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

The solution for y^'=(2x)/(1+2y),y(0)=2 is y=(-1+sqrt(4x^2+25))/2

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

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

y = \frac{- 1 + 4 x ^{2} + 25}{2}

Solution steps

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

Solve separable ODE:

y = \frac{- 1 + 4 x ^{2} + 25}{2}

y = \frac{- 1 + 4 x ^{2} + 25}{2}

\int sec (\frac{x}{5}) d x

\frac{1}{s ( s ^{2} + 5 s + 6 )}

x \to 0 lim (x sin (\frac{1}{x ^{2}}))

f (x) = \frac{x}{x - 1}

f (x) = 6^{x}

What is the solution for y^'=(2x)/(1+2y),y(0)=2 ?
The solution for y^'=(2x)/(1+2y),y(0)=2 is y=(-1+sqrt(4x^2+25))/2