1. The shape of the equation
An equation of the form $a(x) y'' + b(x) y' + c(x) y = g(x)$ in which $y$, $y'$, and $y''$ appear only to the first power and are not multiplied together. "Second-order" because the highest derivative is $y''$; "linear" because $y$ and its derivatives enter linearly.
For the rest of this topic we focus on the constant-coefficient case, where $a$, $b$, $c$ are numbers, not functions of $x$:
$$ a y'' + b y' + c y = g(x), \qquad a \neq 0. $$Two further pieces of vocabulary set the agenda:
- Homogeneous means $g(x) = 0$. The equation describes an unforced, isolated system — a spring with no one pushing it.
- Inhomogeneous means $g(x) \neq 0$. The right-hand side is a forcing term — an outside agent driving the system.
We solve the homogeneous case first because the inhomogeneous case is built on top of it.
2. Linearity and superposition
The reason second-order linear ODEs are so tractable is that solutions add. If $y_1$ and $y_2$ both satisfy the homogeneous equation, so does $c_1 y_1 + c_2 y_2$ for any constants $c_1, c_2$. This is the superposition principle, and it has a precise consequence:
The set of solutions to a second-order linear homogeneous ODE forms a two-dimensional vector space. Pick any two linearly independent solutions $y_1, y_2$ — together they span every possible solution: $y = c_1 y_1 + c_2 y_2$.
So the entire game is to find any two linearly independent solutions. The constants $c_1$ and $c_2$ are then pinned down by two initial conditions, typically $y(x_0)$ and $y'(x_0)$ — one for each dimension of the solution space.
3. The characteristic equation
For the homogeneous equation $ay'' + by' + cy = 0$, try the educated guess $y = e^{rx}$. Differentiating once and twice:
$$ y' = r e^{rx}, \qquad y'' = r^2 e^{rx}. $$Substitute and factor out the (nonzero) $e^{rx}$:
$$ (ar^2 + br + c)\, e^{rx} = 0 \quad\Longrightarrow\quad ar^2 + br + c = 0. $$The quadratic $ar^2 + br + c = 0$ obtained by substituting $y = e^{rx}$ into the homogeneous ODE. Its roots determine every solution.
From here we just read off the roots using the quadratic formula:
$$ r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}. $$The discriminant $\Delta = b^2 - 4ac$ decides which of three regimes we're in. Everything else follows.
4. Three regimes of the homogeneous solution
One quadratic, three personalities. Each regime corresponds to a sign of the discriminant.
| Discriminant | Roots | General solution |
|---|---|---|
| $\Delta > 0$ distinct real |
$r_1, r_2 \in \mathbb{R}$, $r_1 \neq r_2$ | $y = c_1 e^{r_1 x} + c_2 e^{r_2 x}$ |
| $\Delta = 0$ repeated |
$r$ (double root) | $y = (c_1 + c_2 x)\, e^{r x}$ |
| $\Delta < 0$ complex conjugate |
$r = \alpha \pm \beta i$ | $y = e^{\alpha x}\bigl(c_1 \cos\beta x + c_2 \sin\beta x\bigr)$ |
Distinct real roots
Both $e^{r_1 x}$ and $e^{r_2 x}$ solve the equation, and they're linearly independent (different exponential rates), so they span the solution space. Done.
Repeated root
With only one root, we have only one exponential — and the solution space is two-dimensional. The trick: multiply by $x$. You can verify by substitution that $x e^{rx}$ also satisfies $ay'' + by' + cy = 0$ when $r$ is a double root. The pair $\{e^{rx}, x e^{rx}\}$ is linearly independent and spans the space.
Complex roots
The roots come as a conjugate pair $\alpha \pm \beta i$. The complex-exponential solutions $e^{(\alpha + \beta i) x}$ and $e^{(\alpha - \beta i) x}$ are perfectly valid, but Euler's formula $e^{i\theta} = \cos\theta + i\sin\theta$ lets us repackage them as real functions:
$$ e^{(\alpha + \beta i) x} = e^{\alpha x}(\cos\beta x + i\sin\beta x). $$Linear combinations of $e^{(\alpha + \beta i) x}$ and $e^{(\alpha - \beta i) x}$ produce $e^{\alpha x}\cos\beta x$ and $e^{\alpha x}\sin\beta x$ — two real solutions that span the same space. The pattern: the real part $\alpha$ controls amplitude growth or decay; the imaginary part $\beta$ controls the oscillation frequency.
The discriminant is doing the work. Run a slider over $b$ in $y'' + by' + y = 0$: at $b > 2$ the system overshoots toward zero from both directions (two real exponentials); at $b = 2$ the curves meet (repeated root); at $b < 2$ they peel off into spirals (complex roots). The qualitative behavior of the solution flips at $\Delta = 0$.
5. Physical picture: damped oscillation
The cleanest place to see the three regimes is the spring-mass-damper system. A mass $m$ on a spring of stiffness $k$, sliding against friction proportional to velocity with coefficient $c$, obeys Newton's second law:
$$ m\, y'' + c\, y' + k\, y = 0. $$Here $y$ is the displacement from equilibrium. The characteristic equation is $m r^2 + c r + k = 0$ and its discriminant is $\Delta = c^2 - 4mk$. The sign of $\Delta$ has a name in physics:
| Sign of $\Delta$ | Name | Motion |
|---|---|---|
| $c^2 > 4mk$ | Overdamped | Returns to equilibrium without oscillating — slow and sluggish. |
| $c^2 = 4mk$ | Critically damped | Returns to equilibrium as fast as possible without overshooting. |
| $c^2 < 4mk$ | Underdamped | Oscillates with exponentially shrinking amplitude — a ringing-down decay. |
Critical damping is what engineers usually want: a door closer that doesn't slam and doesn't drift; a car suspension that absorbs a bump without bouncing. Too little damping and the system rings; too much and it crawls.
Three solutions of $my'' + cy' + ky = 0$ released from rest at $y(0) = 1$, $y'(0) = 0$. Same starting point, three temperaments — set by the sign of the discriminant.
6. Inhomogeneous equations: $y = y_h + y_p$
Now turn the forcing back on: $ay'' + by' + cy = g(x)$ with $g \neq 0$. The structure theorem extends in a beautiful way.
Every solution of the inhomogeneous equation has the form $y = y_h + y_p$, where $y_h$ is the general solution of the associated homogeneous equation and $y_p$ is any one particular solution of the inhomogeneous equation.
The argument is pure linearity: if $y_p$ satisfies the inhomogeneous equation and $\tilde y$ is another solution, then their difference $\tilde y - y_p$ satisfies the homogeneous equation. So $\tilde y - y_p$ is some $y_h$, which rearranges to $\tilde y = y_h + y_p$.
This splits the problem cleanly:
- Solve the homogeneous equation — you already know how. This gives $y_h = c_1 y_1 + c_2 y_2$.
- Find any particular solution $y_p$ of the inhomogeneous equation.
- Add them: $y = y_h + y_p$. Use initial conditions to nail down $c_1, c_2$.
The two standard ways to find $y_p$ are undetermined coefficients (a guess) and variation of parameters (a recipe).
7. Method 1 — Undetermined coefficients
When the forcing $g(x)$ is built from polynomials, exponentials, sines, and cosines — and products of those — the answer "looks like" $g$ does. The method: guess a form for $y_p$ with unknown coefficients, plug it into the equation, and match coefficients to solve for them.
The guess table
| If $g(x)$ is… | Try $y_p =$ |
|---|---|
| $P_n(x)$ (polynomial of degree $n$) | $A_n x^n + A_{n-1} x^{n-1} + \cdots + A_0$ |
| $e^{\alpha x}$ | $A e^{\alpha x}$ |
| $\cos\beta x$ or $\sin\beta x$ | $A \cos\beta x + B\sin\beta x$ |
| $P_n(x)\, e^{\alpha x}\cos\beta x$ or sine version | $\bigl(\text{degree-}n\text{ poly}\bigr) e^{\alpha x}\cos\beta x + \bigl(\text{degree-}n\text{ poly}\bigr) e^{\alpha x}\sin\beta x$ |
Walk-through
Solve $y'' - y = x$.
Step 1 — homogeneous. Characteristic: $r^2 - 1 = 0$, roots $\pm 1$. So $y_h = c_1 e^x + c_2 e^{-x}$.
Step 2 — guess. $g(x) = x$ is a degree-1 polynomial. Try $y_p = Ax + B$.
Step 3 — substitute. $y_p'' = 0$, so the equation reads $0 - (Ax + B) = x$. Matching coefficients: $-A = 1$ and $-B = 0$. So $A = -1$, $B = 0$, giving $y_p = -x$.
Step 4 — combine. $y = c_1 e^x + c_2 e^{-x} - x$.
If your guessed form already appears in $y_h$, plugging it in gives $0$ on the left — and you can't solve $0 = g$. The fix: multiply the guess by $x$ (or by $x^2$ for a double root collision). For instance, $y'' + y = \cos x$ would normally suggest $y_p = A\cos x + B\sin x$, but those are already homogeneous solutions because $r = \pm i$. Promote to $y_p = x(A\cos x + B\sin x)$, then solve.
This case has a physical name: resonance. When the forcing frequency matches the system's natural frequency, the response grows without bound — the extra factor of $x$ is the mathematics announcing the runaway.
8. Method 2 — Variation of parameters
What if $g(x)$ is $\tan x$, or $\ln x$, or $1/(1 + x^2)$? The guess table doesn't help — no finite combination of polynomials, exponentials, and sinusoids reproduces these. Variation of parameters is the universal fallback. It works for any continuous $g$, at the cost of an integral.
Start from the homogeneous solution $y_h = c_1 y_1 + c_2 y_2$. Instead of constants, allow the coefficients to vary:
$$ y_p = u_1(x)\, y_1(x) + u_2(x)\, y_2(x). $$That's two unknown functions for one equation, so we have one degree of freedom to spend. Spend it on a convenient constraint: $u_1' y_1 + u_2' y_2 = 0$. Substituting into the ODE (after dividing through by $a$ to put it in the form $y'' + py' + qy = g/a$) leaves a clean system:
$$ \begin{aligned} u_1' y_1 + u_2' y_2 &= 0, \\ u_1' y_1' + u_2' y_2' &= g(x)/a. \end{aligned} $$Solve for $u_1'$ and $u_2'$ via Cramer's rule. The denominator is the Wronskian $W = y_1 y_2' - y_2 y_1'$, which is nonzero precisely because $y_1, y_2$ are linearly independent:
$$ u_1' = -\frac{y_2\, g/a}{W}, \qquad u_2' = \frac{y_1\, g/a}{W}. $$Integrate to recover $u_1, u_2$, and assemble $y_p$.
Undetermined coefficients is faster — when it applies. Variation of parameters always applies but generates integrals that may not have closed forms. Use undetermined coefficients first; reach for variation of parameters only when $g(x)$ is irregular.
9. Application: the spring-mass-damper, forced
Putting both halves together, the full forced spring-mass-damper system is:
$$ m\, y'' + c\, y' + k\, y = F(t). $$The homogeneous part $y_h$ describes the system's natural "ringing down" — overdamped, critically damped, or underdamped depending on $c^2$ vs $4mk$. Because all three regimes have negative real parts in their exponents (for $c > 0$), $y_h \to 0$ as $t \to \infty$. This is the transient: it depends on how the system was started, but it dies out.
The particular solution $y_p$ describes the long-term response to the driving force. If $F(t)$ is itself sinusoidal — say $F(t) = F_0 \cos\omega t$ — then $y_p$ is also sinusoidal at frequency $\omega$, but with a different amplitude and phase. This is the steady state.
Two pieces, two roles:
- Transient ($y_h$). Set by initial conditions, dies away over a time scale governed by the damping.
- Steady state ($y_p$). Set by the forcing, persists as long as the forcing does. Resonance peaks when $\omega$ matches the natural frequency $\sqrt{k/m}$.
This decomposition is exactly why a guitar string both rings briefly when plucked (transient) and sings at the driver's frequency when bowed (steady state). Same equation, same split.
10. Common pitfalls
$\{e^{rx}\}$ alone is one-dimensional — you need a partner. The partner is $x e^{rx}$, not $e^{rx}$ a second time. Verify by substituting into the ODE that $x e^{rx}$ really does solve it when $r$ is a double root.
When the roots are $\alpha \pm \beta i$, the answer $C_1 e^{(\alpha + \beta i) x} + C_2 e^{(\alpha - \beta i) x}$ is technically correct but unhelpful. Convert it to the real form $e^{\alpha x}(c_1 \cos\beta x + c_2 \sin\beta x)$ using Euler's formula. Real-world systems have real-valued answers.
If your undetermined-coefficients guess overlaps with $y_h$, you must multiply by $x$ (or $x^2$ for a double overlap). Forgetting this gives the absurdity $0 = g(x)$ and a wasted page of algebra.
The formulas $u_1' = -y_2 g/(aW)$ and $u_2' = y_1 g/(aW)$ assume the ODE has been divided through so the coefficient of $y''$ is $1$ (or equivalently, that you carry the $1/a$ factor explicitly). Mixing up the leading coefficient is the single most common mistake in variation-of-parameters problems.
11. Worked examples
Example 1 · Distinct real roots: $y'' - 3y' + 2y = 0$
Characteristic equation: $r^2 - 3r + 2 = 0$, factoring as $(r - 1)(r - 2) = 0$. Roots $r_1 = 1$, $r_2 = 2$.
General solution:
$$ y = c_1 e^{x} + c_2 e^{2x}. $$This is overdamped territory: two decaying exponentials (well, here they grow, but the point is that they're real, with no oscillation).
Example 2 · Repeated root: $y'' - 2y' + y = 0$
Characteristic equation: $r^2 - 2r + 1 = (r - 1)^2 = 0$. Double root $r = 1$.
General solution:
$$ y = (c_1 + c_2 x)\, e^{x}. $$The factor of $x$ is the second linearly independent solution that the repeated root would otherwise have denied you.
Example 3 · Complex roots: $y'' + 2y' + 5y = 0$
Characteristic equation: $r^2 + 2r + 5 = 0$. Discriminant $\Delta = 4 - 20 = -16$. Roots:
$$ r = \frac{-2 \pm \sqrt{-16}}{2} = -1 \pm 2i. $$So $\alpha = -1$ and $\beta = 2$. General solution:
$$ y = e^{-x}\bigl(c_1 \cos 2x + c_2 \sin 2x\bigr). $$An underdamped oscillator: it rings at frequency $2$ while its amplitude decays like $e^{-x}$.
Example 4 · Undetermined coefficients: $y'' - y = x$
Homogeneous. $r^2 - 1 = 0$, roots $\pm 1$, so $y_h = c_1 e^x + c_2 e^{-x}$.
Guess. $g(x) = x$ is a degree-1 polynomial — try $y_p = Ax + B$. Then $y_p'' = 0$.
Substituting: $0 - (Ax + B) = x$, so $-A = 1$ and $-B = 0$. Hence $y_p = -x$.
General solution.
$$ y = c_1 e^{x} + c_2 e^{-x} - x. $$Example 5 · Resonance: $y'' + y = \cos x$
Homogeneous. $r^2 + 1 = 0$, roots $\pm i$, so $y_h = c_1 \cos x + c_2 \sin x$. The natural frequency is $1$.
Naïve guess. $y_p = A\cos x + B\sin x$ — but this is exactly $y_h$. Bumping by $x$:
$$ y_p = x(A\cos x + B\sin x). $$Compute $y_p''$ (chain rule and product rule), substitute, and matching coefficients of $\cos x$ and $\sin x$ on both sides gives $A = 0$, $B = \tfrac{1}{2}$. So:
$$ y_p = \tfrac{1}{2}\, x \sin x. $$That factor of $x$ in the answer is the signature of resonance: amplitude grows linearly forever when an undamped oscillator is driven at its own frequency.
Example 6 · Variation of parameters: $y'' - y = x$ revisited
Same equation as Example 4, this time with the universal method. From the homogeneous solution: $y_1 = e^x$, $y_2 = e^{-x}$.
Wronskian. $W = y_1 y_2' - y_2 y_1' = e^x(-e^{-x}) - e^{-x}(e^x) = -1 - 1 = -2$.
Compute $u_1', u_2'$. Here $a = 1$ and $g = x$:
$$ u_1' = -\frac{y_2\, g}{W} = -\frac{e^{-x}\, x}{-2} = \tfrac{1}{2} x e^{-x}, $$ $$ u_2' = \frac{y_1\, g}{W} = \frac{e^{x}\, x}{-2} = -\tfrac{1}{2} x e^{x}. $$Integrate (by parts, once each):
$$ u_1 = -\tfrac{1}{2}(x + 1) e^{-x}, \qquad u_2 = -\tfrac{1}{2}(x - 1) e^{x}. $$Assemble.
$$ y_p = u_1 y_1 + u_2 y_2 = -\tfrac{1}{2}(x+1) - \tfrac{1}{2}(x-1) = -x. $$Same $y_p$ as Example 4 — a useful sanity check that the two methods agree when both apply.