Question:
$y=x^{2}+2 x+\mathrm{C} \quad: y^{\prime}-2 x-2=0$
Solution:
$y=x^{2}+2 x+\mathrm{C}$
Differentiating both sides of this equation with respect to x, we get:
$y^{\prime}=\frac{d}{d x}\left(x^{2}+2 x+\mathrm{C}\right)$
$\Rightarrow y^{\prime}=2 x+2$
Substituting the value of $y^{\prime}$ in the given differential equation, we get:
L.H.S. $=y^{\prime}-2 x-2=2 x+2-2 x-2=0=$ R.H.S.
Hence, the given function is the solution of the corresponding differential equation.
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