Case 2 - repeated real root — reduction of order: Exercises

This page is your self-test track for Case 2: repeated real roots. Every problem has a fully worked solution hidden inside a collapsible callout — try it first, then reveal. Levels climb from recognising a repeated root to building whole solution families from scratch.

Level 1 — Recognition

Goal: read off whether a root is repeated, and what the root is, without solving anything hard.

Recall Solution 1.1

The discriminant $b^{2} - 4 a c$ answers one question: are the two roots equal? A repeated root needs it to be exactly $0$ .

(a) $a = 1, b = - 10, c = 25$ : $b^{2} - 4 a c = 100 - 100 = 0$ . Repeated — double root at $r = - b /2 a = 10/2 = 5$ . (b) $a = 1, b = 2, c = 5$ : $4 - 20 = - 16 < 0$ . Not repeated — complex roots (see Case 3 complex roots). (c) $a = 9, b = - 6, c = 1$ : $36 - 36 = 0$ . Repeated — double root at $r = - b /2 a = 6/18 = \frac{1}{3}$ .

Recall Solution 1.2

(C) is correct.

(A) collapses to $(C_{1} + C_{2}) e^{- 2 x}$ — a single function, a 1-dimensional family. Too small for a 2nd-order ODE.
(B) treats $C_{2}$ (a bare constant) as a solution, but a constant only solves the ODE when $c = 0$ . Here it does not.
(C) attaches the required factor $x$ to build the genuinely new partner $x e^{- 2 x}$ .

Level 2 — Application

Goal: solve a repeated-root ODE end to end, including odd leading coefficients.

Recall Solution 2.1

Characteristic: $r^{2} - 14 r + 49 = 0$ . This is a perfect square: $(r - 7)^{2} = 0$ , so $r = 7$ (double). Check: discriminant $196 - 196 = 0$ . ✓
Solutions: $e^{7 x}$ and $x e^{7 x}$ .
General solution: $y = (C_{1} + C_{2} x) e^{7 x}$ .

Recall Solution 2.2

$16 r^{2} + 8 r + 1 = 0$ . As a perfect square: $(4 r + 1)^{2} = 0$ , so $4 r + 1 = 0 \Rightarrow r = - \frac{1}{4}$ (double). Cross-check with $r = - b /2 a = - 8/32 = - \frac{1}{4}$ . ✓
General solution: $y = (C_{1} + C_{2} x) e^{- x /4}$ .

Recall Solution 2.3

$r^{2} + 12 r + 3 = 0$ . Discriminant $b^{2} - 4 a c = 12 - 12 = 0$ — repeated. Root $r = - \frac{12}{2} = - \frac{2 3}{2} = - 3$ .
General solution: $y = (C_{1} + C_{2} x) e^{- 3 x}$ .

Level 3 — Analysis

Goal: initial-value problems, boundary problems, and building the ODE from its solution.

Recall Solution 3.1

$r^{2} - 2 r + 1 = (r - 1)^{2} = 0 \Rightarrow r = 1$ (double).
$y = (C_{1} + C_{2} x) e^{x}$ .
$y (0) = 2$ : at $x = 0$ , $e^{0} = 1$ and the $x$ -term dies, so $C_{1} = 2$ .
Differentiate: $y^{'} = C_{2} e^{x} + (C_{1} + C_{2} x) e^{x} = (C_{1} + C_{2} + C_{2} x) e^{x}$ . At $x = 0$ : $y^{'} (0) = C_{1} + C_{2} = 5 \Rightarrow C_{2} = 5 - 2 = 3$ .
Answer: $y = (2 + 3 x) e^{x}$ .

Recall Solution 3.2

$r^{2} + 6 r + 9 = (r + 3)^{2} = 0 \Rightarrow r = - 3$ (double).
$y = (C_{1} + C_{2} x) e^{- 3 x}$ .
$y (0) = 0 \Rightarrow C_{1} = 0$ .
$y^{'} = C_{2} e^{- 3 x} - 3 (C_{1} + C_{2} x) e^{- 3 x} = (C_{2} - 3 C_{1} - 3 C_{2} x) e^{- 3 x}$ . At $x = 0$ : $y^{'} (0) = C_{2} - 3 C_{1} = 4 \Rightarrow C_{2} = 4$ .
Answer: $y = 4 x e^{- 3 x}$ .

Recall Solution 3.3

The solution form tells us the root is $r = 5$ , repeated. The characteristic equation is therefore $(r - 5)^{2} = 0$ , i.e. $r^{2} - 10 r + 25 = 0$ . Matching to $a r^{2} + b r + c$ with $a = 1$ : $b = - 10, c = 25.$ ODE: $y^{''} - 10 y^{'} + 25 y = 0$ .

Level 4 — Synthesis

Goal: combine reduction of order, the Wronskian, and long-run behaviour.

Recall Solution 4.1

Set $y = v (x) e^{- x}$ . Then $y^{'} = e^{- x} (v^{'} - v), y^{''} = e^{- x} (v^{''} - 2 v^{'} + v) .$ Substitute into $y^{''} + 2 y^{'} + y = 0$ and drop the common $e^{- x} \neq = 0$ : $(v^{''} - 2 v^{'} + v) + 2 (v^{'} - v) + v = 0.$ Group: $v^{''} + (- 2 + 2) v^{'} + (1 - 2 + 1) v = 0 \Rightarrow v^{''} = 0$ . Both lower terms vanish (the hallmark of a repeated root). So $v = C_{1} + C_{2} x$ , and the new piece is $v = x$ , giving $y_{2} = x e^{- x}$ . ✓

Recall Solution 4.2

The Wronskian $W = y_{1} y_{2}^{'} - y_{1}^{'} y_{2}$ asks: is one solution a constant multiple of the other? If $W \neq = 0$ anywhere, no — they are independent.

$y_{1} = e^{- x}$ , $y_{1}^{'} = - e^{- x}$ .
$y_{2} = x e^{- x}$ , $y_{2}^{'} = e^{- x} - x e^{- x} = e^{- x} (1 - x)$ . $W = e^{- x} \cdot e^{- x} (1 - x) - (- e^{- x}) \cdot x e^{- x} = e^{- 2 x} [(1 - x) + x] = e^{- 2 x} .$ Since $e^{- 2 x} > 0$ everywhere, they are independent. At $x = 1$ : $W (1) = e^{- 2} \approx 0.135$ .

Recall Solution 4.3

Limit: exponential decay $e^{- x}$ beats the linear growth $x$ , so $x e^{- x} \to 0$ as $x \to + \infty$ (and generally $(C_{1} + C_{2} x) e^{- x} \to 0$ ). See the figure below.

Maximum of $y = x e^{- x}$ : set $y^{'} = e^{- x} (1 - x) = 0 \Rightarrow x = 1$ . Then $y (1) = 1 \cdot e^{- 1} = e^{- 1} \approx 0.368$ . This is a maximum (a hump), matching the shape in the figure.

Figure — Case 2 - repeated real root — reduction of order

Level 5 — Mastery

Goal: extend the pattern to higher multiplicity and non-standard forms.

Recall Solution 5.1

Characteristic: $r^{3} - 9 r^{2} + 27 r - 27 = 0$ . Recognise the binomial expansion: $(r - 3)^{3} = r^{3} - 9 r^{2} + 27 r - 27$ . So $r = 3$ with multiplicity 3. By the pattern for repeated roots, a root of multiplicity $m$ contributes $e^{r x}, x e^{r x}, \dots, x^{m - 1} e^{r x}$ . Here $m = 3$ : $y = (C_{1} + C_{2} x + C_{3} x^{2}) e^{3 x} .$

Recall Solution 5.2

Repeated root needs discriminant $k^{2} - 4 (1) (9) = 0 \Rightarrow k^{2} = 36 \Rightarrow k = \pm 6$ . Taking $k > 0$ : $k = 6$ . Then $r = - k /2 = - 3$ (double), and $y = (C_{1} + C_{2} x) e^{- 3 x}$ .

Recall Solution 5.3

Characteristic: $r^{4} + 8 r^{2} + 16 = 0$ . Let $u = r^{2}$ : $u^{2} + 8 u + 16 = (u + 4)^{2} = 0 \Rightarrow u = - 4$ (double). So $r^{2} = - 4 \Rightarrow r = \pm 2 i$ , each repeated (multiplicity 2). A repeated complex pair $\pm β i$ gives solutions $cos β x, sin β x$ and $x cos β x, x sin β x$ (the repeated-root rule applied inside Case 3 complex roots). With $β = 2$ : $y = (C_{1} + C_{2} x) cos 2 x + (C_{3} + C_{4} x) sin 2 x .$

Recall One-line summary of the whole ladder

Spot the double root $r = - b /2 a$ ::: then the general solution is $(C_{1} + C_{2} x) e^{r x}$ ; for multiplicity $m$ climb to $x^{m - 1} e^{r x}$ .

Connections

Case 2 - repeated real root — reduction of order (index 4.6.12) — the parent theory these drills exercise.
Characteristic equation of linear ODEs — every problem starts here.
Case 1 distinct real roots / Case 3 complex roots — the neighbouring cases (Ex 1.1b, 5.3).
Reduction of order (general method) — Exercise 4.1 does it by hand.
Wronskian and linear independence — Exercise 4.2.
Higher-order repeated roots — Exercises 5.1 and 5.3.