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

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

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

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

y = tan (x + πn) - x + 2

Solution steps

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

Solve using substitution:

y = tan (x + πn) - x + 2

y = tan (x + πn) - x + 2

\int_{- 10}^{10} 100 - x^{2} d x

cos (y) y^{^{'}} + \frac{1}{x} sin (y) = 1

y = 2 x, y = 2, x = 0

50

y^{^{'}} = \frac{y ^{3}}{e ^{x}}, y (0) = - 1

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