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 — December 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, Cauchy–Euler equations, surface/flux integrals, Stokes' theorem; Stewart, Calculus: Early Transcendentals (9th ed., Cengage) — lines and planes in space, volumes of revolution, tangent-plane linear approximation; 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 $\sec 2x$ 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 $\cos 2x,\sin 2x$. (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. $r^2+4=0\Rightarrow r=\pm2i$, so $y_1=\cos2x,\ y_2=\sin2x$, and
$$W=y_1y_2'-y_2y_1'=2\cos^22x+2\sin^22x=2.$$
(a) Variation-of-parameters integrals. With $f(x)=\sec2x$,
$$\int\frac{y_2f}{W}dx=\int\frac{\sin2x\sec2x}{2}dx=\frac12\int\tan2x\,dx=-\frac14\ln|\cos2x|,$$
$$\int\frac{y_1f}{W}dx=\int\frac{\cos2x\sec2x}{2}dx=\int\frac12dx=\frac{x}{2}.$$
(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,
$$y_{p2}''+y_{p2}'-6y_{p2}=5De^{2x}\ \Rightarrow\ D=\tfrac15.$$
$$y_{p2}=\tfrac15xe^{2x}.$$
$$y(x)=\boxed{C_1e^{-3x}+C_2e^{2x}-\tfrac12x^2-\tfrac16x-\tfrac{7}{36}+\tfrac15xe^{2x}}$$