What is the solution for y^'=4+e^{y-4x+1},y(0)=-1 ?

The solution for y^'=4+e^{y-4x+1},y(0)=-1 is y=-ln(-x+1)+4x-1

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y^'=4+e^{y-4x+1},y(0)=-1

y^{^{'}} = 4 + e^{y - 4 x + 1}, y (0) = - 1

y = - ln (- x + 1) + 4 x - 1

Solution steps

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

Solve using substitution:

y = - ln (- x + 1) + 4 x - 1

y = - ln (- x + 1) + 4 x - 1

38 (e^{x^{2}} + 2 e^{x^{2}} x^{2})

\frac{0.5}{( s ^{2} + s + 1 ) \cdot s}

y = \frac{1}{1 - x}, x = 3

\frac{d y}{d x} = 6 y (10 - y)

\frac{d y}{d x} = \frac{1}{8} y cos^{2} (y)

What is the solution for y^'=4+e^{y-4x+1},y(0)=-1 ?
The solution for y^'=4+e^{y-4x+1},y(0)=-1 is y=-ln(-x+1)+4x-1