What is the solution for y^'-(4x-y+1)^2=0 ?

The solution for y^'-(4x-y+1)^2=0 is y=-(-2e^{4x+4c_{1}}-2)/(1-e^{4(c_{1)+x)}}+4x+1

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y^'-(4x-y+1)^2=0

y^{^{'}} - (4 x - y + 1)^{2} = 0

y = - \frac{- 2 e ^{4 x + 4 c_{1}} - 2}{1 - e ^{4 (c_{1} + x)}} + 4 x + 1

Solution steps

y^{'} (x) - (4 x - y + 1)^{2} = 0

Solve using substitution:

y = - \frac{- 2 e ^{4 x + 4 c_{1}} - 2}{1 - e ^{4 (c_{1} + x)}} + 4 x + 1

y = - \frac{- 2 e ^{4 x + 4 c_{1}} - 2}{1 - e ^{4 (c_{1} + x)}} + 4 x + 1

\frac{d}{d x} (\frac{1}{0})

\int sin (2) d x

\frac{d y}{d x} + \frac{3 y}{x} + 2 = 3 x

12 x^{2} y + y^{^{'}} = 5 x^{2}

f (x) = cos (2 x)

What is the solution for y^'-(4x-y+1)^2=0 ?
The solution for y^'-(4x-y+1)^2=0 is y=-(-2e^{4x+4c_{1}}-2)/(1-e^{4(c_{1)+x)}}+4x+1