Verify that the given function $y = x^{2} + 2x + C$ is a solution of the differential equation $y' - 2x - 2 = 0$.
(N/A) Given function: $y = x^{2} + 2x + C$
Differentiating both sides of this equation with respect to $x$,we get:
$y' = \frac{d}{dx}(x^{2} + 2x + C)$
$y' = 2x + 2$
Now,substitute the value of $y'$ into the given differential equation $y' - 2x - 2 = 0$:
$L.H.S. = y' - 2x - 2$
$L.H.S. = (2x + 2) - 2x - 2$
$L.H.S. = 2x - 2x + 2 - 2 = 0$
$L.H.S. = R.H.S.$
Since the $L.H.S.$ equals the $R.H.S.$,the given function is indeed a solution to the differential equation.
Differentiating both sides of this equation with respect to $x$,we get:
$y' = \frac{d}{dx}(x^{2} + 2x + C)$
$y' = 2x + 2$
Now,substitute the value of $y'$ into the given differential equation $y' - 2x - 2 = 0$:
$L.H.S. = y' - 2x - 2$
$L.H.S. = (2x + 2) - 2x - 2$
$L.H.S. = 2x - 2x + 2 - 2 = 0$
$L.H.S. = R.H.S.$
Since the $L.H.S.$ equals the $R.H.S.$,the given function is indeed a solution to the differential equation.
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