What is the solution for (dy)/(dx)=1-e^{x-y+8} ?

The solution for (dy)/(dx)=1-e^{x-y+8} is y=ln(-x-c_{1})+x+8

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(dy)/(dx)=1-e^{x-y+8}

\frac{d y}{d x} = 1 - e^{x - y + 8}

y = ln (- x - c_{1}) + x + 8

Solution steps

\frac{d y}{d x} = 1 - e^{x - y + 8}

Solve using substitution:

y = ln (- x - c_{1}) + x + 8

y = ln (- x - c_{1}) + x + 8

- y^{^{'''}} + 2 y^{^{''}} + 3 y^{^{'}} + 3 = 0

h (x) = \frac{1}{x ^{10}}

\int \frac{1}{x ^{- 1} + 1} d x

\frac{d}{d x} (ln (x + 2 y))

f (x) = 3 x^{5} - 11 x^{3}, at x = 1

What is the solution for (dy)/(dx)=1-e^{x-y+8} ?
The solution for (dy)/(dx)=1-e^{x-y+8} is y=ln(-x-c_{1})+x+8