What is the solution for y^'=(x+y-3)^2 ?

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

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y^'=(x+y-3)^2

y^{^{'}} = (x + y - 3)^{2}

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

Solution steps

y^{'} (x) = (x + y - 3)^{2}

Solve using substitution:

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

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

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