Verify that the given functions (explicit or implicit) is a solution of the corresponding differential equation:
$y = x^{2} + 2 x + C : y^{'} - 2 x - 2 = 0$ .

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Solution

Let,
$y = x^{2} + 2 x + C$
Differentiating both sides w.r.t. $x,$ then we get
$y^{'} = (x^{2} + 2 x + c)^{'}$
$y^{'} = 2 x + 2 + 0$
$y^{'} = 2 x + 2$
Taking LHS
$y^{'} - 2 x - 2 = 2 x + 2 - 2 x - 2$
$y^{'} - 2 x - 2 = 0$
Thus $L H S = R H S$
Hence verified.

Final answer:
Hence, the function $y = x^{2} + 2 x + C$ is a solution of the differential equation $y^{'} - 2 x - 2 = 0.$

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