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

The solution for y^'=(x+y-4)^2 is y=tan(x+c_{1})-x+4

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

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

y = tan (x + c_{1}) - x + 4

Solution steps

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

Solve using substitution:

y = tan (x + c_{1}) - x + 4

y = tan (x + c_{1}) - x + 4

x \to 1 lim (cos (π \cdot x))

\int_{2}^{\infty} x e^{- x^{2}} d x

\frac{d}{d x} (4 - x)

\int \frac{sinh ( x )}{1 + cosh ( x )} d x

\frac{d y}{d x} = \frac{2 y}{x ^{2} - 1}

What is the solution for y^'=(x+y-4)^2 ?
The solution for y^'=(x+y-4)^2 is y=tan(x+c_{1})-x+4