What is the solution for y^'=1-e^{x-y+8} ?

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

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y^'=1-e^{x-y+8}

y^{^{'}} = 1 - e^{x - y + 8}

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

Solution steps

y^{'} (x) = 1 - e^{x - y + 8}

Solve using substitution:

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

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

\frac{5}{8 s}

x \to 3 + lim ((x^{2} - 9)^{x - 3})

\int \frac{t ^{2} + 8}{t ^{2} - 5 t + 6} d t

3 x^{2} + 4 x

\int ∣ x - 5 ∣ d x

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