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

The solution for y^'=(x+y-1)^2,y(0)=0 is y=tan(x+(3pi)/4+pin)-x+1

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

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

y = tan (x + \frac{3 π}{4} + πn) - x + 1

Solution steps

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

Solve using substitution:

y = tan (x + \frac{3 π}{4} + πn) - x + 1

y = tan (x + \frac{3 π}{4} + πn) - x + 1

What is the solution for y^'=(x+y-1)^2,y(0)=0 ?
The solution for y^'=(x+y-1)^2,y(0)=0 is y=tan(x+(3pi)/4+pin)-x+1