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04-BS-1 · May 2013

Question 2 of 8: General Solutions of Three First/Second-Order ODEs

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Notes on this paper

National Exams — May 2013 — 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, systems of ODEs, line/surface integrals, Stokes' theorem; Stewart, Calculus: Early Transcendentals (9th ed., Cengage) — partial derivatives/tangent planes, surface area, multiple integrals; Strang, Introduction to Linear Algebra (6th ed.) — eigenvalues and eigenvectors.

Question 2: General Solutions of Three First/Second-Order ODEs (a) 7, (b) 7, (c) 6 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. Three independent ODEs: (a) first-order linear, (b) first-order separable (Bernoulli with $n=2$), (c) second-order linear homogeneous, constant coefficients.

Find. The general solution of each.

Approach. (a) integrating factor; (b) direct separation of variables; (c) characteristic equation with complex roots.

  1. (a) Integrating factor. Standard form $y'+2xy=2xe^{-x^2}$ has integrating factor $\mu=e^{\int 2x\,dx}=e^{x^2}$. Multiplying through, $$\left(e^{x^2}y\right)'=2xe^{x^2}e^{-x^2}=2x \;\Rightarrow\; e^{x^2}y=x^2+C.$$ $$\Rightarrow\quad y=\boxed{(x^2+C)e^{-x^2}}$$
  2. (b) Separation of variables. Rewrite $\dfrac{dy}{dx}=-2xy^2$. For $y\ne0$, separate: $$\int \frac{dy}{y^2}=-\int 2x\,dx \;\Rightarrow\; -\frac{1}{y}=-x^2+C_1 \;\Rightarrow\; \frac{1}{y}=x^2-C_1.$$ Relabelling the constant, $$\Rightarrow\quad y=\boxed{\dfrac{1}{x^2+C}}\quad\text{(plus the trivial solution } y\equiv 0\text{)}.$$
  3. (c) Characteristic equation. $r^2-2r+3=0\Rightarrow r=\dfrac{2\pm\sqrt{4-12}}{2}=1\pm i\sqrt2$. Complex-conjugate roots give $$\Rightarrow\quad y=\boxed{e^{x}\left(C_1\cos(\sqrt2\,x)+C_2\sin(\sqrt2\,x)\right)}$$
PartGeneral solution
(a)$y=(x^2+C)e^{-x^2}$
(b)$y=\dfrac{1}{x^2+C}$ (plus $y\equiv0$)
(c)$y=e^{x}\left(C_1\cos(\sqrt2\,x)+C_2\sin(\sqrt2\,x)\right)$