What is the solution for y^'=1+e^{y-x+5} ?

The solution for y^'=1+e^{y-x+5} is y=-ln(-x-c_{1})+x-5

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y^'=1+e^{y-x+5}

y^{^{'}} = 1 + e^{y - x + 5}

y = - ln (- x - c_{1}) + x - 5

Solution steps

y^{'} (x) = 1 + e^{y - x + 5}

Solve using substitution:

y = - ln (- x - c_{1}) + x - 5

y = - ln (- x - c_{1}) + x - 5

\int (x^{2} - 4)^{2} (2 x) d x

\int 13 sin^{3} (x co) s^{2} x d x

\int \frac{arctan ( x )}{x} d x

f (x) = tan^{4} (x^{3})

x \to 0 lim (\frac{5}{x ^{2}})

What is the solution for y^'=1+e^{y-x+5} ?
The solution for y^'=1+e^{y-x+5} is y=-ln(-x-c_{1})+x-5