What is the solution for (dy)/(dx)=3+e^{(-3x+y+4)} ?

The solution for (dy)/(dx)=3+e^{(-3x+y+4)} is y=-ln(-x-c_{1})+3x-4

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(dy)/(dx)=3+e^{(-3x+y+4)}

\frac{d y}{d x} = 3 + e^{(- 3 x + y + 4)}

y = - ln (- x - c_{1}) + 3 x - 4

Solution steps

\frac{d y}{d x} = 3 + e^{(- 3 x + y + 4)}

Solve using substitution:

y = - ln (- x - c_{1}) + 3 x - 4

y = - ln (- x - c_{1}) + 3 x - 4

\frac{\partial}{\partial x} (x^{2} + x y^{3})

\frac{d}{d x} (\frac{4 2}{3} x^{\frac{3}{2}} - 1)

\frac{d}{d x} (- \frac{1}{sin ( x )})

y^{^{''}} + 3 y^{^{'}} = sin (t), y (0) = 1, y^{^{'}} (0) = 2

\int sec (x) + tan (x) d x

What is the solution for (dy)/(dx)=3+e^{(-3x+y+4)} ?
The solution for (dy)/(dx)=3+e^{(-3x+y+4)} is y=-ln(-x-c_{1})+3x-4