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

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

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(dy)/(dt)=(2t)/(1+2y),y(2)=0

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

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

Solution steps

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

Solve separable ODE:

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

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

y^{^{'}} = (1 - y)^{2}

\frac{d}{d x} ((y - 3)^{2}) = 2 - (x - 1)^{2}

\frac{d}{df} (H (f)) = 3 u (f)^{2} f

x^{^{'}} = 0.0175 - 0.3821 x

2 \frac{d x}{d t} + \frac{d x}{d t} - 2 x = 1

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