What is the solution for y^'=e^{-y}(2x-4),y(0)=1 ?

The solution for y^'=e^{-y}(2x-4),y(0)=1 is y=ln(x^2-4x+e)

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y^'=e^{-y}(2x-4),y(0)=1

y^{^{'}} = e^{- y} (2 x - 4), y (0) = 1

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

Solution steps

y^{'} (x) = e^{- y} (2 x - 4)

Solve separable ODE:

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

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

y^{^{''}} + 6 y^{^{'}} + 9 y = 4 e^{- 3 t}

y = ln (x^{3} e^{2 x})

\frac{x ^{3} + 2}{3}

(4 + x) \frac{d y}{d x} = y^{3}

y = 2 x^{3} + 3 x^{2} - 12 x + 9

What is the solution for y^'=e^{-y}(2x-4),y(0)=1 ?
The solution for y^'=e^{-y}(2x-4),y(0)=1 is y=ln(x^2-4x+e)