What is the solution for (dy)/(dx)=4+e^{y-4x+5} ?

The solution for (dy)/(dx)=4+e^{y-4x+5} is y=-ln(-x-c_{1})+4x-5

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(dy)/(dx)=4+e^{y-4x+5}

\frac{d y}{d x} = 4 + e^{y - 4 x + 5}

y = - ln (- x - c_{1}) + 4 x - 5

Solution steps

\frac{d y}{d x} = 4 + e^{y - 4 x + 5}

Solve using substitution:

y = - ln (- x - c_{1}) + 4 x - 5

y = - ln (- x - c_{1}) + 4 x - 5

\int x^{2} \cdot e^{3 x} d x

\frac{d}{d x} (4 3 x^{2} - 5 x + 2)

y^{^{''}} + 2 y^{^{'}} + y = 4 e^{- t}, y (0) = 2, y^{^{'}} (0) = - 1

x^{^{'}} = x^{2} + 1

y^{^{''}} + 6 y^{^{'}} - 5 y = 0

What is the solution for (dy)/(dx)=4+e^{y-4x+5} ?
The solution for (dy)/(dx)=4+e^{y-4x+5} is y=-ln(-x-c_{1})+4x-5