FE Practice Exam

This is a first-order linear differential equation with constant coefficients, so the solution is the homogeneous solution plus a particular solution.

Homogeneous part: the characteristic equation of $y' + 2y = 0$ is $r + 2 = 0$, so $y_h = Ce^{-2x}$.

Particular part: the forcing term is a constant, so try a constant $y_p = A$. Then $y_p' = 0$ and $2A = 6$, giving $A = 3$. This is also the steady-state value the solution approaches as $x \to \infty$.

General solution:

Apply the initial condition: $y(0) = 3 + C = 1$, so $C = -2$. The exponential term is subtracted because the solution starts below its steady-state value and rises toward it.

Evaluate at $x = 0.5$:

y(0.5) ≈ 2.26 (option B)

The most tempting wrong answer is C (3.74). It comes from taking $C = +2$, which would make the solution start above the steady-state value and decay toward it; the initial value 1 is below 3, so the sign must be negative. Option A is the homogeneous solution alone with the particular solution dropped, and option D uses 6 instead of $6/2 = 3$ as the steady-state value.

Replace the distributed load by its resultant so that moments can be taken easily: $wL = 4 \times 6 = 24\,\text{kN}$, acting at the centre of the loaded length, 3 m from A.

Take moments about A, because the unknown reaction at A passes through A and drops out of the equation. With counter-clockwise positive and $R_B$ assumed upward:

Vertical force equilibrium then gives the reaction at A:

Both reactions come out positive, so both act upward as assumed. The reaction at A is the larger one because the 12 kN load sits closer to A than to B, while the distributed load is shared equally.

$R_B$ = 16 kN, acting upward

(a) Convert every cash flow into an equivalent uniform amount per year over the 6-year life, then add the costs and subtract the receipts.

The first cost is a single present amount, so it is spread over the life with the capital-recovery factor:

The salvage value is a single future receipt at year 6, so it is converted with the sinking-fund factor and credited against the costs:

The operating and maintenance cost is already a uniform annual amount, so it is added directly:

EUAC ≈ $14,290 per year

The factors come from the formulas with $(1.08)^6 = 1.5869$: $(A/P, 8\%, 6) = \dfrac{0.08(1.5869)}{1.5869 - 1} = 0.2163$ and $(A/F, 8\%, 6) = (A/P, 8\%, 6) - i = 0.1363$.

(b) The EUAC is a cost per year and the output is packages per year, so dividing the two gives the equivalent cost per package:

About $0.71 per package