Congruency in Calculus.

Let $q(x) = e^{\int p(x) dx}$ so that $q'(x) = p(x) \cdot e^{\int p(x) dx} = p(x)q(x)$ . If we multiply the above differential equation through by $q(x)$ , we now obtain:

$q(x)f''(x) + [p(x)q(x)]f'(x) - q(x)f(x) = 0$ ;

or $q(x)f''(x) + q'(x)f'(x) - q(x)f(x) = 0$ ;

or $[q(x)f'(x)]' - q(x)f(x) = 0$ (i)

Knowing that $f(x)$ contains two zeros, namely $f(a) = f(b) = 0$ ( $a \neq b$ ), one can multiply (i) through by $f(x)$ to produce:

$[q(x)f'(x)]' \cdot f(x) - q(x)f(x)^2 = 0$ (ii)

and integrate (ii) over the above zero-boundary conditions to yield:

$\int_{a}^{b} [q(x)f'(x)]' \cdot f(x) dx - \int_{a}^{b} q(x)f(x)^2 dx = 0$ ;

or $[q(x)f'(x)] \cdot f(x)|_{a}^{b} - \int_{a}^{b} q(x)f'(x)^2 dx - \int_{a}^{b} q(x)f(x)^2 dx = 0$ ;

or $[[q(b)f'(b)]f(b) - [q(a)f'(a)]f(a)] - \int_{a}^{b} q(x)f'(x)^2 dx - \int_{a}^{b} q(x)f(x)^2 dx = 0$ (iii)

Applying the zeros of $f(x)$ reduces (iii) to:

$- \int_{a}^{b} q(x)f'(x)^2 dx - \int_{a}^{b} q(x)f(x)^2 dx = 0$ (iv)

Since $q(x) > 0$ for all $x \in \mathbb{R}$ , the only function $f$ that satisfies (iv) is $f(x) = 0$ . Hence, $f(2018!) = \boxed{0}.$