Question 1 of 8: Two Linear ODEs — Variation of Parameters with a Secant Forcing, and a Resonant Polynomial+Exponential Forcing
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
National Exams — May 2014 — 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, Laplace transforms/unit-step forcing, surface/flux integrals, Stokes' theorem; Stewart, Calculus: Early Transcendentals (9th ed., Cengage) — Lagrange multipliers, planes/tangent lines in space, volumes of revolution; Strang, Introduction to Linear Algebra (6th ed.) — eigenvalues/eigenvectors and linear systems of ODEs.
Question 1: Two Linear ODEs — Variation of Parameters with a Secant Forcing, and a Resonant Polynomial+Exponential Forcing (a) 10, (b) 10 marks
Given. (a) A linear, constant-coefficient ODE with a $\sec 3x$ forcing term (not a polynomial-times-exponential, so undetermined coefficients does not apply). (b) A linear, constant-coefficient ODE forced by a sum $3x^2+e^{-2x}$, where the exponential term's rate matches a homogeneous root.
Find. (a) The general solution $y(x)$. (b) The general solution $y(x)$.
Approach. (a) Solve the homogeneous equation, then use variation of parameters (the only method that handles $\sec$ forcing) with the Wronskian of $\cos 3x,\sin 3x$. (b) Split the forcing by superposition: a polynomial trial for $3x^2$, and — because $e^{-2x}$ duplicates the homogeneous root $r=-2$ — a resonance trial $Dxe^{-2x}$ for the exponential piece.
(a) Homogeneous solution and Wronskian. $r^2+9=0\Rightarrow r=\pm3i$, so $y_1=\cos3x,\ y_2=\sin3x$, and
$$W=y_1y_2'-y_2y_1'=3\cos^23x+3\sin^23x=3.$$
(a) Variation-of-parameters integrals. With $f(x)=\sec3x$,
$$\int\frac{y_2f}{W}dx=\int\frac{\sin3x\sec3x}{3}dx=\frac13\int\tan3x\,dx=-\frac19\ln|\cos3x|,$$
$$\int\frac{y_1f}{W}dx=\int\frac{\cos3x\sec3x}{3}dx=\int\frac13dx=\frac{x}{3}.$$
(b) Homogeneous solution. $r^2-r-6=0=(r-3)(r+2)\Rightarrow r=3,-2$, so $y_h=C_1e^{3x}+C_2e^{-2x}$.
(b) Particular solution for $3x^2$. Try $y_{p1}=Ax^2+Bx+C$. Substituting and matching coefficients of $x^2,x^1,x^0$:
$$-6A=3,\quad -2A-6B=0,\quad 2A-B-6C=0\ \Rightarrow\ A=-\tfrac12,\ B=\tfrac16,\ C=-\tfrac{7}{36}.$$
$$y_{p1}=-\tfrac12x^2+\tfrac16x-\tfrac{7}{36}.$$
(b) Particular solution for $e^{-2x}$ — resonance. Since $r=-2$ is already a homogeneous root, the plain trial $Ae^{-2x}$ fails; use $y_{p2}=Dxe^{-2x}$. Substituting,
$$y_{p2}''-y_{p2}'-6y_{p2}=-5De^{-2x}\ \Rightarrow\ D=-\tfrac15.$$
$$y_{p2}=-\tfrac15xe^{-2x}.$$
$$y(x)=\boxed{C_1e^{3x}+C_2e^{-2x}-\tfrac12x^2+\tfrac16x-\tfrac{7}{36}-\tfrac15xe^{-2x}}$$