For the following question verify that the given function (explicit or implicit) is a solution of the corresponding differential equation.

$y = x^{2} + 2 x + C a n d y^{'} - 2 x - 2 = 0$

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Solution

Given, $y = x^{2} + 2 x + C$
On differentiating both sides w.r.t. x, we get
$\Rightarrow \frac{d y}{d x} = y^{'} = \frac{d}{d x} (x^{2} + 2 x + C) \Rightarrow y^{'} = 2 x + 2$
But we have to verify, y'-2x-2=0 ....(i)
On substituting the value of y' in Eq. (i), we get
LHS=y'-2x-2=2x+2-2x-2=0=RHS
Hence, $y = x^{2} + 2 x + C$ is a solution of the given differential equation.

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For the following question verify that the given function (explicit or implicit) is a solution of the corresponding differential equation. y=x2+2x+C and y′−2x−2=0

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For the following question verify that the given function (explicit or implicit) is a solution of the corresponding differential equation.

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