Visual walkthrough — Case 2 - repeated real root — reduction of order

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Step 1 — What is the machine we are solving?

WHAT. We have an equation that ties together a function $y(x)$, its slope $y^{\prime }$, and the way its slope bends $y^{\prime \prime }$: $\underset{\text{stiffness}}{\underbrace{a}}\underset{\text{bending}}{\underbrace{y^{\prime \prime }}}+\underset{\text{damping}}{\underbrace{b}}\underset{\text{slope}}{\underbrace{y^{\prime }}}+\underset{\text{spring}}{\underbrace{c}}\underset{\text{height}}{\underbrace{y}}=0.$

Read every symbol:

• $y$ is the height of a curve at position $x$ — imagine a mass on a spring, $y$ is how far it is displaced.
• $y^{\prime }$ (say "y-prime") is the slope — how fast the height is changing. It is the ordinary rate-of-change you already know as speed.
• $y^{\prime \prime }$ ("y-double-prime") is the bending — how fast the slope itself changes. It curls the graph up or down.
• $a,b,c$ are fixed numbers (they never change with $x$) — that is what "constant-coefficient" means.

WHY. A second-order equation (the highest tick is $y^{\prime \prime }$, two primes) needs two independent starting facts to pin down a unique curve — usually the starting height $y(0)$ and starting slope $y^{\prime }(0)$. So we will need two independent building-block solutions. Hold that number: two.

PICTURE. Below, one wobbling curve $y(x)$ with its height, slope arrow, and bending arc marked.

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Step 2 — The trial $e^{rx}$ and the characteristic equation

WHAT. We guess that the solution is an exponential $y=e^{rx}$, where $r$ is a number we must find. The letter $e\approx 2.718$ is the special base whose exponential is its own slope.

Why is this guess so good? Because for $y=e^{rx}$: $y^{\prime }=r{e}^{rx},y^{\prime \prime }=r^2{e}^{rx}.$ Each derivative just pulls down a factor of $r$ and leaves $e^{rx}$ untouched. So differentiating never produces a new kind of function — everything stays proportional to $e^{rx}$.

Substitute into $a{y}^{\prime \prime }+b{y}^{\prime }+cy=0$ and factor out the common $e^{rx}$ (which is never zero): $e^{rx}\left(\underset{\text{must be }0}{\underbrace{a{r}^2+br+c}}\right)=0\Rightarrow \boxed{a{r}^2+br+c=0.}$

This last line is the characteristic equation. See Characteristic equation of linear ODEs.

WHY this tool (a quadratic)? Because it converts a calculus problem (derivatives) into a plain algebra problem (find roots of a quadratic). That is a massive simplification, and it works only because exponentials reproduce themselves under differentiation.

PICTURE. The exponential curve and the "root-machine" turning derivatives into powers of $r$.

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Step 3 — The repeated root: where the two roots collide

WHAT. The quadratic $a{r}^2+br+c=0$ has roots given by the familiar formula $r=\frac{-b\pm \sqrt{b^2-4ac}}{2a}.$ The quantity under the root, $b^2-4ac$, is the discriminant $\Delta$. It decides how many roots we get:

• $\Delta >0$ → two different roots (Case 1 distinct real roots),
• $\Delta <0$ → two complex roots (Case 3 complex roots),
• $\Delta =0$ → the $\pm \sqrt{\Delta }$ vanishes, so the two roots merge into one: $r=-\frac{b}{2a}.$

WHY this matters. When $\Delta =0$ the two roots are the same number. The trial $y=e^{rx}$ therefore hands us only one building block — but Step 1 insisted we need two. This is the whole crisis of Case 2: we are one solution short.

PICTURE. A number line with two roots sliding toward each other as $\Delta \to 0$, fusing into a single dot.

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Step 4 — Reduction of order manufactures the partner

WHAT. We already own one solution $y_1=e^{rx}$. Guess the second as that same solution scaled by an unknown wobble $v(x)$: $y_2=v(x)e^{rx}.$ Here $v(x)$ is a mystery function we will solve for. This is Reduction of order (general method).

Differentiate with the product rule (each term below is labelled by where it comes from): $y_2^{\prime }=e^{rx}(\underset{\text{wobble changes}}{\underbrace{v^{\prime }}}+\underset{\text{exp changes}}{\underbrace{rv}}),$ $y_2^{\prime \prime }=e^{rx}(\underset{\text{wobble bends}}{\underbrace{v^{\prime \prime }}}+\underset{\text{cross term}}{\underbrace{2r{v}^{\prime }}}+\underset{\text{exp bends}}{\underbrace{r^{2}v}}).$

Substitute into $a{y}^{\prime \prime }+b{y}^{\prime }+cy=0$, divide out $e^{rx}$, and gather by $v^{\prime \prime },v^{\prime },v$: $a{v}^{\prime \prime }+\underset{\text{coeff of }{v}^{\prime }}{\underbrace{(2ar+b)}}{v}^{\prime }+\underset{\text{coeff of }v}{\underbrace{(a{r}^2+br+c)}}v=0.$

Now the magic of the repeated root fires twice:

• $a{r}^2+br+c=0$ because $r$ is a root → the whole $v$-term dies.
• $2ar+b=2a\left(-\frac{b}{2a}\right)+b=0$ because $r=-b/2a$ → the whole $v^{\prime }$-term dies too.

Two coefficients vanish at once — only possible for a repeated root. What survives is $a{v}^{\prime \prime }=0\Rightarrow v^{\prime \prime }=0.$

WHY this tool (reduction of order)? Because we already knew one solution and wanted to use it as leverage. Assuming $y_2=v{y}_1$ guarantees the leftover equation for $v$ is simpler ("order reduced"). For a distinct root, $2ar+b\ne 0$, the $v^{\prime }$-term lives, and $v$ comes out exponential — recovering the other root instead. Only the collision case leaves the ultra-simple $v^{\prime \prime }=0$.

PICTURE. The three grouped terms with the two doomed ones struck out, leaving $a{v}^{\prime \prime }=0$.

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Step 5 — Solving $v^{\prime \prime }=0$ and reading the new block

WHAT. "$v^{\prime \prime }=0$" says the wobble has zero bending — a straight line: $v^{\prime \prime }=0\Rightarrow v^{\prime }=C_2\Rightarrow v=C_1+C_{2}x.$ Multiply back by $e^{rx}$: $y=v{e}^{rx}=(C_1+C_{2}x)e^{rx}=\underset{\text{old block}}{\underbrace{C_1{e}^{rx}}}+\underset{\text{brand-new block}}{\underbrace{C_{2}x{e}^{rx}}}.$

The genuinely new function is $==x{e}^{rx}==$ — the old exponential with an $x$ stapled on the front.

WHY it counts as a second block: it must be linearly independent of $e^{rx}$ — not just a rescaling of it. We test with the Wronskian $W=y_1{y}_2^{\prime }-y_1^{\prime }{y}_2$. With $y_2^{\prime }=e^{rx}(1+rx)$: $W=e^{rx}\cdot e^{rx}(1+rx)-r{e}^{rx}\cdot x{e}^{rx}=e^{2rx}[(1+rx)-rx]=e^{2rx}\ne 0.$ Never zero → truly independent everywhere. ✓

PICTURE. $e^{rx}$ (a pure curve) and $x{e}^{rx}$ (rises, peaks, then the exponential wins) plotted together — visibly different shapes, not scalings.

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Step 6 — Why the $x$? The coalescing-roots picture (degenerate limit)

WHAT. Pretend for a second the two roots are almost equal: $r$ and $r+\epsilon$ for a tiny $\epsilon$. Then $e^{rx}$ and $e^{(r+\epsilon )x}$ are both solutions, so their difference divided by $\epsilon$ is too: $\frac{e^{(r+\epsilon )x}-e^{rx}}{\epsilon }.$ This is exactly the definition of a derivative — but with respect to the root $r$, not $x$. Let $\epsilon \to 0$ (the roots fuse): ${\lim }_{\epsilon \to 0}\frac{e^{(r+\epsilon )x}-e^{rx}}{\epsilon }=\frac{\partial }{\partial r}{e}^{rx}=x{e}^{rx}.$

WHY show this? Because it explains where the $x$ physically comes from: it is the slope of the solution family as you nudge the root. The instant two roots collide, that slope direction — $x{e}^{rx}$ — becomes the new second solution. The algebra of Step 4 and this limit agree perfectly.

PICTURE. Two nearby exponentials, their scaled difference, and the limiting $x{e}^{rx}$ they converge to.

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The one-picture summary

The full pipeline compressed: the ODE → its quadratic → the discriminant collapsing to zero → one root → reduction of order killing two terms → $v^{\prime \prime }=0$ → the two blocks $\{e^{rx},x{e}^{rx}\}$.

Recall Feynman retelling — the whole walkthrough in plain words

We're solving a spring-like equation that mixes a curve's height, slope, and bend. A neat trick: guess an exponential $e^{rx}$, because exponentials keep their shape when you take slopes — every derivative just multiplies by $r$. That turns the calculus into a plain quadratic $a{r}^2+br+c=0$. Usually the quadratic has two answers → two building blocks → done. But when the two answers crash into the same number (discriminant zero), we only get one block, and a second-order equation demands two. So we take our one block $e^{rx}$, multiply it by an unknown wobble $v(x)$, and shove it back in. Because the root is repeated, two whole terms cancel and we're left with the baby equation "$v$ has no bend," i.e. $v$ is a straight line $C_1+C_{2}x$. Multiplying back gives the second block: the old exponential with an $x$ glued on, $x{e}^{rx}$. The Wronskian confirms it's genuinely new. And if you're wondering why an $x$ appears — imagine the two roots almost-but-not-quite equal and let them slide together; the gap between their two solutions, rescaled, becomes exactly $x{e}^{rx}$. That's the missing partner, and now you can build any solution.

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Connections

• Case 2 - repeated real root — reduction of order — parent topic (the result this page draws).
• Characteristic equation of linear ODEs — Step 2, where $r$ is born.
• Case 1 distinct real roots — the $\Delta >0$ sibling (roots stay apart).
• Case 3 complex roots — the $\Delta <0$ sibling.
• Reduction of order (general method) — the engine of Step 4.
• Wronskian and linear independence — the independence check in Step 5.
• Higher-order repeated roots — repeat the $x$-trick: $e^{rx},x{e}^{rx},x^2{e}^{rx},\dots$