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

This page is the "no case left behind" companion to Second-order Linear ODEs — Superposition Principle & General Theory. We built the machinery there; here we run it against every kind of input a second-order linear ODE can hand you.

The scenario matrix

Read this as a checklist. Every cell below is covered by at least one worked example.

Cell	What makes it special	Solution shape	Example
A — real distinct roots ( $r_{1} \neq = r_{2}$ )	discriminant $> 0$	$c_{1} e^{r_{1} x} + c_{2} e^{r_{2} x}$	Ex 1
B — repeated real root ( $r_{1} = r_{2}$ )	discriminant $= 0$ , degenerate	$(c_{1} + c_{2} x) e^{r x}$	Ex 2
C — complex roots ( $α \pm i β$ )	discriminant $< 0$ , oscillation	$e^{α x} (c_{1} cos β x + c_{2} sin β x)$	Ex 3
D — non-homogeneous, ordinary forcing	RHS $\neq = 0$ , no overlap	$y_{h} + y_{p}$	Ex 4
E — resonance (forcing = homogeneous soln)	naive guess fails, multiply by $x$	$y_{h} + x \cdot (\dots)$	Ex 5
F — variable coefficients (Euler)	$p, q$ depend on $x$	power-law solutions	Ex 6
G — sign of discriminant at the boundary	limiting behaviour as case B→A/C	—	Ex 7
H — real-world word problem	spring / RLC circuit	case C in disguise	Ex 8

Cells to watch for degeneracy/limits: B (roots collide), E (particular guess collides with homogeneous), G (what happens between cases).

Ex 1 — Cell A: real distinct roots

Worked example Solve

$y^{''} + 5 y^{'} + 6 y = 0$ , with $y (0) = 1, y^{'} (0) = 0$ . Forecast: guess the two roots in your head before reading on — factor $r^{2} + 5 r + 6$ .

Step 1. Substitute the trial $y = e^{r x}$ . Then $y^{'} = r e^{r x}$ , $y^{''} = r^{2} e^{r x}$ , and the equation becomes $(r^{2} + 5 r + 6) e^{r x} = 0$ . Why this step? $e^{r x}$ is the one function whose derivatives are just copies of itself, so calculus collapses into the algebra of Characteristic equation — constant coefficient ODEs. Since $e^{r x} \neq = 0$ , we may cancel it.

Step 2. Solve $r^{2} + 5 r + 6 = (r + 2) (r + 3) = 0 \Rightarrow r = - 2, r = - 3$ . Why this step? Two distinct real roots means two genuinely different exponential solutions.

Step 3. Fundamental set $y_{1} = e^{- 2 x}, y_{2} = e^{- 3 x}$ ; check independence with the Wronskian: $W = e^{- 2 x} - 2 e^{- 2 x} e^{- 3 x} - 3 e^{- 3 x} = - 3 e^{- 5 x} - (- 2 e^{- 5 x}) = - e^{- 5 x} \neq = 0.$ Why this step? $W \neq = 0$ (anywhere) certifies a fundamental set, so $y = c_{1} e^{- 2 x} + c_{2} e^{- 3 x}$ is every solution.

Step 4. Apply ICs. $y (0) = c_{1} + c_{2} = 1$ . $y^{'} = - 2 c_{1} e^{- 2 x} - 3 c_{2} e^{- 3 x}$ , so $y^{'} (0) = - 2 c_{1} - 3 c_{2} = 0$ . Solving: $c_{2} = - 2$ , $c_{1} = 3$ . Why this step? Two initial conditions pin the two constants uniquely (Existence & Uniqueness).

Answer: $y = 3 e^{- 2 x} - 2 e^{- 3 x} .$

Verify: $y (0) = 3 - 2 = 1$ ✔; $y^{'} (0) = - 6 + 6 = 0$ ✔. And $y^{''} + 5 y^{'} + 6 y$ : each term $e^{- 2 x}$ contributes $(4 - 10 + 6) = 0$ , each $e^{- 3 x}$ contributes $(9 - 15 + 6) = 0$ . ✔

Ex 2 — Cell B: repeated (degenerate) root

Worked example Solve

$y^{''} - 4 y^{'} + 4 y = 0$ , with $y (0) = 1, y^{'} (0) = 5$ . Forecast: the discriminant is exactly zero here. What second solution do you use when the two roots merge into one?

Step 1. Characteristic equation: $r^{2} - 4 r + 4 = (r - 2)^{2} = 0 \Rightarrow r = 2$ (twice). Why this step? Discriminant $b^{2} - 4 a c = 16 - 16 = 0$ : a double root. This is the degenerate case where the naive method gives only one function $e^{2 x}$ .

Step 2. We need a second independent solution. Claim: $y_{2} = x e^{2 x}$ works. Check by substitution: $y_{2}^{'} = (1 + 2 x) e^{2 x}$ , $y_{2}^{''} = (4 + 4 x) e^{2 x}$ , so $y_{2}^{''} - 4 y_{2}^{'} + 4 y_{2} = e^{2 x} [(4 + 4 x) - 4 (1 + 2 x) + 4 x] = e^{2 x} \cdot 0 = 0.✔$ Why this step? When roots collide, Variation of parameters (or a limiting argument) tells us to multiply the repeated solution by $x$ — it is independent of $e^{2 x}$ because $x$ is not a constant.

Step 3. General solution $y = (c_{1} + c_{2} x) e^{2 x}$ . Apply ICs: $y (0) = c_{1} = 1$ . $y^{'} = (c_{2} + 2 c_{1} + 2 c_{2} x) e^{2 x}$ , so $y^{'} (0) = c_{2} + 2 c_{1} = 5 \Rightarrow c_{2} = 3$ . Why this step? Same "two ICs → two constants" logic; the shape just changed.

Answer: $y = (1 + 3 x) e^{2 x} .$

Verify: $y (0) = 1$ ✔; $y^{'} (0) = 3 + 2 = 5$ ✔.

Ex 3 — Cell C: complex roots (oscillation)

Worked example Solve

$y^{''} + 2 y^{'} + 5 y = 0$ , with $y (0) = 0, y^{'} (0) = 4$ . Forecast: discriminant negative → the roots are complex. Do you expect the answer to decay, grow, or oscillate? Both, actually.

Step 1. Characteristic: $r^{2} + 2 r + 5 = 0$ . By the quadratic formula $r = \frac{- 2 \pm 4 - 20}{2} = \frac{- 2 \pm - 16}{2} = - 1 \pm 2 i$ . Why this step? $- 16 = 4 i$ where $i^{2} = - 1$ (the imaginary unit). Negative discriminant forces a complex conjugate pair $α \pm i β$ with $α = - 1, β = 2$ .

Step 2. Convert to real form. $e^{(- 1 + 2 i) x} = e^{- x} (cos 2 x + i sin 2 x)$ by Euler's formula. Taking real and imaginary parts gives two real solutions $e^{- x} cos 2 x$ and $e^{- x} sin 2 x$ . So $y = e^{- x} (c_{1} cos 2 x + c_{2} sin 2 x) .$ Why this step? We want real functions for a real physical quantity. $α = - 1$ is the decay rate; $β = 2$ is the oscillation frequency. See the picture below: the exponential envelope squeezes a sine.

Step 3. ICs. $y (0) = c_{1} = 0$ . Differentiate: $y^{'} = e^{- x} [(- c_{1} + 2 c_{2}) cos 2 x + (- c_{2} - 2 c_{1}) sin 2 x]$ . At $x = 0$ : $y^{'} (0) = - c_{1} + 2 c_{2} = 4$ . With $c_{1} = 0$ , $c_{2} = 2$ . Why this step? Standard two-IC pinning; the derivative mixes the sine and cosine, which is why both must be present.

Answer: $y = 2 e^{- x} sin 2 x .$

Verify: $y (0) = 0$ ✔; $y^{'} (0) = 2 \cdot 1 \cdot 1 = 2 cos 0 = 2$ … wait — $y^{'} = 2 e^{- x} (- sin 2 x + 2 cos 2 x)$ , so $y^{'} (0) = 2 (0 + 2) = 4$ ✔.

Ex 4 — Cell D: non-homogeneous, ordinary forcing

Worked example Solve

$y^{''} + y = 3 x$ (general solution). Forecast: the forcing $3 x$ is a straight line. Guess the shape of one particular solution before you compute.

Step 1. Homogeneous part: $r^{2} + 1 = 0 \Rightarrow r = \pm i$ , so $y_{h} = c_{1} cos x + c_{2} sin x$ . Why this step? By the structure theorem, every solution is $y_{h} + y_{p}$ . Build $y_{h}$ first.

Step 2. Particular solution via Method of undetermined coefficients: the RHS is degree-1 polynomial, so try $y_{p} = A x + B$ . Then $y_{p}^{''} = 0$ , and $0 + (A x + B) = 3 x$ forces $A = 3, B = 0$ . Why this step? $L$ maps polynomials to polynomials of the same degree here, so a linear guess is closed under the operation — no need for anything fancier. Crucially, $x$ is not a homogeneous solution (those are sines/cosines), so no resonance.

Step 3. Assemble: $y = c_{1} cos x + c_{2} sin x + 3 x .$ Why this step? Structure theorem: general = homogeneous + one particular.

Verify: $y^{''} = - c_{1} cos x - c_{2} sin x$ , so $y^{''} + y = (- c_{1} cos x - c_{2} sin x) + (c_{1} cos x + c_{2} sin x + 3 x) = 3 x$ ✔.

Ex 5 — Cell E: resonance (the guess collides)

Worked example Solve

$y^{''} + y = cos x$ (general solution). Forecast: try the "obvious" guess $y_{p} = A cos x + B sin x$ — and watch it explode into nonsense. Why?

Step 1. Homogeneous: $y_{h} = c_{1} cos x + c_{2} sin x$ (same as Ex 4).

Step 2. The forcing $cos x$ is already a homogeneous solution. Plugging $A cos x + B sin x$ into $L = \frac{d ^{2}}{d x ^{2}} + 1$ gives exactly $0$ — you can never produce $cos x$ on the right. This is resonance. Why this step? When the input frequency matches a natural frequency, the naive family lies in the kernel of $L$ . The fix (from Method of undetermined coefficients) is to multiply the trial by $x$ .

Step 3. Try $y_{p} = x (A cos x + B sin x)$ . Differentiate: $y_{p}^{'} = A cos x + B sin x + x (- A sin x + B cos x),$ $y_{p}^{''} = - 2 A sin x + 2 B cos x + x (- A cos x - B sin x) .$ Then $y_{p}^{''} + y_{p} = - 2 A sin x + 2 B cos x$ . Set equal to $cos x$ : $- 2 A = 0, 2 B = 1 \Rightarrow A = 0, B = \frac{1}{2}$ . Why this step? The $x$ -multiplied terms cancel against $y_{p}$ , leaving only the derivative-of-envelope terms — exactly what survives to match the forcing.

Answer: $y = c_{1} cos x + c_{2} sin x + \frac{x}{2} sin x .$

Verify: with $y_{p} = \frac{1}{2} x sin x$ : $y_{p}^{'} = \frac{1}{2} sin x + \frac{1}{2} x cos x$ , $y_{p}^{''} = cos x - \frac{1}{2} x sin x$ , so $y_{p}^{''} + y_{p} = cos x$ ✔. The amplitude grows linearly with $x$ — the physical signature of resonance.

Ex 6 — Cell F: variable coefficients (Euler equation)

Worked example Solve

$x^{2} y^{''} - 2 x y^{'} + 2 y = 0$ for $x > 0$ . Forecast: here $p, q$ depend on $x$ , so exponentials are the wrong guess. What power of $x$ could work?

Step 1. The coefficients scale so that each term becomes degree-matched if $y = x^{m}$ : then $x^{2} y^{''} = m (m - 1) x^{m}$ , $- 2 x y^{'} = - 2 m x^{m}$ , $2 y = 2 x^{m}$ . Why this step? $x^{r}$ is to Euler equations what $e^{r x}$ is to constant-coefficient ones — each derivative lowers the power by one and multiplies by a factor, so $x^{2} \cdot (stuff)$ restores the same power $x^{m}$ . Everything collapses to algebra again.

Step 2. Factor out $x^{m}$ : $m (m - 1) - 2 m + 2 = m^{2} - 3 m + 2 = (m - 1) (m - 2) = 0 \Rightarrow m = 1, 2$ . Why this step? Two distinct real exponents → two independent power-law solutions $x^{1}$ and $x^{2}$ .

Step 3. General solution $y = c_{1} x + c_{2} x^{2}$ . Independence check via Wronskian: $W = x 1 x^{2} 2 x = 2 x^{2} - x^{2} = x^{2} \neq = 0 (for x > 0) .✔$ Why this step? Nonzero Wronskian on the interval $x > 0$ confirms a fundamental set (compare Abel's theorem, which explains why $W$ can't sneakily vanish).

Answer: $y = c_{1} x + c_{2} x^{2} .$

Verify: for $x^{2}$ : $x^{2} (2) - 2 x (2 x) + 2 x^{2} = 2 x^{2} - 4 x^{2} + 2 x^{2} = 0$ ✔. For $x$ : $x^{2} (0) - 2 x (1) + 2 x = 0$ ✔.

Ex 7 — Cell G: the boundary between cases (limiting behaviour)

Worked example What happens to

$y^{''} + 2 ε y^{'} + y = 0$ as the damping $ε$ crosses $1$ ? Forecast: at exactly $ε = 1$ you sit on the fence between oscillation (Cell C) and pure decay (Cell A). Predict which side is which.

Step 1. Characteristic roots: $r = - ε \pm ε^{2} - 1$ . Why this step? The whole story is the sign of the discriminant $ε^{2} - 1$ .

Step 2. Three regimes:

$ε < 1$ (Cell C): $ε^{2} - 1 < 0$ , roots $- ε \pm i 1 - ε^{2}$ → decaying oscillation.
$ε = 1$ (Cell B): discriminant exactly $0$ , double root $r = - 1$ → critically damped $(c_{1} + c_{2} x) e^{- x}$ .
$ε > 1$ (Cell A): $ε^{2} - 1 > 0$ , two real negative roots → overdamped, no oscillation.

Why this step? This is exactly the limit where Cell C degenerates into Cell B and then splits into Cell A. As $ε \to 1^{-}$ , the frequency $1 - ε^{2} \to 0$ : the oscillation slows to a stop, and the two complex roots slide together onto the real axis.

Verify (at $ε = 1$ ): double root $r = - 1$ from $r^{2} + 2 r + 1 = (r + 1)^{2}$ ; solution $(c_{1} + c_{2} x) e^{- x}$ satisfies the ODE (same check pattern as Ex 2). As $ε = 0.5$ : roots $- 0.5 \pm i 0.866$ , frequency $\approx 0.866 > 0$ (oscillates). As $ε = 2$ : roots $- 2 \pm 3$ , both real negative (decays, no wiggle). ✔

Ex 8 — Cell H: real-world spring/RLC word problem

Worked example A mass on a spring obeys

$m \overset{u}{¨} + c \overset{u}{˙} + k u = 0$ with $m = 1 kg$ , $c = 2 kg/s$ , $k = 5 N/m$ . The mass starts at $u (0) = 0.1 m$ , at rest $\overset{u}{˙} (0) = 0$ . Find $u (t)$ . Forecast: damping $c = 2$ , stiffness $k = 5$ — is this over-, critically-, or under-damped? Guess before computing.

Step 1. Divide by $m = 1$ : $\overset{u}{¨} + 2 \overset{u}{˙} + 5 u = 0$ . Characteristic $r^{2} + 2 r + 5 = 0 \Rightarrow r = - 1 \pm 2 i$ (this is exactly Ex 3's equation!). Why this step? Newton's law for a damped spring is a constant-coefficient second-order linear ODE — same machine, physical meaning attached. Discriminant $4 - 20 < 0$ ⇒ underdamped (it will oscillate as it decays).

Step 2. General solution $u (t) = e^{- t} (c_{1} cos 2 t + c_{2} sin 2 t)$ . Decay rate $1 s^{- 1}$ , angular frequency $2 rad/s$ . Why this step? Real form is what a displacement (a real, measurable metres value) demands.

Step 3. ICs. $u (0) = c_{1} = 0.1$ . $\overset{u}{˙} = e^{- t} [(- c_{1} + 2 c_{2}) cos 2 t + (- c_{2} - 2 c_{1}) sin 2 t]$ , so $\overset{u}{˙} (0) = - c_{1} + 2 c_{2} = 0 \Rightarrow c_{2} = c_{1} /2 = 0.05$ . Why this step? Starting from rest ( $\overset{u}{˙} (0) = 0$ ) and a known displacement gives the two data needed.

Answer: $u (t) = e^{- t} (0.1 cos 2 t + 0.05 sin 2 t) m .$

Verify: units — every term is in metres (coefficients metres, trig/exp dimensionless), $t$ in seconds inside dimensionless $e^{- t}$ and $cos 2 t$ ✔. $u (0) = 0.1$ m ✔. $\overset{u}{˙} (0) = - 0.1 + 2 (0.05) = 0$ ✔. Physically: amplitude shrinks by factor $e^{- 1} \approx 0.37$ each second while wobbling — a plausible damped spring.

Recall Which root case gives which shape?

Two real distinct roots → sum of exponentials ::: $c_{1} e^{r_{1} x} + c_{2} e^{r_{2} x}$ (Cell A) Repeated real root $r$ → ::: $(c_{1} + c_{2} x) e^{r x}$ (Cell B, multiply by $x$ ) Complex $α \pm i β$ → ::: $e^{α x} (c_{1} cos β x + c_{2} sin β x)$ (Cell C) Forcing equals a homogeneous solution → ::: resonance, multiply trial by $x$ (Cell E)

Back to the parent: Second-order Linear ODEs — Superposition Principle & General Theory. For the existence guarantee behind every "this captures all solutions", see Existence and uniqueness theorems for ODEs; for the vector-space framing of the two-constant family, Linear algebra — vector spaces and bases.