Case 2 - repeated real root — reduction of order

For the constant-coefficient ODE $ay'' + by' + cy = 0$, when the characteristic equation has a repeated real root $r$ (i.e. discriminant $b^2 - 4ac = 0$), the two independent solutions are $e^{rx}$ and $==xe^{rx}==$.

The Setup

WHY $xe^{rx}$? — Two derivations from scratch

Derivation A: Reduction of Order (the honest method)

Start with $ay'' + by' + cy = 0$ and $y = v e^{rx}$.

Compute derivatives (product rule): $y' = e^{rx}(v' + r v),\qquad y'' = e^{rx}(v'' + 2r v' + r^2 v).$

Why this step? We need $y',y''$ in terms of $v$ so we can substitute.

Substitute and factor out $e^{rx}\neq 0$: $a(v'' + 2rv' + r^2 v) + b(v' + rv) + c v = 0.$

Group by $v'', v', v$: $a\,v'' + (2ar + b)\,v' + (ar^2 + br + c)\,v = 0.$

Why this step? We isolate the coefficients to use what we know about $r$.

Now use the two facts that define this case:

• $r$ is a root: $==ar^2 + br + c = 0==$ → the $v$-term vanishes.
• $r$ is the repeated root $r = -b/2a$, so $2ar + b = 2a(-b/2a)+b = ==0==$ → the $v'$-term vanishes.

We are left with just $a\,v'' = 0 \quad\Rightarrow\quad v'' = 0.$

Derivation B: The limit / coalescing roots (intuition booster)

Independence check (so it's a valid second solution)

With $y_1 = e^{rx},\ y_2 = xe^{rx}$: $y_2' = e^{rx} + rxe^{rx} = e^{rx}(1+rx).$ $W = e^{rx}\cdot e^{rx}(1+rx) - re^{rx}\cdot xe^{rx} = e^{2rx}\big[(1+rx) - rx\big] = ==e^{2rx}==\neq 0.$

Good — independent everywhere. ✓

Worked Examples

Common Mistakes

Flashcards

Connections

• Characteristic equation of linear ODEs — where the root $r$ comes from.
• Case 1 distinct real roots — contrast: two roots, two clean exponentials.
• Case 3 complex roots — when discriminant $<0$.
• Reduction of order (general method) — the technique used here, works for any known $y_1$.
• Wronskian and linear independence — proves $e^{rx}, xe^{rx}$ are independent.
• Higher-order repeated roots — root of multiplicity $m$ gives $e^{rx},xe^{rx},\dots,x^{m-1}e^{rx}$.

Concept Map

Hinglish (regional understanding)