What is the solution for (dy)/(dt)=e^{-y}(2t-4) ?

The solution for (dy)/(dt)=e^{-y}(2t-4) is y=ln(t^2-4t+c_{1})

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(dy)/(dt)=e^{-y}(2t-4)

\frac{d y}{d t} = e^{- y} (2 t - 4)

y = ln (t^{2} - 4 t + c_{1})

Solution steps

\frac{d y}{d t} = e^{- y} (2 t - 4)

Solve separable ODE:

y = ln (t^{2} - 4 t + c_{1})

y = ln (t^{2} - 4 t + c_{1})

y e^{- x} y^{^{'}} = x

y^{^{'}} = \frac{sin ( x )}{sin ( y )}

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

y^{^{'}} - e^{- y} sec^{2} (x) = 0

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

What is the solution for (dy)/(dt)=e^{-y}(2t-4) ?
The solution for (dy)/(dt)=e^{-y}(2t-4) is y=ln(t^2-4t+c_{1})