Find the differential equation for the family of curves given by eliminating arbitrary constants $a$ and $b$ from the equation: $\frac{x}{a} + \frac{y}{b} = 1$.
(D) Given equation: $\frac{x}{a} + \frac{y}{b} = 1$.
Differentiating both sides with respect to $x$,we get:
$\frac{1}{a} + \frac{1}{b} \frac{dy}{dx} = 0$
Differentiating again with respect to $x$,we get:
$0 + \frac{1}{b} \frac{d^2y}{dx^2} = 0$
Since $b \neq 0$,we must have $\frac{d^2y}{dx^2} = 0$.
Thus,the required differential equation is $y'' = 0$.
Differentiating both sides with respect to $x$,we get:
$\frac{1}{a} + \frac{1}{b} \frac{dy}{dx} = 0$
Differentiating again with respect to $x$,we get:
$0 + \frac{1}{b} \frac{d^2y}{dx^2} = 0$
Since $b \neq 0$,we must have $\frac{d^2y}{dx^2} = 0$.
Thus,the required differential equation is $y'' = 0$.
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