Notes on this paper
National Exams — December 2018 — 04-BS-1 Mathematics. Three-hour, closed-book exam (one double-sided aid sheet permitted, approved calculator allowed). Format: eight questions offered; any five constitute a complete paper and only the first five appearing in the answer book are marked. All eight are solved below for completeness.
Reference texts: Kreyszig, Advanced Engineering Mathematics (10th ed., Wiley) — ODEs, forced oscillations, tangent planes, line/surface integrals, Stokes'/divergence theorems; Stewart, Calculus: Early Transcendentals (9th ed., Cengage) — vectors, gradients, constrained optimization.
Question 1: General Solutions of Two First–/Second–Order ODEs (a) 10, (b) 10 marks
Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.
Given. (a) The linear first-order ODE $y'+xy=2xe^{-x^2}$. (b) The linear constant-coefficient second-order ODE $y''+y'-6y=0$.
Find. The general solution $y(x)$ in each case.
Approach. (a) Use an integrating factor $\mu=e^{\int x\,dx}$. (b) Solve the characteristic equation for the constant-coefficient homogeneous ODE.
(a) Integrating factor. $\mu(x)=e^{\int x\,dx}=e^{x^2/2}$. Multiplying through,
$$\left(e^{x^2/2}y\right)'=2xe^{-x^2}\cdot e^{x^2/2}=2xe^{-x^2/2}.$$(a) Integrate. $\displaystyle\int 2xe^{-x^2/2}\,dx=-2e^{-x^2/2}+K$, so
$$e^{x^2/2}y=-2e^{-x^2/2}+K\ \Longrightarrow\ y=-2e^{-x^2}+Ke^{-x^2/2}.$$(b) Characteristic equation. $r^2+r-6=0=(r+3)(r-2)\Rightarrow r=-3,\,2$, so
$$y(x)=C_1e^{-3x}+C_2e^{2x}.$$
$$\text{(a)}\quad y(x)=\boxed{Ke^{-x^2/2}-2e^{-x^2}}\qquad\qquad\text{(b)}\quad y(x)=\boxed{C_1e^{-3x}+C_2e^{2x}}$$
Quantity Result (a) Integrating factor $e^{x^2/2}$ (a) $y(x)$ $Ke^{-x^2/2}-2e^{-x^2}$ (b) Characteristic roots $r=-3,\,2$ (b) $y(x)$ $C_1e^{-3x}+C_2e^{2x}$
Concept & Theory
Topic. Integrating-factor method for linear first-order ODEs; characteristic-equation method for constant-coefficient homogeneous ODEs.
Key relations. For $y'+p(x)y=q(x)$, the factor $\mu=e^{\int p\,dx}$ makes the left side $(\mu y)'$; then $\mu y=\int\mu q\,dx$. For $ay''+by'+cy=0$, the trial $y=e^{rx}$ gives the auxiliary equation $ar^2+br+c=0$; distinct real roots give $y=C_1e^{r_1x}+C_2e^{r_2x}$.
Why this works. In (a) the chosen $\mu=e^{x^2/2}$ exactly cancels the $e^{-x^2}$ forcing down to a bare exponential $e^{-x^2/2}$, so the resulting integral is elementary — this cancellation is a deliberate feature of the coefficient $x$ matching the forcing's Gaussian rate. In (b) each exponential term $e^{rx}$ makes the ODE's operator act as multiplication by the characteristic polynomial, so real roots directly give an independent pair of solutions.
Common pitfall. In (a), forgetting to multiply the forcing term by $\mu$ as well as the left side before integrating; in (b), mis-factoring the quadratic (e.g. sign errors on $-3,2$).
Source. Kreyszig, Advanced Engineering Mathematics , Ch. 1 (§1.5 Linear ODEs, Integrating Factor) and Ch. 2 (§2.2 Homogeneous Linear ODEs of Second Order).
