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

The solution for y^'=(x+y-4)^2,y(0)=0 is y=tan(x-1.32581…+pin)-x+4

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

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

y = tan (x - 1.32581 \dots + πn) - x + 4

Solution steps

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

Solve using substitution:

y = tan (x - 1.32581 \dots + πn) - x + 4

y = tan (x - 1.32581 \dots + πn) - x + 4

\frac{10}{( 2 s + 3 )}

\frac{x ^{4}}{4} + \frac{1}{8 x ^{2}}

\frac{d}{d x} ((x^{2} - x + 1)^{- 7})

\int e^{- 3 x} sin (3 x) d x

\frac{\partial}{\partial y} (sin (x - y))

What is the solution for y^'=(x+y-4)^2,y(0)=0 ?
The solution for y^'=(x+y-4)^2,y(0)=0 is y=tan(x-1.32581…+pin)-x+4