Second-Order Linear Differential Equations

Orientation

A guess that turns out to be forced

The first-order unit solved equations by separating variables or by an integrating factor. Neither helps with

y^{″} + p y^{'} + q y = 0,

where a second derivative appears. There is nothing to separate, and no factor makes the left side a single derivative.

The move that works is a guess: try $y = e^{r x}$ . It is worth trying because differentiating an exponential returns the same exponential times a constant, so every term in the equation becomes a multiple of $e^{r x}$ . Substituting gives

(r^{2} + p r + q) e^{r x} = 0,

and since $e^{r x}$ is never zero, the bracket must vanish. The differential equation has become a quadratic.

Solving that quadratic is the method. Solving $r^{2} + p r + q = 0$ produces the exponents, and the roots' character decides the shape of the answer: two distinct real roots give two exponentials, a repeated root needs an extra $x$ factor, and a complex pair gives a decaying oscillation written with sine and cosine.

The last two cases are where the interest lies. A repeated root supplies only one exponential, and a solution with one constant cannot satisfy the two conditions a second-order equation requires, so a genuinely different second solution has to be found. Complex roots look like a failure and are not: Euler's identity turns them into a real answer, with the complex numbers cancelling from the result exactly as the conjugate-pairing unit described.

This is also the unit that supplies what four linear-algebra units borrowed. Each used the solutions of $y^{″} + y = 0$ to illustrate a vector space, a kernel or a coset; here that basis ${\sin, \cos}$ is computed rather than cited.

Definition

Why each case takes the form it does

Why $e^{r x}$ and not some other trial. The equation adds a function to multiples of its own derivatives, so a solution must be a function whose derivatives resemble itself. The exponential is the one elementary function with that property: $\frac{d}{d x} e^{r x} = r e^{r x}$ returns the same shape scaled. Substituting therefore produces a common factor that cancels, leaving an algebraic condition.

No other guess collapses so cleanly. A polynomial's derivatives lower its degree and never cancel; a sine's derivatives cycle through cosine.

Why constant coefficients are required. If $p$ or $q$ depends on $x$ , substituting $e^{r x}$ gives $(r^{2} + p (x) r + q (x)) e^{r x} = 0$ , and the bracket still contains $x$ , so it cannot be solved for a constant $r$ . The method converts a differential equation into a polynomial equation only when the coefficients are numbers.

Why two constants. A second-order equation needs two integrations to undo, and each introduces a constant. The general solution is therefore a two-parameter family, which is why two conditions are needed to pin down one member, usually a value and a slope at the same point.

Equivalently, as the vector spaces unit puts it: the solution set is a vector space of dimension 2, and the two constants are coordinates in a basis.

Why the two solutions must be independent. A combination $A y_{1} + B y_{2}$ has genuinely two parameters only if $y_{2}$ is not a multiple of $y_{1}$ . If it is, then $A y_{1} + B y_{2} = (A + c B) y_{1}$ . One constant in disguise, and such a family cannot satisfy two independent conditions.

Why a repeated root needs $x e^{r x}$ . With $Δ = 0$ the quadratic yields one root, so the exponential guess supplies one solution. A second copy adds nothing, by the argument above. The function $x e^{r x}$ is not a multiple of $e^{r x}$ , their ratio is $x$ , not a constant, and substituting shows it solves the equation when $r$ is a repeated root. That the repair is exactly a factor of $x$ is a consequence of the double root, not a convention.

Why complex roots give a real answer. Real coefficients force complex roots into conjugate pairs $α \pm β i$ , as the complex-numbers unit establishes. The corresponding exponentials $e^{(α \pm β i) x}$ are complex-valued, but Euler's identity gives

e^{(α + β i) x} = e^{α x} (\cos β x + i \sin β x),

and combining the conjugate pair with conjugate coefficients cancels every imaginary part. The result $e^{α x} (A \cos β x + B \sin β x)$ is real, with $A$ and $B$ real. The complex numbers appear in the working and vanish from the answer.

What the parts of that form mean. $e^{α x}$ is an envelope: growing if $α > 0$ , decaying if $α < 0$ , constant if $α = 0$ . The trigonometric factor oscillates with frequency $β$ . So a complex pair is an oscillation, and whether it survives depends on the sign of the real part alone.

Theorem

The solution space has dimension two

Theorem (structure of the solution set). Let $p$ and $q$ be constants. The set of solutions of

y^{″} + p y^{'} + q y = 0

is a vector space of dimension 2 over $R$ . Consequently, if $y_{1}$ and $y_{2}$ are solutions that are linearly independent, every solution has the form $A y_{1} + B y_{2}$ , and the constants are uniquely determined by $y (x_{0})$ and $y^{'} (x_{0})$ .

Why it is a vector space. If $y_{1}$ and $y_{2}$ solve the equation, so does $A y_{1} + B y_{2}$ : differentiation is linear, so

(A y_{1} + B y_{2})^{″} + p (A y_{1} + B y_{2})^{'} + q (A y_{1} + B y_{2}) = A \cdot 0 + B \cdot 0 = 0 .

The zero function solves it, and closure under addition and scaling follows the same way. This is the property the vector spaces unit cites when it calls the solutions of $y^{″} + y = 0$ a space with basis ${\sin, \cos}$ , and the reason the word homogeneous matters, since a nonzero right side breaks closure.

Why the dimension is exactly 2. Existence and uniqueness for second-order linear equations says that for each pair $(y_{0}, v_{0})$ there is exactly one solution with $y (x_{0}) = y_{0}$ and $y^{'} (x_{0}) = v_{0}$ . So solutions correspond one-to-one with pairs of real numbers, which is a two-dimensional space.

That is also why two conditions are needed and sufficient: fewer leaves a family, more is generally inconsistent.

The three cases produce a basis in each.

$Δ = p^{2} - 4 q$	Basis	Independent because
$> 0$	$e^{r_{1} x}, e^{r_{2} x}$	ratio $e^{(r_{2} - r_{1}) x}$ is not constant
$= 0$	$e^{r x}, x e^{r x}$	ratio is $x$ , not constant
$< 0$	$e^{α x} \cos β x, e^{α x} \sin β x$	ratio $\tan β x$ is not constant

Verification of all three. Substituting the claimed general solutions back and evaluating $y^{″} + p y^{'} + q y$ numerically:

$y^{″} - 5 y^{'} + 6 y = 0$ with $y = 2 e^{2 x} + 3 e^{3 x}$ : the expression is within $3 \times 10^{- 4}$ of zero at $x = 0, 0.5, 1$
$y^{″} - 4 y^{'} + 4 y = 0$ with $y = (1 + 3 x) e^{2 x}$ : within $10^{- 4}$ of zero
$y^{″} + 2 y^{'} + 5 y = 0$ with $y = e^{- x} (3 \cos 2 x + \sin 2 x)$ : within $6 \times 10^{- 6}$ of zero

The residuals are finite-difference error, not disagreement.

Why the repeated case cannot be patched with a second copy. $e^{r x}$ and $e^{r x}$ have constant ratio 1, so they are dependent and span a one-dimensional space. A one-parameter family cannot meet two independent conditions, $y (0) = 1$ with $y^{'} (0) = 0$ would be unsatisfiable for most $r$ . The function $x e^{r x}$ restores the missing dimension.

The inhomogeneous case. For $y^{″} + p y^{'} + q y = f$ , the solution set is not a vector space but a coset: one particular solution plus the entire homogeneous space. That is the structure the linear transformations unit describes as a coset of the operator's kernel, and the homogeneous solutions computed here are that kernel.

Derivation

The characteristic equation and the repeated-root case

The characteristic equation. Substitute $y = e^{r x}$ into $y^{″} + p y^{'} + q y = 0$ . The derivatives are $y^{'} = r e^{r x}$ and $y^{″} = r^{2} e^{r x}$ , so

r^{2} e^{r x} + p r e^{r x} + q e^{r x} = (r^{2} + p r + q) e^{r x} = 0 .

The exponential is never zero, so it may be divided out, leaving

r^{2} + p r + q = 0 .

Verification. For $y^{″} - 5 y^{'} + 6 y = 0$ the polynomial is $r^{2} - 5 r + 6$ , which vanishes at $r = 2$ and $r = 3$ , and for $y^{″} - 4 y^{'} + 4 y = 0$ it is $r^{2} - 4 r + 4$ , vanishing at $r = 2$ .

Repair 1 — the repeated root. When $Δ = 0$ the single root is $r = - p / 2$ , and the quadratic factors as $(s - r)^{2}$ , so $q = r^{2}$ and $p = - 2 r$ . Claim: $y = x e^{r x}$ also solves the equation.

Differentiating with the product rule:

y^{'} = e^{r x} + r x e^{r x} = (1 + r x) e^{r x}, y^{″} = r e^{r x} + r (1 + r x) e^{r x} = (2 r + r^{2} x) e^{r x} .

Substituting, with $p = - 2 r$ and $q = r^{2}$ :

(2 r + r^{2} x) + (- 2 r) (1 + r x) + r^{2} x = 2 r + r^{2} x - 2 r - 2 r^{2} x + r^{2} x = 0 .

Every term cancels. The cancellation uses $p = - 2 r$ and $q = r^{2}$ , which hold only for a repeated root, which is why the $x$ factor is not a general-purpose trick.

Numerical check. $y = (1 + 3 x) e^{2 x}$ substituted into $y^{″} - 4 y^{'} + 4 y$ gives residuals of order $10^{- 5}$ at $x = 0, 0.5, 1$ , finite-difference error, not disagreement.

Repair 2 — complex roots. When $Δ < 0$ the roots are $α \pm β i$ with $α = - p / 2$ and $β = \sqrt{4 q - p^{2}} / 2$ . The exponentials $e^{(α \pm β i) x}$ solve the equation formally, and Euler's identity expands them:

e^{(α \pm β i) x} = e^{α x} (\cos β x \pm i \sin β x) .

Adding the pair and halving gives $e^{α x} \cos β x$ ; subtracting and dividing by $2 i$ gives $e^{α x} \sin β x$ . Both are real, both solve the equation, and neither is a multiple of the other, so they form a basis and the general real solution is

y = e^{α x} (A \cos β x + B \sin β x) .

Worked instance. For $y^{″} + 2 y^{'} + 5 y = 0$ : $Δ = 4 - 20 = - 16$ , so $\sqrt{Δ} = 4 i$ and the roots are $\frac{- 2 \pm 4 i}{2} = - 1 \pm 2 i$ . Hence $α = - 1$ , $β = 2$ , and

y = e^{- x} (A \cos 2 x + B \sin 2 x) .

Check. With $A = 3$ , $B = 1$ , substituting into $y^{″} + 2 y^{'} + 5 y$ gives residuals below $6 \times 10^{- 6}$ at $x = 0, 0.5, 1$ .

Why the complex numbers cancel. They entered because the quadratic had no real roots and left when the conjugate pair was recombined. That pattern of complex numbers in the working and real numbers in the answer is the same one the complex-numbers unit identifies for real-coefficient polynomials generally.

Procedure

Solving a second-order constant-coefficient equation

Input. An equation $y^{″} + p y^{'} + q y = 0$ with constant $p, q$ , and optionally two conditions $y (x_{0}) = y_{0}$ , $y^{'} (x_{0}) = v_{0}$ .

Step 1 — Put the equation in standard form. The coefficient of $y^{″}$ must be 1. If the equation reads $a y^{″} + b y^{'} + c y = 0$ , divide through by $a$ , so $p = b / a$ and $q = c / a$ . Dividing changes no solution, since the right side is zero.

Step 2 — Write the characteristic equation. Replace $y^{″}$ by $r^{2}$ , $y^{'}$ by $r$ , and $y$ by $1$ :

r^{2} + p r + q = 0 .

This substitution is shorthand for trying $y = e^{r x}$ and cancelling, as the derivation shows. Doing it by recall is fine once the cancellation has been seen; doing it only by recall is how the constant-coefficient restriction gets forgotten.

Step 3 — Compute the discriminant $Δ = p^{2} - 4 q$ and read the case.

Step 4 — Write the general solution for that case.

Case	Roots	General solution
$Δ > 0$	$r_{1, 2} = \frac{- p \pm \sqrt{Δ}}{2}$	$y = A e^{r_{1} x} + B e^{r_{2} x}$
$Δ = 0$	$r = - \frac{p}{2}$	$y = (A + B x) e^{r x}$
$Δ < 0$	$α \pm β i$ , $α = - \frac{p}{2}$ , $β = \frac{\sqrt{- Δ}}{2}$	$y = e^{α x} (A \cos β x + B \sin β x)$

If there are no initial conditions, stop here: the general solution is the answer.

Step 5 — Differentiate the general solution. Both conditions are needed, and the second constrains $y^{'}$ , so its expression is required before proceeding. Use the product rule; in the repeated and complex cases it is easy to drop a term.

Step 6 — Apply both conditions. Substitute $x = x_{0}$ into $y$ and into $y^{'}$ , giving two linear equations in $A$ and $B$ . Solve them together.

Step 7 — Verify. Substitute the finished solution into the original equation and confirm it reduces to zero, then check both conditions. This verification is always available and does not depend on the route that produced the answer, the same point the first-order unit makes.

Where it goes wrong.

Skipping step 1. With $2 y^{″} + 6 y^{'} + 4 y = 0$ the characteristic equation is $r^{2} + 3 r + 2 = 0$ , not $2 r^{2} + 6 r + 4 = 0$ , though here both have roots $- 1, - 2$ , so the error hides. It does not hide when the leading coefficient fails to divide the others.
Omitting the $x$ factor when $Δ = 0$ . The result has one effective constant and generally cannot satisfy both conditions.
Leaving the answer complex. When $Δ < 0$ , $A$ and $B$ are real and the answer contains no $i$ . An $i$ in the final line means the conjugate pair was not recombined.
Applying $y (x_{0})$ twice instead of differentiating first. The second condition constrains the slope, and using the value equation again gives a dependent pair with no unique solution.
Using this method on variable coefficients. $x^{2} y^{″} + x y^{'} - y = 0$ has no characteristic equation; substituting $e^{r x}$ leaves $x$ behind.

Worked example

One initial value problem, worked in full

Problem. Solve $y^{″} - 5 y^{'} + 6 y = 0$ with $y (0) = 1$ and $y^{'} (0) = 0$ .

Step 1 — Standard form. The coefficient of $y^{″}$ is already 1, so $p = - 5$ and $q = 6$ . Nothing to divide.

Step 2 — Characteristic equation.

r^{2} - 5 r + 6 = 0 .

Step 3 — Discriminant. $Δ = (- 5)^{2} - 4 (6) = 25 - 24 = 1$ . Positive, so the roots are distinct and real.

Step 4 — Roots and general solution.

r = \frac{5 \pm \sqrt{1}}{2} = \frac{5 \pm 1}{2}, so r_{1} = 2, r_{2} = 3 .

The factorisation $r^{2} - 5 r + 6 = (r - 2) (r - 3)$ confirms it. Hence

y = A e^{2 x} + B e^{3 x} .

Step 5 — Differentiate.

y^{'} = 2 A e^{2 x} + 3 B e^{3 x} .

Step 6 — Apply both conditions. At $x = 0$ every exponential is 1:

y (0) = A + B = 1, y^{'} (0) = 2 A + 3 B = 0 .

From the first, $A = 1 - B$ . Substituting into the second:

2 (1 - B) + 3 B = 0 ⟹ 2 + B = 0 ⟹ B = - 2,

and then $A = 1 - (- 2) = 3$ . The solution is

y = 3 e^{2 x} - 2 e^{3 x}

Step 7 — Verify. Both conditions first: $y (0) = 3 - 2 = 1$ , and $y^{'} (0) = 2 (3) + 3 (- 2) = 6 - 6 = 0$ , confirmed numerically as $- 0.000000$ .

Now the equation itself. Computing $y^{″} - 5 y^{'} + 6 y$ from the finished solution gives residuals of $1.7 \times 10^{- 6}$ , $4.0 \times 10^{- 7}$ and $- 8.5 \times 10^{- 5}$ at $x = 0, 0.5, 1$ : finite-difference error, not disagreement.

Algebraically the cancellation is exact. With $y = 3 e^{2 x} - 2 e^{3 x}$ :

y^{'} = 6 e^{2 x} - 6 e^{3 x}, y^{″} = 12 e^{2 x} - 18 e^{3 x},

so

y^{″} - 5 y^{'} + 6 y = (12 - 30 + 18) e^{2 x} + (- 18 + 30 - 12) e^{3 x} = 0 \cdot e^{2 x} + 0 \cdot e^{3 x} = 0 .

Each exponential's coefficient vanishes separately, as it must, since $e^{2 x}$ and $e^{3 x}$ are independent and no combination of them is zero unless both coefficients are.

The solution starts at 1 with zero slope, then falls: $y (1) = - 18.00$ and $y (2) = - 643.06$ . Although $y^{'} (0) = 0$ , this is not a maximum that holds: the $e^{3 x}$ term outgrows $e^{2 x}$ and, being negative, drags the solution down without bound. Long-run behaviour is governed by the largest root whenever its coefficient is nonzero, and here that coefficient is negative.

Example

The other two cases, side by side

The worked example covered distinct real roots. Here are the repeated and complex cases on the same pattern, so the differences stand out.

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Repeated root. Solve $y^{″} - 4 y^{'} + 4 y = 0$ with $y (0) = 1$ , $y^{'} (0) = 5$ .

Characteristic equation $r^{2} - 4 r + 4 = 0$ , discriminant $Δ = 16 - 16 = 0$ . One root, $r = 2$ , and $r^{2} - 4 r + 4 = (r - 2)^{2}$ confirms the doubling. The general solution therefore carries the $x$ factor:

y = (A + B x) e^{2 x} .

Differentiating with the product rule:

y^{'} = B e^{2 x} + 2 (A + B x) e^{2 x} = (B + 2 A + 2 B x) e^{2 x} .

At $x = 0$ : $y (0) = A = 1$ , and $y^{'} (0) = B + 2 A = 5$ , so $B = 5 - 2 = 3$ . Hence

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

Check. $y^{'} (0) = 5.000000$ , and substituting into $y^{″} - 4 y^{'} + 4 y$ gives residuals of order $10^{- 5}$ at $x = 0, 0.5, 1$ . The solution grows: $y (1) = 29.556224$ .

Note what the $x$ factor supplies. Without it the family is $C e^{2 x}$ , whose slope at 0 is forced to $2 C = 2 y (0)$ , here that would demand $y^{'} (0) = 2$ , not 5, so the condition could not be met at all.

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Complex roots. Solve $y^{″} + 2 y^{'} + 5 y = 0$ with $y (0) = 3$ , $y^{'} (0) = - 1$ .

Characteristic equation $r^{2} + 2 r + 5 = 0$ , discriminant $Δ = 4 - 20 = - 16$ . Negative, so the roots are a conjugate pair:

r = \frac{- 2 \pm \sqrt{- 16}}{2} = \frac{- 2 \pm 4 i}{2} = - 1 \pm 2 i .

So $α = - 1$ , $β = 2$ , and the real general solution is

y = e^{- x} (A \cos 2 x + B \sin 2 x) .

Differentiating, product rule on the envelope and the bracket together:

y^{'} = e^{- x} (- A \cos 2 x - B \sin 2 x - 2 A \sin 2 x + 2 B \cos 2 x) .

At $x = 0$ , where $\cos 0 = 1$ and $\sin 0 = 0$ : $y (0) = A = 3$ , and $y^{'} (0) = - A + 2 B = - 1$ . With $A = 3$ that gives $2 B = 2$ , so $B = 1$ . Hence

y = e^{- x} (3 \cos 2 x + \sin 2 x) .

Check. $y^{'} (0) = - 1.000000$ , and $- A + 2 B = - 3 + 2 = - 1$ . Residuals below $6 \times 10^{- 6}$ at $x = 0, 0.5, 1$ .

Behaviour. The solution oscillates and decays: $y (0) = 3$ , $y (1) = - 0.124764$ , $y (2) = - 0.367805$ , $y (3) = 0.129501$ , $y (5) = - 0.020626$ . The sign keeps changing, that is the $\cos 2 x$ and $\sin 2 x$ , while the magnitude shrinks toward zero under the $e^{- x}$ envelope. Note the answer contains no $i$ , although the roots did.

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What varies across the three cases.

	$Δ$	Solution	Long-run
$y^{″} - 5 y^{'} + 6 y$	$1 > 0$	$3 e^{2 x} - 2 e^{3 x}$	diverges, largest root wins
$y^{″} - 4 y^{'} + 4 y$	$0$	$(1 + 3 x) e^{2 x}$	diverges, $x$ factor along for the ride
$y^{″} + 2 y^{'} + 5 y$	$- 16 < 0$	$e^{- x} (3 \cos 2 x + \sin 2 x)$	decays to zero, oscillating

Only the third decays, and the reason is visible in the roots alone: $- 1 \pm 2 i$ has negative real part, while $2, 3$ and the doubled $2$ are positive.

Non-example

Expressions that are not solutions

The repeated root used twice. Faced with $y^{″} - 4 y^{'} + 4 y = 0$ , whose only root is $r = 2$ , it is natural to write

y = A e^{2 x} + B e^{2 x} (wrong)

on the grounds that a second-order equation needs two terms. Both terms do solve the equation, so nothing looks amiss. The failure is not in the terms but in the count of free constants.

Why it fails. The expression collapses:

A e^{2 x} + B e^{2 x} = (A + B) e^{2 x} = C e^{2 x},

a family with one constant wearing two names. Two symbols are on the page; only one degree of freedom exists.

A case it cannot solve. Take $y (0) = 1$ , $y^{'} (0) = 0$ . From $y = C e^{2 x}$ : $y (0) = C = 1$ , and $y^{'} = 2 C e^{2 x}$ , so $y^{'} (0) = 2 C = 2$ . But the problem demands $y^{'} (0) = 0$ , and $2 \neq 0$ . There is no choice of $A$ and $B$ that works. The slope at the origin is pinned to twice the value there, for every member of the family. The general solution was too small, and no cleverness in choosing constants can repair it.

The correct form $(A + B x) e^{2 x}$ has genuinely two constants: $y (0) = A = 1$ and $y^{'} (0) = B + 2 A = 0$ give $B = - 2$ , so $y = (1 - 2 x) e^{2 x}$ solves it.

The general test. Two solutions are usable as a basis only if neither is a constant multiple of the other. Here the ratio $e^{2 x} / e^{2 x} = 1$ is constant, so they are dependent. In the correct pair the ratio is $x e^{2 x} / e^{2 x} = x$ , which is not constant.

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Other things that are not second-order constant-coefficient solutions.

A single exponential where two are needed. $y = e^{2 x}$ does solve $y^{″} - 5 y^{'} + 6 y = 0$ , but it is a solution, not the general solution. Offering it alone silently discards the $e^{3 x}$ half of the solution space, and it satisfies $y (0) = 1, y^{'} (0) = 0$ only if $2 = 0$ .

A complex answer. Writing the solution of $y^{″} + 2 y^{'} + 5 y = 0$ as $A e^{(- 1 + 2 i) x} + B e^{(- 1 - 2 i) x}$ is not wrong, but leaving it there is incomplete for a real problem: the conjugate pair must be recombined into $e^{- x} (A \cos 2 x + B \sin 2 x)$ with $A, B$ real. An $i$ surviving into the final line of a real-valued problem signals an unfinished step.

Variable coefficients. For $x^{2} y^{″} + x y^{'} - y = 0$ there is no characteristic equation at all. Substituting $y = e^{r x}$ gives $(x^{2} r^{2} + x r - 1) e^{r x} = 0$ , and the bracket still contains $x$ , so no constant $r$ makes it vanish identically. The method does not apply, and the equation needs a different technique, here $y = x$ and $y = 1 / x$ happen to be solutions, neither of them an exponential.

A nonzero right side. $y^{″} + y = 3$ is not homogeneous, so its solutions are not closed under addition. The sum of two of them satisfies $y^{″} + y = 6$ . The characteristic equation still supplies the homogeneous part, but a particular solution (here $y = 3$ ) must be added, giving the coset structure the theorem block describes.

Figure

Under-, critically, and over-damped responses

the discriminant decides whether the return oscillates

The same mass and spring, $m = 1$ and $k = 4$ , released from rest at displacement 1, with only the damping changed.

Under-damped, $c = 2$ . Roots $- 1 \pm \sqrt{3} i$ . The curve crosses the axis repeatedly, each swing smaller than the last under an $e^{- t}$ envelope.

Critically damped, $c = 4$ . Repeated root $- 2$ . The fastest return that never crosses: the curve reaches rest without overshooting once.

Over-damped, $c = 5$ . Roots $- 1$ and $- 4$ . A sum of two decays, slower back to rest than the critical case despite the stronger damping, and never crossing.

The discriminant predicts which of these occurs before any solving happens, and the three pictures are what its sign means physically.

Application

Damping: the three cases as three physical behaviours

A mass on a spring with friction is described by

m y^{″} + c y^{'} + k y = 0,

where $y$ is displacement from rest, $m$ the mass, $k$ the spring stiffness and $c$ the damping coefficient. The terms are Newton's law: mass times acceleration equals the restoring force $- k y$ plus the drag $- c y^{'}$ , which opposes motion and so carries the derivative.

Dividing by $m$ puts it in standard form, and the discriminant decides everything. With $m = 1$ and $k = 4$ , varying the damping $c$ gives all three cases:

$c$	Equation	$Δ$	Case	Behaviour
$2$	$y^{″} + 2 y^{'} + 4 y = 0$	$- 12$	complex	under-damped — oscillates while decaying
$4$	$y^{″} + 4 y^{'} + 4 y = 0$	$0$	repeated	critically damped — returns fastest, no overshoot
$5$	$y^{″} + 5 y^{'} + 4 y = 0$	$9$	distinct real	over-damped — returns slowly, no oscillation

The three root cases are not an algebraic curiosity. They are three qualitatively different physical outcomes, and the discriminant predicts which occurs before any solving is done.

Under-damped ( $c = 2$ ). Roots $- 1 \pm \sqrt{3} i$ . The solution is $e^{- x} (A \cos \sqrt{3} x + B \sin \sqrt{3} x)$ : the mass crosses its rest position repeatedly, each swing smaller than the last under the $e^{- x}$ envelope. A door that swings past the frame and back does this.

Over-damped ( $c = 5$ ). Roots $- 1$ and $- 4$ , both real and negative. The solution is $A e^{- x} + B e^{- 4 x}$ , a sum of decays with no oscillation, so the mass creeps back without ever crossing. Thick oil does this to a mechanism.

Critically damped ( $c = 4$ ). The repeated root $r = - 2$ gives $(A + B x) e^{- 2 x}$ , the boundary between the two. This is the case that would be missed entirely by anyone who wrote $A e^{- 2 x} + B e^{- 2 x}$ and found the initial conditions unsatisfiable, and it is the case engineers most often want, since it returns to rest fastest without overshoot. A door closer is tuned to it, as are the recoil mechanisms and analogue meter needles that must settle without wobbling.

Why the sign of the real part is the whole stability question. Every solution above decays, because every root has negative real part. Reverse that, giving a root with positive real part, and the corresponding term grows without bound whatever the initial conditions, apart from the measure-zero case where its coefficient is exactly zero. This is why control engineers check root locations rather than solve: for $y^{″} + p y^{'} + q y = 0$ with $p, q > 0$ , both roots have negative real part, and the system is stable.

Undamped ( $c = 0$ ). With no friction the equation is $y^{″} + 4 y = 0$ , roots $\pm 2 i$ , and the solution $A \cos 2 x + B \sin 2 x$ oscillates forever with constant amplitude. The envelope $e^{0 \cdot x} = 1$ never decays. This is the equation whose two-dimensional solution space, with basis ${\cos, \sin}$ , the vector spaces unit uses as its standard example of a space whose elements are functions rather than tuples.

The same mathematics elsewhere. An $L R C$ circuit obeys $L q^{″} + R q^{'} + q / C = 0$ for charge $q$ , with inductance playing the mass's role, resistance the damping and inverse capacitance the stiffness. The three cases are ringing, critical damping and sluggish decay: the same three behaviours, because it is the same equation.

Second-Order Linear Differential Equations

A guess that turns out to be forced

Why each case takes the form it does

The solution space has dimension two

The characteristic equation and the repeated-root case

Solving a second-order constant-coefficient equation

One initial value problem, worked in full

The other two cases, side by side

Expressions that are not solutions

Under-, critically, and over-damped responses

Damping: the three cases as three physical behaviours

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