02.06.E1 · analysis / transcendental

Ordinary differential equations exercise pack (Apostol Vol. 2 Ch. 6-7 supplement)

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Formal definition of the pack Intermediate

Apostol Vol. 2 Chapters 6-7 treat linear ordinary differential equations through the linear-algebra lens. Chapter 6 develops the first-order linear equation $y^{'} + p (x) y = q (x)$ via the integrating factor, then the $n$ -th-order constant-coefficient equation $L y = 0$ whose solution space is an $n$ -dimensional vector space governed by the characteristic polynomial of $L = p (D)$ ; it adds the Wronskian test for independence, and the inhomogeneous solution by undetermined coefficients and variation of parameters. Chapter 7 generalises to systems $\overset{x}{˙} = A x$ , solved by the matrix exponential $e^{t A}$ , computed through diagonalisation or Jordan form.

This pack collects ten problems from those chapters — three easy, four medium, three hard — each with a hint and a complete worked solution. It tests operational competence: solving a first-order linear equation, building the general solution of a constant-coefficient equation from roots of the characteristic polynomial, handling resonance in the forced case, testing independence with the Wronskian, exponentiating a matrix, and producing a power-series solution.

Conventions follow the prerequisite units: $D = d / d x$ is the differentiation operator; $L = p (D)$ a constant-coefficient operator with characteristic polynomial $p (r)$ ; the Wronskian of $y_{1}, \dots, y_{n}$ is $W = det (y_{j}^{(i - 1)})$ ; $e^{t A} = \sum_{k \geq 0} t^{k} A^{k} / k!$ the matrix exponential.

Key theorem with full solution Intermediate

Before the pack proper, we work one problem in full as an exemplar of the format. The remaining nine follow the same structure (problem, hint, full answer in <details> blocks).

Lead problem. Forced harmonic oscillator with resonance. Solve $y^{''} + ω^{2} y = cos (ω t)$ with $y (0) = 0$ , $y^{'} (0) = 0$ , where the forcing frequency matches the natural frequency $ω$ .

Solution. The homogeneous equation $y^{''} + ω^{2} y = 0$ has characteristic polynomial $r^{2} + ω^{2} = 0$ , roots $r = \pm iω$ , so $y_{h} = c_{1} cos (ω t) + c_{2} sin (ω t)$ .

For a particular solution, the forcing $cos (ω t)$ coincides with a homogeneous solution — this is resonance. Undetermined coefficients then requires multiplying the trial form by $t$ : try $y_{p} = t (A cos ω t + B sin ω t)$ . Differentiate: $$ y_p' = A\cos\omega t + B\sin\omega t + t(-A\omega\sin\omega t + B\omega\cos\omega t), $$ $$ y_p'' = -2A\omega\sin\omega t + 2B\omega\cos\omega t + t(-A\omega^2\cos\omega t - B\omega^2\sin\omega t). $$ Then $y_{p}^{''} + ω^{2} y_{p} = - 2 A ω sin ω t + 2 B ω cos ω t$ (the $t$ -terms cancel). Matching $cos (ω t)$ : $2 B ω = 1$ , so $B = \frac{1}{2 ω}$ ; matching $sin (ω t)$ : $A = 0$ . Thus $y_{p} = \frac{t}{2 ω} sin (ω t)$ .

General solution: $y = c_{1} cos ω t + c_{2} sin ω t + \frac{t}{2 ω} sin ω t$ . Apply initial conditions. $y (0) = c_{1} = 0$ . Then $y^{'} = c_{2} ω cos ω t + \frac{1}{2 ω} sin ω t + \frac{t}{2} cos ω t$ , and $y^{'} (0) = c_{2} ω = 0$ , so $c_{2} = 0$ .

Final: $y (t) = \frac{t}{2 ω} sin (ω t)$ . $□$

The amplitude grows linearly in $t$ — the signature of resonance. Driving an undamped oscillator at its natural frequency feeds energy in coherently, and the response is unbounded. This is the canonical Apostol Ch. 6 resonance example.

Exercises Intermediate

Exercise 4 (medium, symbolic). Repeated root.

Solve $y^{''} - 4 y^{'} + 4 y = 0$ and explain the form of the second solution.

Hint

The characteristic polynomial has a double root. When $r$ is a root of multiplicity $2$ , the solution space is spanned by $e^{r x}$ and $x e^{r x}$ .

Answer

Characteristic polynomial: $r^{2} - 4 r + 4 = (r - 2)^{2}$ , a double root $r = 2$ . A single exponential $e^{2 x}$ gives only a one-dimensional family, but the equation's solution space is two-dimensional, so a second independent solution is needed.

For a root $r$ of multiplicity $m$ , the functions $e^{r x}, x e^{r x}, \dots, x^{m - 1} e^{r x}$ all solve $L y = 0$ . Here $m = 2$ , so the general solution is $$ y = (c_1 + c_2 x)e^{2x}. $$ Verification that $x e^{2 x}$ works: writing $L = (D - 2)^{2}$ , we have $(D - 2) (x e^{2 x}) = e^{2 x}$ and $(D - 2)^{2} (x e^{2 x}) = (D - 2) e^{2 x} = 0$ . The factor $x$ appears precisely because the operator has the repeated factor $(D - 2)$ .

Exercise 6 (medium, proof). Wronskian and Abel's identity.

For $y^{''} + p (x) y^{'} + q (x) y = 0$ with solutions $y_{1}, y_{2}$ , prove the Wronskian $W = y_{1} y_{2}^{'} - y_{2} y_{1}^{'}$ satisfies $W^{'} = - p (x) W$ , hence $W (x) = W (x_{0}) exp (- \int_{x_{0}}^{x} p)$ .

Hint

Differentiate $W$ directly. The $y_{i}^{''}$ terms can be replaced using the ODE, and most terms cancel.

Answer

Differentiate $W = y_{1} y_{2}^{'} - y_{2} y_{1}^{'}$ : $$ W' = y_1' y_2' + y_1 y_2'' - y_2' y_1' - y_2 y_1'' = y_1 y_2'' - y_2 y_1''. $$ Each solution satisfies $y_{i}^{''} = - p y_{i}^{'} - q y_{i}$ . Substitute: $$ W' = y_1(-p y_2' - q y_2) - y_2(-p y_1' - q y_1) = -p(y_1 y_2' - y_2 y_1') - q(y_1 y_2 - y_2 y_1) = -p W, $$ since the $q$ -term vanishes. This first-order linear equation $W^{'} = - p W$ integrates to $W (x) = W (x_{0}) exp (- \int_{x_{0}}^{x} p (t) d t)$ (Abel's formula).

Consequence: the Wronskian is either identically zero or never zero on the interval. So $y_{1}, y_{2}$ are linearly independent iff $W \neq = 0$ at a single point.

Exercise 7 (medium, numeric). Matrix exponential by diagonalisation.

Compute $e^{t A}$ for $A = (1023)$ , and write the solution of $\overset{x}{˙} = A x$ with $x (0) = (1, 1)$ .

Hint

Diagonalise $A = P D P^{- 1}$ ; then $e^{t A} = P e^{t D} P^{- 1}$ with $e^{t D}$ diagonal. The eigenvalues are the diagonal entries (triangular matrix).

Answer

Eigenvalues $1, 3$ (triangular). For $λ = 1$ : $(A - I) = (0022)$ , eigenvector $(1, 0)$ . For $λ = 3$ : $(A - 3 I) = (- 2 0 20)$ , eigenvector $(1, 1)$ . So $$ P = \begin{pmatrix} 1 & 1 \ 0 & 1\end{pmatrix}, \quad P^{-1} = \begin{pmatrix} 1 & -1 \ 0 & 1\end{pmatrix}, \quad e^{tD} = \begin{pmatrix} e^t & 0 \ 0 & e^{3t}\end{pmatrix}. $$ Then $$ e^{tA} = P e^{tD} P^{-1} = \begin{pmatrix} e^t & e^{3t} - e^t \ 0 & e^{3t}\end{pmatrix}. $$ With $x (0) = (1, 1)$ : $x (t) = e^{t A} (1, 1)^{T} = (e^{t} + e^{3 t} - e^{t}, e^{3 t}) = (e^{3 t}, e^{3 t})$ . (Consistent: $(1, 1)$ is the eigenvector for $λ = 3$ , so the trajectory is pure $e^{3 t}$ growth along it.)

Exercise 9 (hard, symbolic). Power-series solution.

Find the power-series solution of $y^{''} - x y = 0$ (the Airy equation) about $x = 0$ with $y (0) = 1$ , $y^{'} (0) = 0$ , giving the recurrence and the first few terms.

Hint

Write $y = \sum_{n \geq 0} a_{n} x^{n}$ , substitute, and match coefficients of $x^{n}$ to get a recurrence relating $a_{n + 2}$ to $a_{n - 1}$ .

Answer

Let $y = \sum_{n \geq 0} a_{n} x^{n}$ . Then $y^{''} = \sum_{n \geq 0} (n + 2) (n + 1) a_{n + 2} x^{n}$ and $x y = \sum_{n \geq 1} a_{n - 1} x^{n}$ . The equation $y^{''} - x y = 0$ gives, matching $x^{n}$ : $$ (n+2)(n+1)a_{n+2} = a_{n-1} \quad (n \geq 1), \qquad 2a_2 = 0 ;(n = 0). $$ So $a_{2} = 0$ , and $a_{n + 2} = \frac{a _{n - 1}}{( n + 2 ) ( n + 1 )}$ .

Initial conditions: $a_{0} = y (0) = 1$ , $a_{1} = y^{'} (0) = 0$ . The recurrence steps by $3$ . Since $a_{1} = 0$ , the chain $a_{1}, a_{4}, a_{7}, \dots$ all vanish; $a_{2} = 0$ kills $a_{2}, a_{5}, a_{8}, \dots$ . Only the $a_{0}$ -chain survives: $$ a_3 = \frac{a_0}{3\cdot 2} = \frac{1}{6}, \qquad a_6 = \frac{a_3}{6\cdot 5} = \frac{1}{180}, \ldots $$ giving $y (x) = 1 + \frac{x ^{3}}{6} + \frac{x ^{6}}{180} + \dots$ — the Airy function $Ai$ -type even-index solution normalised to $y (0) = 1$ , $y^{'} (0) = 0$ .

Exercise 10 (hard, proof). Reduction to a first-order system and uniqueness.

Show that the second-order equation $y^{''} + p (x) y^{'} + q (x) y = 0$ is equivalent to a first-order linear system $\overset{u}{˙} = A (x) u$ , and use this to argue that a solution with $y (x_{0}) = 0$ , $y^{'} (x_{0}) = 0$ must be identically zero.

Hint

Set $u = (y, y^{'})$ . Express $u^{'}$ in terms of $u$ . Then invoke uniqueness for linear first-order systems with a given initial condition.

Answer

Set $u_{1} = y$ , $u_{2} = y^{'}$ , so $u = (u_{1}, u_{2})$ . Then $u_{1}^{'} = u_{2}$ and $u_{2}^{'} = y^{''} = - p y^{'} - q y = - q u_{1} - p u_{2}$ . In matrix form, $$ u' = A(x)u, \qquad A(x) = \begin{pmatrix} 0 & 1 \ -q(x) & -p(x)\end{pmatrix}. $$ This is a first-order linear system, equivalent to the original equation: any solution $y$ gives $u = (y, y^{'})$ solving the system, and conversely the first component of any system solution solves the second-order equation.

For uniqueness: if $y (x_{0}) = 0$ and $y^{'} (x_{0}) = 0$ , then $u (x_{0}) = (0, 0)$ . The constant zero function $u \equiv 0$ solves $u^{'} = A (x) u$ with this initial value. With $p, q$ continuous, $A (x)$ is continuous, so the linear system satisfies the Picard-Lindelöf hypotheses and its solution with given initial data is unique. Therefore $u \equiv 0$ , hence $y \equiv 0$ . This is why the solution space is exactly two-dimensional: a solution is pinned down by the pair $(y (x_{0}), y^{'} (x_{0}))$ .

Exercise pack. Apostol Vol. 2 Chapters 6-7 supplement: first-order linear and separable equations, constant-coefficient equations and the characteristic polynomial, the Wronskian, series solutions, and systems via the matrix exponential.

Prerequisites

02.06.02 pending
02.06.03 pending
02.08.01 pending
02.08.02 pending

Tier anchors

beginner
intermediate: Apostol Vol. 2 exercises (Ch. 6 §6.8-6.24, Ch. 7 §7.5-7.16): first-order linear/separable ODEs, n-th-order constant-coefficient ODEs, the Wronskian, systems and the matrix exponential, series solutions
master

References

Apostol — Calculus, Volume II: Multi-Variable Calculus and Linear Algebra · Ch. 6 exercises (first-order linear ODEs, n-th-order constant-coefficient equations, undetermined coefficients, variation of parameters, the Wronskian) and Ch. 7 exercises (systems of linear ODEs, the matrix exponential)
Tenenbaum-Pollard — Ordinary Differential Equations · Lessons 11-24, 64-65 — first/second-order linear equations, series solutions, systems
Boyce-DiPrima — Elementary Differential Equations and Boundary Value Problems · Ch. 2-3, 5, 7 — first/second-order linear ODEs, series solutions, linear systems

Estimated time

intermediate: 4h