What is the solution for y^'=(x+y-5)^2 ?

The solution for y^'=(x+y-5)^2 is y=tan(x+c_{1})-x+5

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y^'=(x+y-5)^2

y^{^{'}} = (x + y - 5)^{2}

y = tan (x + c_{1}) - x + 5

Solution steps

y^{'} (x) = (x + y - 5)^{2}

Solve using substitution:

y = tan (x + c_{1}) - x + 5

y = tan (x + c_{1}) - x + 5

\frac{d}{d x} (\frac{ln ( x )}{x + 2})

\frac{d}{d x} (x^{3} sin (5 x))

\frac{d}{d x} (x^{4} - 4 x^{3} + 4 x^{2})

2 y = 5 x, y = 4, 2 y + 3 x = 8

\int (1 + x + x^{2}) ln (x) d x

What is the solution for y^'=(x+y-5)^2 ?
The solution for y^'=(x+y-5)^2 is y=tan(x+c_{1})-x+5