}{7} x^{6} y^{5}

y^{^{'}} = (sin^{5} (x) cos^{2} (x)) (y^{2} + 1), y (0) = 0

What is the solution for y^'=(2x)/(1+2y),y(3)=0 ?
The solution for y^'=(2x)/(1+2y),y(3)=0 is y=(-1+sqrt(4x^2-35))/2