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

The solution for y^'=((2x))/(1+2y),y(2)=1 is y=(-1+sqrt(4x^2-7))/2

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

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

y = \frac{- 1 + 4 x ^{2} - 7}{2}

Solution steps

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

Solve separable ODE:

y = \frac{- 1 + 4 x ^{2} - 7}{2}

y = \frac{- 1 + 4 x ^{2} - 7}{2}

x \to 0 lim (ln (2))

n = 1 \sum \infty \frac{1}{n ^{2} + 1}

2 x y d x + (x^{2} + 1) d y = 0

\frac{d}{d x} (e^{x} - 2 x)

What is the solution for y^'=((2x))/(1+2y),y(2)=1 ?
The solution for y^'=((2x))/(1+2y),y(2)=1 is y=(-1+sqrt(4x^2-7))/2