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

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

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

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

x^{2}

Solution steps

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

Constant function results in a constant

y (0) = 2

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

Expand

(x - 2 + 2)^{2} : x^{2}

= x^{2}

x \to 8 + lim (8 x - 8)

\frac{d}{d t} (\frac{cos ( t )}{- sin ( t )})

\frac{d}{d x} (- e^{- x})

\frac{1}{1 - 2 x}, 0

\frac{d y}{d x} = 5 y + 3