What is the solution for y^'=4+e^{y-4x+1} ?

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

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y^'=4+e^{y-4x+1}

y^{^{'}} = 4 + e^{y - 4 x + 1}

y = - ln (- x - c_{1}) + 4 x - 1

Solution steps

y^{'} (x) = 4 + e^{y - 4 x + 1}

Solve using substitution:

y = - ln (- x - c_{1}) + 4 x - 1

y = - ln (- x - c_{1}) + 4 x - 1

\frac{400}{s ^{2} + 12 s + 400}

f (x) = x^{2} + 4, (- 3, 13)

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\frac{\partial}{\partial x} (3 x y + 4 x z + 5 yz - 3)

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