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

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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