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

The solution for (dy)/(dx)=(2x)/(1+2y),y(2)=0 is y=(-1+sqrt(4x^2-15))/2

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(dy)/(dx)=(2x)/(1+2y),y(2)=0

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

y = \frac{- 1 + 4 x ^{2} - 15}{2}

Solution steps

\frac{d y}{d x} = \frac{2 x}{1 + 2 y}

Solve separable ODE:

y = \frac{- 1 + 4 x ^{2} - 15}{2}

y = \frac{- 1 + 4 x ^{2} - 15}{2}

n = 1 \sum \infty (- 1)^{n - 1} \frac{2 ^{n}}{n ^{4}}

x \to 4 lim (2^{x})

\int \frac{sin ( 5 x )}{16 + cos ^{2} ( 5 x )} d x

\frac{\partial}{\partial x} (6 x + 5 y)

\int \frac{24 t + 6 t ^{3}}{8 t ^{2} + t ^{4}} d t

What is the solution for (dy)/(dx)=(2x)/(1+2y),y(2)=0 ?
The solution for (dy)/(dx)=(2x)/(1+2y),y(2)=0 is y=(-1+sqrt(4x^2-15))/2