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04-BS-1 · December 2019

Question 1 of 8: Forced Second-Order IVP with Complex Roots

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National Exams — December 2019 — 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, multiple integrals.

Question 1: Forced Second-Order IVP with Complex Roots (20 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. $y''-12y'+45y=18\cos(3t)$, $y(0)=0$, $y'(0)=0$.

Find. The particular solution $y(t)$ satisfying the initial conditions.

Approach. Solve the homogeneous equation via the characteristic equation (complex roots), find a particular solution by undetermined coefficients, superpose, then fit the two initial conditions.

  1. Homogeneous solution. $r^2-12r+45=0\Rightarrow r=\dfrac{12\pm\sqrt{144-180}}2=6\pm3i$, so $$y_h(t)=e^{6t}(C_1\cos3t+C_2\sin3t).$$ The real part $6\ne0$, so even though the forcing frequency $3$ equals the imaginary part of the roots, this is not resonance.
  2. Particular solution. Try $y_p=A\cos3t+B\sin3t$. Substituting: $y_p''-12y_p'+45y_p=(36A-36B)\cos3t+(36A+36B)\sin3t$. Matching to $18\cos3t+0\sin3t$: $$36A-36B=18,\qquad 36A+36B=0\ \Longrightarrow\ A=\tfrac14,\ B=-\tfrac14.$$ $$y_p(t)=\tfrac14\cos3t-\tfrac14\sin3t.$$
  3. General solution. $y(t)=e^{6t}(C_1\cos3t+C_2\sin3t)+\tfrac14\cos3t-\tfrac14\sin3t$.
  4. Apply $y(0)=0$. $C_1+\tfrac14=0\Rightarrow C_1=-\tfrac14$.
  5. Apply $y'(0)=0$. $y'(t)=6e^{6t}(C_1\cos3t+C_2\sin3t)+e^{6t}(-3C_1\sin3t+3C_2\cos3t)-\tfrac34\sin3t-\tfrac34\cos3t$. At $t=0$: $6C_1+3C_2-\tfrac34=0\Rightarrow6(-\tfrac14)+3C_2=\tfrac34\Rightarrow3C_2=\tfrac94\Rightarrow C_2=\tfrac34$.

$$y(t)=\boxed{e^{6t}\left(-\tfrac14\cos3t+\tfrac34\sin3t\right)+\tfrac14\cos3t-\tfrac14\sin3t}$$

QuantityResult
Roots$6\pm3i$
Particular solution$\tfrac14\cos3t-\tfrac14\sin3t$
$C_1,\,C_2$$-\tfrac14,\ \tfrac34$
$y(t)$$e^{6t}(-\tfrac14\cos3t+\tfrac34\sin3t)+\tfrac14\cos3t-\tfrac14\sin3t$
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