Second-order linear ODEs — superposition principle, general theory: Exercises

Before we start, four words we will lean on constantly, each pinned to a picture:

Figure — Second-order linear ODEs — superposition principle, general theory

Read the four-anchor figure above like a control panel, left to right. The leftmost cyan box is the operator $L$ : a function $y$ enters on its left wire and the transformed function $L [y]$ leaves on its right — this is the "machine" of anchor 1. The upper branch shows the homogeneous case, where the output wire reads $L [y] = 0$ (nothing pushes the system). The lower branch shows the forcing term $g (x)$ arriving as an amber input arrow and appearing on the output as $L [y] = g$ — the external push of anchor 3. The right panel draws the Wronskian as the little $2 \times 2$ determinant of $y_{1}, y_{2}$ stacked over their derivatives $y_{1}^{'}, y_{2}^{'}$ : the amber "area" it measures is nonzero exactly when the two solution-arrows are genuinely different directions (anchor 4). Keep this panel in mind — every exercise below pokes at one of these four boxes.

Intuition One theory fact we will need at L5 — repeated roots

For a constant-coefficient homogeneous ODE $a y^{''} + b y^{'} + cy = 0$ , the trick is to try $y = e^{r x}$ , turning calculus into the algebraic characteristic equation $a r^{2} + b r + c = 0$ (see Characteristic equation — constant coefficient ODEs).

Distinct real roots $r_{1} \neq = r_{2}$ : fundamental set ${e^{r_{1} x}, e^{r_{2} x}}$ — two clearly different directions.
Repeated root $r_{1} = r_{2} = r$ : the exponential gives only one solution $e^{r x}$ . The second independent solution is $x e^{r x}$ — an extra factor of $x$ . Fundamental set ${e^{r x}, x e^{r x}}$ .

WHY the extra $x$ : when the two roots collide there is only one exponential direction left, so the solution space would seem one-dimensional — but a second-order ODE always has a two-dimensional solution space. The missing direction reappears as $x e^{r x}$ . L5.3 is exactly this case in its cleanest form ( $r = 0$ ).

Level 1 — Recognition

Can you name the type and read off the pieces?

L1.1

Classify each equation as (a) linear or nonlinear, (b) homogeneous or non-homogeneous. Give the coefficient functions $p (x), q (x)$ and forcing $g (x)$ where linear. $(i) y^{''} + 4 y^{'} + 3 y = 0 (ii) y^{''} + x^{2} y = e^{x} (iii) y^{''} + y y^{'} = sin x$

Recall Solution

WHAT we do: compare each to the standard form $y^{''} + p (x) y^{'} + q (x) y = g (x)$ . WHY this answers everything at once: the standard form is a template with named slots — the moment an equation matches it, "linear?" is answered (does $y$ enter only through those slots, first power, no products/nonlinear functions?), and "homogeneous?" is answered by reading the right-hand slot $g (x)$ (zero or not). One comparison settles all three cousin questions.

(i) Linear (each of $y, y^{'}, y^{''}$ appears once, first power). Homogeneous ( $g = 0$ ). Read off $p = 4$ , $q = 3$ , $g = 0$ .

(ii) Linear — the $x^{2}$ multiplies $y$ , and $x$ is the input variable, not $y$ ; linearity only cares how $y$ enters. Non-homogeneous, $g = e^{x}$ . Here $p = 0$ , $q = x^{2}$ , $g = e^{x}$ .

(iii) Nonlinear — the term $y y^{'}$ is a product of the unknown with its derivative, which no slot of the template allows. No $p, q, g$ apply.

L1.2

Which of the following pairs could form a fundamental set for some second-order homogeneous linear ODE (i.e. are they linearly independent)? Just answer yes/no with a one-line reason. $(a) {e^{x}, e^{2 x}} (b) {e^{x}, 5 e^{x}} (c) {sin x, cos x}$

Recall Solution

WHAT we do: ask "is one a constant multiple of the other?" WHY this is the right test: a fundamental set must span a two-dimensional solution space; if one function is just a scaled copy of the other, together they point in a single direction and can only produce a one-parameter family — too few to meet two initial conditions. So "constant multiple?" directly decides fundamental-set eligibility.

(a) Yes, independent — $e^{2 x} / e^{x} = e^{x}$ is not constant.

(b) No — $5 e^{x}$ is a constant multiple of $e^{x}$ . They span only one direction, one parameter.

(c) Yes, independent — $tan x$ is not constant, so $sin x$ is not a fixed multiple of $cos x$ .

Level 2 — Application

Run the standard machinery.

L2.1

Verify that $y_{1} = e^{- x}$ and $y_{2} = e^{- 3 x}$ each solve $y^{''} + 4 y^{'} + 3 y = 0$ , then write the general solution.

Recall Solution

WHAT & WHY — substitute each candidate. We plug in because "is a solution" means the equation holds after substitution.

For $y_{1} = e^{- x}$ : $y_{1}^{'} = - e^{- x}$ , $y_{1}^{''} = e^{- x}$ . Then $e^{- x} + 4 (- e^{- x}) + 3 e^{- x} = (1 - 4 + 3) e^{- x} = 0. ✓$ For $y_{2} = e^{- 3 x}$ : $y_{2}^{'} = - 3 e^{- 3 x}$ , $y_{2}^{''} = 9 e^{- 3 x}$ . Then $9 e^{- 3 x} + 4 (- 3 e^{- 3 x}) + 3 e^{- 3 x} = (9 - 12 + 3) e^{- 3 x} = 0. ✓$ They are independent (ratio $e^{- 3 x} / e^{- x} = e^{- 2 x}$ not constant), so $y = c_{1} e^{- x} + c_{2} e^{- 3 x} .$

L2.2

Compute the Wronskian $W (y_{1}, y_{2})$ for $y_{1} = e^{- x}, y_{2} = e^{- 3 x}$ and confirm it is never zero.

Recall Solution

WHAT we do: $W = y_{1} y_{2}^{'} - y_{2} y_{1}^{'}$ . WHY: this determinant is the exact condition for "no nonzero $(c_{1}, c_{2})$ kills both $y$ and $y^{'}$ simultaneously" — i.e. independence. $W = e^{- x} (- 3 e^{- 3 x}) - e^{- 3 x} (- e^{- x}) = - 3 e^{- 4 x} + e^{- 4 x} = - 2 e^{- 4 x} .$ Since $e^{- 4 x} > 0$ for every $x$ , $W = - 2 e^{- 4 x} \neq = 0$ everywhere. Independent — confirms L2.1.

L2.3

Solve the IVP $y^{''} + 4 y^{'} + 3 y = 0$ , $y (0) = 2$ , $y^{'} (0) = - 4$ .

Recall Solution

Step 1. General solution (from L2.1): $y = c_{1} e^{- x} + c_{2} e^{- 3 x}$ .

Step 2 — apply conditions. WHY: a second-order ODE integrates in "twice," producing two constants; two initial conditions pin them down uniquely (guaranteed by existence–uniqueness). $y (0) = c_{1} + c_{2} = 2.$ $y^{'} = - c_{1} e^{- x} - 3 c_{2} e^{- 3 x}$ , so $y^{'} (0) = - c_{1} - 3 c_{2} = - 4$ .

Step 3 — solve the pair. Add $c_{1} + c_{2} = 2$ to $- c_{1} - 3 c_{2} = - 4$ : $- 2 c_{2} = - 2 \Rightarrow c_{2} = 1$ , hence $c_{1} = 1$ . $y = e^{- x} + e^{- 3 x} .$ Check: $y (0) = 1 + 1 = 2$ ✓; $y^{'} (0) = - 1 - 3 = - 4$ ✓.

Level 3 — Analysis

Reason about structure, not just crank the handle.

L3.1

Using Abel's theorem, find the Wronskian of any two solutions of $y^{''} + \frac{1}{x} y^{'} + q (x) y = 0 (x > 0)$ up to a constant, without knowing the solutions. Then explain why $W$ cannot vanish at a single isolated point.

Recall Solution

WHAT & WHY. Abel's theorem says $W^{'} = - p (x) W$ , so $W (x) = W (x_{0}) exp (- \int p)$ . This lets us know $W$ 's shape from $p$ alone, no solving required.

Here $p (x) = 1/ x$ . So $W (x) = C exp (- \int \frac{1}{x} d x) = C e^{- l n x} = \frac{C}{x} .$ For $x > 0$ : if $C = 0$ then $W \equiv 0$ ; if $C \neq = 0$ then $W = C / x \neq = 0$ for every $x > 0$ . There is no way for $W$ to be zero at one isolated point and nonzero elsewhere — the exponential is never zero, so $W$ is all-or-nothing. That is why checking one convenient point suffices.

L3.2

Two functions $f (x) = x^{2}$ and $g (x) = x ∣ x ∣$ are given on $I = (- 1, 1)$ . Show $W (f, g) \equiv 0$ everywhere, yet $f, g$ are linearly independent. Why does this NOT contradict the Wronskian test?

Recall Solution

Set up piecewise. For $x \geq 0$ , $g = x^{2}$ ; for $x < 0$ , $g = - x^{2}$ .

Region $x \geq 0$ : $f = x^{2}, g = x^{2}$ , so $g = f$ there and $W = f g^{'} - g f^{'} = x^{2} (2 x) - x^{2} (2 x) = 0$ .

Region $x < 0$ : $f = x^{2}, g = - x^{2}$ , so $g^{'} = - 2 x$ . Then $W = x^{2} (- 2 x) - (- x^{2}) (2 x) = - 2 x^{3} + 2 x^{3} = 0$ .

So $W \equiv 0$ on all of $I$ .

Independence. Suppose $c_{1} x^{2} + c_{2} x ∣ x ∣ = 0$ for all $x \in I$ . "For all $x$ " means the equation must hold at every point we care to plug in — so we are free to pick convenient points, and each choice gives one true equation about $c_{1}, c_{2}$ . We only need enough equations to force $c_{1} = c_{2} = 0$ ; here two well-chosen points suffice because they yield two independent linear equations in two unknowns.

At $x = 1$ (where $x ∣ x ∣ = 1$ ): $c_{1} + c_{2} = 0$ .
At $x = - 1$ (where $x ∣ x ∣ = - 1$ ): $c_{1} - c_{2} = 0$ .

Adding gives $2 c_{1} = 0 \Rightarrow c_{1} = 0$ , then $c_{2} = 0$ . The only constants that work are both zero, so $f, g$ are linearly independent. (We did not need to verify the identity everywhere else; showing the only possible constants are zero already settles independence.)

Why no contradiction: the clean " $W \equiv 0 \Rightarrow$ dependent" converse holds only for solutions of the same linear ODE with continuous coefficients. Here $f, g$ are not both solutions of one such ODE — $g = x ∣ x ∣$ fails to be twice-differentiable in a way consistent with a nice $p, q$ at $x = 0$ . So the Wronskian test remains one-directional in general: $W \neq = 0$ at a point $\Rightarrow$ independent, always; the reverse needs the ODE context (via Abel's theorem).

L3.3

Suppose $y_{p}$ is one solution of $L [y] = g$ and $\tilde{y}_{p}$ is another. Prove $y_{p} - \tilde{y}_{p}$ solves the homogeneous equation, and use this to argue that the general solution has the form $y_{h} + y_{p}$ .

Recall Solution

WHAT we do — apply $L$ to the difference. Because $L$ is linear (superposition): $L [y_{p} - \tilde{y}_{p}] = L [y_{p}] - L [\tilde{y}_{p}] = g - g = 0.$ So any two particular solutions differ by a homogeneous solution.

General form. Let $y$ be any solution of $L [y] = g$ and fix one particular $y_{p}$ . Then $L [y - y_{p}] = g - g = 0$ , so $y - y_{p}$ is some homogeneous solution $y_{h}$ . Rearranged, $y = y_{h} + y_{p} = c_{1} y_{1} + c_{2} y_{2} + y_{p} .$ Because $y$ was arbitrary, every solution has this shape — nothing is missed.

Level 4 — Synthesis

Combine several tools into one full solution.

L4.1

Solve completely: $y^{''} - y = - 2$ , $y (0) = 3$ , $y^{'} (0) = 1$ . (Homogeneous part known: $y_{h} = c_{1} e^{x} + c_{2} e^{- x}$ .)

Recall Solution

Step 1 — particular solution. The forcing $- 2$ is a constant. WHY guess a constant? Feeding a constant $A$ through $L$ : $A^{''} = 0$ , so $L [A] = - A$ ; matching $- A = - 2$ gives $A = 2$ . Try $y_{p} = 2$ : indeed $0 - 2 = - 2$ ✓. (This is the simplest case of Method of undetermined coefficients.)

Step 2 — assemble. By the structure theorem, $y = y_{h} + y_{p} = c_{1} e^{x} + c_{2} e^{- x} + 2.$

Step 3 — apply conditions. $y (0) = c_{1} + c_{2} + 2 = 3 \Rightarrow c_{1} + c_{2} = 1.$ $y^{'} = c_{1} e^{x} - c_{2} e^{- x}$ , so $y^{'} (0) = c_{1} - c_{2} = 1$ . Adding: $2 c_{1} = 2 \Rightarrow c_{1} = 1$ , then $c_{2} = 0$ . $y = e^{x} + 2.$ Check: $y^{''} = e^{x}$ , $y^{''} - y = e^{x} - (e^{x} + 2) = - 2$ ✓; $y (0) = 1 + 2 = 3$ ✓; $y^{'} (0) = 1$ ✓.

L4.2

Solve $y^{''} + 4 y^{'} + 3 y = 9 x$ (general solution only). Use the homogeneous set ${e^{- x}, e^{- 3 x}}$ from Level 2.

Recall Solution

Step 1 — particular solution. Forcing $9 x$ is a degree-1 polynomial, so try $y_{p} = a x + b$ (a general degree-1 polynomial). WHY this form: derivatives of a linear polynomial stay polynomial of degree $\leq 1$ , so the equation can balance term-by-term. $y_{p}^{'} = a, y_{p}^{''} = 0.$ $y_{p}^{''} + 4 y_{p}^{'} + 3 y_{p} = 0 + 4 a + 3 (a x + b) = 3 a x + (4 a + 3 b) .$ Match to $9 x + 0$ : coefficient of $x$ gives $3 a = 9 \Rightarrow a = 3$ ; constant gives $4 a + 3 b = 0 \Rightarrow 12 + 3 b = 0 \Rightarrow b = - 4$ . So $y_{p} = 3 x - 4$ .

Step 2 — assemble. $y = c_{1} e^{- x} + c_{2} e^{- 3 x} + 3 x - 4.$ Check the particular part: $y_{p}^{''} = 0, y_{p}^{'} = 3$ , $0 + 12 + 3 (3 x - 4) = 12 + 9 x - 12 = 9 x$ ✓.

Level 5 — Mastery

Prove, generalise, or handle a degenerate case.

L5.1

Prove the superposition principle fails to allow free scaling for non-homogeneous equations, and state exactly the one linear combination of two particular solutions that IS a solution.

Recall Solution

Let $L [y_{1}] = g$ and $L [y_{2}] = g$ . By linearity, $L [c_{1} y_{1} + c_{2} y_{2}] = c_{1} L [y_{1}] + c_{2} L [y_{2}] = (c_{1} + c_{2}) g .$ This equals $g$ iff $c_{1} + c_{2} = 1$ . So free scaling is not allowed: only combinations whose coefficients sum to $1$ (affine combinations) stay solutions. In particular $c_{1} = 1, c_{2} = - 1$ gives $L [y_{1} - y_{2}] = 0$ — the difference is homogeneous, which is the seed of the structure theorem. Free scaling is exclusively a homogeneous property, where $g = 0$ makes $(c_{1} + c_{2}) \cdot 0 = 0$ for any $c_{1}, c_{2}$ .

L5.2

Consider $y^{''} - y = 0$ on all of $R$ . You are told $y_{1} = cosh x$ and $y_{2} = sinh x$ are solutions. Show they form a fundamental set, and re-express the general solution in the exponential basis to confirm it is the same solution space.

Recall Solution

Solutions? $(cosh x)^{''} = cosh x = y_{1}$ ✓; $(sinh x)^{''} = sinh x = y_{2}$ ✓.

Independence via Wronskian. $y_{1}^{'} = sinh x, y_{2}^{'} = cosh x$ : $W = cosh x cosh x - sinh x sinh x = cosh^{2} x - sinh^{2} x = 1 \neq = 0.$ Nonzero everywhere ⇒ fundamental set ⇒ general solution $y = A cosh x + B sinh x$ .

Same space as ${e^{x}, e^{- x}}$ . Using $cosh x = \frac{e ^{x} + e ^{- x}}{2}, sinh x = \frac{e ^{x} - e ^{- x}}{2}$ , $A cosh x + B sinh x = \frac{A + B}{2} e^{x} + \frac{A - B}{2} e^{- x} = c_{1} e^{x} + c_{2} e^{- x},$ with $c_{1} = \frac{A + B}{2}, c_{2} = \frac{A - B}{2}$ . Any $(A, B)$ maps to some $(c_{1}, c_{2})$ and vice versa — the two "bases" span the identical two-dimensional solution space. This is exactly a change of basis: same plane, different arrows.

L5.3 (degenerate / limiting case)

Solve $y^{''} = 0$ directly (no characteristic-equation shortcut). What is the fundamental set, what is $W$ , and why is this the "boundary case" where the two roots of the characteristic equation collide?

Recall Solution

Direct integration. The equation says the second derivative is zero everywhere. Integrate once: $y^{'} = c_{1}$ (a constant, since its derivative $y^{''}$ is $0$ ). Integrate again: $y = c_{1} x + c_{2}$ . So a fundamental set is ${y_{1} = 1, y_{2} = x}, general solution y = c_{1} x + c_{2} .$ (Note the two arbitrary constants $c_{1}, c_{2}$ — exactly what a second-order equation must produce.)

Wronskian. With $y_{1} = 1, y_{2} = x$ we have $y_{1}^{'} = 0, y_{2}^{'} = 1$ , so $W = y_{1} y_{2}^{'} - y_{2} y_{1}^{'} = (1) (1) - (x) (0) = 1 \neq = 0.$ Nonzero everywhere ⇒ ${1, x}$ is genuinely independent ⇒ a valid fundamental set.

Why the boundary case. Writing $y = e^{r x}$ turns $y^{''} = 0$ into the characteristic equation $r^{2} = 0$ , a repeated root $r = 0$ . When the two roots coincide, the exponential family delivers only one independent solution, $e^{0 x} = 1$ ; the second must carry the extra factor of $x$ from the repeated-root rule, namely $x e^{0 x} = x$ . That is precisely the ${1, x}$ we found by hand — no coincidence. So $y^{''} = 0$ is the cleanest picture of two distinct exponential roots merging into one, and it is exactly why the repeated-root formula in Characteristic equation — constant coefficient ODEs carries that factor of $x$ .