What is the solution for y^'=(x+y-3)^2,y(0)=0 ?

The solution for y^'=(x+y-3)^2,y(0)=0 is y=tan(x-1.24904…+pin)-x+3

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y^'=(x+y-3)^2,y(0)=0

y^{^{'}} = (x + y - 3)^{2}, y (0) = 0

y = tan (x - 1.24904 \dots + πn) - x + 3

Solution steps

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

Solve using substitution:

y = tan (x - 1.24904 \dots + πn) - x + 3

y = tan (x - 1.24904 \dots + πn) - x + 3

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What is the solution for y^'=(x+y-3)^2,y(0)=0 ?
The solution for y^'=(x+y-3)^2,y(0)=0 is y=tan(x-1.24904…+pin)-x+3