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

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

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(dy)/(dx)=e^{-y}(2x-4)

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

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

Solution steps

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

Solve separable ODE:

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

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

4 y^{^{'}} + 6 y = 120, y (0) = 1

y^{^{'}} = 4 - y^{2}

2 y^{2} y^{^{'}} = 3 e^{2 x}

\frac{d y}{d x} = 0.02 y + 40000, y (0) = 2000000

y (25 - x^{2})^{\frac{1}{2}} d y = (25 + y^{2})^{\frac{1}{2}} d x

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