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

The solution for y^'=(x-y+1)^2+(x-y) is y=-(-e^{3x+3c_{1}}-2)/(1-e^{3(c_{1)+x)}}+x+1

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

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

y = - \frac{- e ^{3 x + 3 c_{1}} - 2}{1 - e ^{3 (c_{1} + x)}} + x + 1

Solution steps

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

Solve using substitution:

y = - \frac{- e ^{3 x + 3 c_{1}} - 2}{1 - e ^{3 (c_{1} + x)}} + x + 1

y = - \frac{- e ^{3 x + 3 c_{1}} - 2}{1 - e ^{3 (c_{1} + x)}} + x + 1

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What is the solution for y^'=(x-y+1)^2+(x-y) ?
The solution for y^'=(x-y+1)^2+(x-y) is y=-(-e^{3x+3c_{1}}-2)/(1-e^{3(c_{1)+x)}}+x+1