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

The solution for (dy)/(dt)=(2t+1)/(2y-2) is y=1+sqrt(t^2+t+c_{1)+1},y=1-sqrt(t^2+t+c_{1)+1}

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(dy)/(dt)=(2t+1)/(2y-2)

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

y = 1 + t^{2} + t + c_{1} + 1, y = 1 - t^{2} + t + c_{1} + 1

Solution steps

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

Solve separable ODE:

y = 1 + t^{2} + t + c_{1} + 1, y = 1 - t^{2} + t + c_{1} + 1

y = 1 + t^{2} + t + c_{1} + 1, y = 1 - t^{2} + t + c_{1} + 1

y^{^{'}} = - \frac{1}{x} y

v \frac{d v}{d t} - t (v (t)^{2} + 1) = 0, v (0) = - 1

\frac{d y}{d x} = \frac{x arctan ( x )}{y}

7 x y d x - 4 x d y = 7 x y d y - 8 y d x

y^{^{'}} = 7 x y

What is the solution for (dy)/(dt)=(2t+1)/(2y-2) ?
The solution for (dy)/(dt)=(2t+1)/(2y-2) is y=1+sqrt(t^2+t+c_{1)+1},y=1-sqrt(t^2+t+c_{1)+1}