Case 1 - two distinct real roots: Exercises — Maths deep dive

How to use this page: try each problem before opening its solution. Every answer sits inside a collapsible [!recall]- callout — click to reveal. Problems climb from recognising the case to building your own ODE to match a required behaviour. All numeric answers are machine-checked.

Everything here rests on one machine: $a y^{''} + b y^{'} + cy = 0 y = e^{r x} a r^{2} + b r + c = 0 b^{2} - 4 a c > 0 y = C_{1} e^{r_{1} x} + C_{2} e^{r_{2} x} .$ See the parent note for why. If you need why the exponential is the natural guess, revisit Exponential function and its derivative; for why we add solutions, Superposition principle.

Level 1 — Recognition

Goal: read the discriminant and name the case. No solving required beyond spotting roots.

Recall Solution 1.1

Read off $a = 1$ each time. (a) $b^{2} - 4 a c = 25 - 24 = 1 > 0$ → Case 1. (b) $16 - 16 = 0$ → Case 2 (repeated root), see Case 2 repeated real root. (c) $4 - 20 = - 16 < 0$ → Case 3 (complex roots), see Case 3 complex conjugate roots. Why the sign is all that matters: $r_{1, 2} = \frac{- b \pm b ^{2} - 4 a c}{2 a}$ . The thing under the square root decides whether the $\pm$ splits into two real numbers ( $> 0$ ), collapses to one ( $= 0$ ), or goes imaginary ( $< 0$ ).

Recall Solution 1.2

Factor: $(r - 2) (r - 5) = 0 \Rightarrow r_{1} = 2, r_{2} = 5$ . Distinct reals, so $y = C_{1} e^{2 x} + C_{2} e^{5 x} .$

Level 2 — Application

Goal: turn an ODE into its general solution.

Recall Solution 2.1

Characteristic equation: $r^{2} + r - 6 = 0$ . Factor: $(r + 3) (r - 2) = 0 \Rightarrow r_{1} = - 3, r_{2} = 2$ . Check: discriminant $= 1 + 24 = 25 > 0$ ✔ Case 1. $y = C_{1} e^{- 3 x} + C_{2} e^{2 x} .$

Recall Solution 2.2

Characteristic: $2 r^{2} - 3 r - 2 = 0$ . Quadratic formula: $r = \frac{3 \pm 9 + 16}{4} = \frac{3 \pm 5}{4}$ . So $r_{1} = \frac{8}{4} = 2$ and $r_{2} = \frac{- 2}{4} = - \frac{1}{2}$ . $y = C_{1} e^{2 x} + C_{2} e^{- x /2} .$ Why keep $a = 2$ inside the quadratic: the $2 a = 4$ in the denominator comes from that leading coefficient. Dropping it silently gives wrong roots.

Recall Solution 2.3

First: $r^{2} - 9 = 0 \Rightarrow r = \pm 3$ . $y = C_{1} e^{3 x} + C_{2} e^{- 3 x}$ . Second: $r^{2} - 3 r = 0 \Rightarrow r (r - 3) = 0 \Rightarrow r_{1} = 0, r_{2} = 3$ . Since $e^{0 \cdot x} = 1$ , $y = C_{1} + C_{2} e^{3 x} .$ The comparison: a root of $r = 0$ is still a perfectly valid distinct real root — it just produces the constant solution $C_{1}$ . Zero is not a degenerate case here; it's an honest member of Case 1.

Level 3 — Analysis

Goal: use initial/boundary conditions to pin the constants, and read long-term behaviour.

Recall Solution 3.1

Roots: $r^{2} - r - 2 = (r - 2) (r + 1) = 0 \Rightarrow r_{1} = 2, r_{2} = - 1$ . General: $y = C_{1} e^{2 x} + C_{2} e^{- x}$ , and $y^{'} = 2 C_{1} e^{2 x} - C_{2} e^{- x}$ . Conditions at $x = 0$ (both exponentials $= 1$ ): $C_{1} + C_{2} = 0, 2 C_{1} - C_{2} = 3.$ Add: $3 C_{1} = 3 \Rightarrow C_{1} = 1$ , then $C_{2} = - 1$ . $y = e^{2 x} - e^{- x} .$

Recall Solution 3.2

Roots: $r^{2} + 3 r + 2 = (r + 1) (r + 2) = 0 \Rightarrow r_{1} = - 1, r_{2} = - 2$ . $y = C_{1} e^{- x} + C_{2} e^{- 2 x}$ , $y^{'} = - C_{1} e^{- x} - 2 C_{2} e^{- 2 x}$ . At $x = 0$ : $C_{1} + C_{2} = 2$ , $- C_{1} - 2 C_{2} = - 5 \Rightarrow C_{1} + 2 C_{2} = 5$ . Subtract first from second: $C_{2} = 3$ , then $C_{1} = - 1$ . $y = - e^{- x} + 3 e^{- 2 x} .$ Long term: both exponents negative → $y \to 0$ as $x \to \infty$ . See figure below.

Figure — Case 1 - two distinct real roots

Recall Solution 3.3

Roots $r = \pm 2$ : $y = C_{1} e^{2 x} + C_{2} e^{- 2 x}$ . $y (0) = 1$ : $C_{1} + C_{2} = 1$ . $y (1) = 0$ : $C_{1} e^{2} + C_{2} e^{- 2} = 0 \Rightarrow C_{1} = - C_{2} e^{- 4}$ . Substitute: $- C_{2} e^{- 4} + C_{2} = 1 \Rightarrow C_{2} (1 - e^{- 4}) = 1 \Rightarrow C_{2} = \frac{1}{1 - e ^{- 4}}$ . Then $C_{1} = \frac{- e ^{- 4}}{1 - e ^{- 4}}$ . Numerically $C_{2} \approx 1.01866$ , $C_{1} \approx - 0.018659$ . $y = \frac{- e ^{- 4}}{1 - e ^{- 4}} e^{2 x} + \frac{1}{1 - e ^{- 4}} e^{- 2 x} .$ Why the method is identical: boundary conditions still give two linear equations in $C_{1}, C_{2}$ . Only where we sample $y$ changes.

Level 4 — Synthesis

Goal: build the ODE, or the solution, backwards from given data.

Recall Solution 4.1

The exponents tell us the roots: $r_{1} = 4, r_{2} = - 1$ . Build the characteristic equation from them: $(r - 4) (r + 1) = r^{2} - 3 r - 4 = 0.$ Read powers of $r$ back to derivatives ( $r^{2} \to y^{''}, r \to y^{'}, 1 \to y$ ): $y^{''} - 3 y^{'} - 4 y = 0 .$ Why this works: the map ODE $\leftrightarrow$ characteristic polynomial is a two-way street. Roots $\to$ factors $\to$ polynomial $\to$ ODE.

Recall Solution 4.2

Constant $1 = e^{0 \cdot x}$ means $r_{1} = 0$ ; $e^{5 x}$ means $r_{2} = 5$ . $(r - 0) (r - 5) = r^{2} - 5 r = 0 \Rightarrow y^{''} - 5 y^{'} = 0 .$ Check: $c = 0$ , matching that a root sits at $r = 0$ (no constant term $\Rightarrow$ zero is a root).

Recall Solution 4.3

$y_{1}^{'} = 2 e^{2 x}$ , $y_{2}^{'} = - 3 e^{- 3 x}$ . $W = e^{2 x} (- 3 e^{- 3 x}) - (2 e^{2 x}) e^{- 3 x} = (- 3 - 2) e^{- x} = - 5 e^{- x} .$ Since $e^{- x} \neq = 0$ for all $x$ , $W \neq = 0$ everywhere → linearly independent. Note $- 5 = r_{2} - r_{1} = - 3 - 2$ , matching the general formula $W = (r_{2} - r_{1}) e^{(r_{1} + r_{2}) x}$ from Wronskian and linear independence.

Level 5 — Mastery

Goal: combine everything — design behaviour, handle a system-like constraint, reason about signs.

Recall Solution 5.1

"Decays to 0" needs both roots negative. "One twice as fast" means one root is twice the other: pick $r_{1} = - 1, r_{2} = - 2$ . $(r + 1) (r + 2) = r^{2} + 3 r + 2 = 0 \Rightarrow y^{''} + 3 y^{'} + 2 y = 0 .$ General solution $y = C_{1} e^{- x} + C_{2} e^{- 2 x}$ ; the $e^{- 2 x}$ mode decays twice as fast as $e^{- x}$ . ✔ Sanity via sign rules: both roots negative $\Leftrightarrow$ sum $= - b < 0$ (so $b > 0$ ) and product $= c > 0$ . Here $b = 3 > 0, c = 2 > 0$ . ✔

Recall Solution 5.2

Roots: $r^{2} - r - 6 = (r - 3) (r + 2) = 0 \Rightarrow r_{1} = 3, r_{2} = - 2$ . $y = C_{1} e^{3 x} + C_{2} e^{- 2 x}$ , $y^{'} = 3 C_{1} e^{3 x} - 2 C_{2} e^{- 2 x}$ . Conditions: $C_{1} + C_{2} = A$ , $3 C_{1} - 2 C_{2} = 0 \Rightarrow C_{1} = \frac{2}{3} C_{2}$ . Then $\frac{2}{3} C_{2} + C_{2} = A \Rightarrow \frac{5}{3} C_{2} = A \Rightarrow C_{2} = \frac{3 A}{5}, C_{1} = \frac{2 A}{5}$ . $y = \frac{2 A}{5} e^{3 x} + \frac{3 A}{5} e^{- 2 x} .$ The $e^{3 x}$ term blows up unless its coefficient is zero: $C_{1} = \frac{2 A}{5} = 0 \Rightarrow A = 0$ . Interpretation: with $r_{1} > 0 > r_{2}$ this is a saddle. Only the exact starting choice $A = 0$ (giving $y \equiv 0$ ) lands on the stable direction; any $A \neq = 0$ excites the growing mode. See figure.

Recall Solution 5.3

Characteristic: $3 r^{2} - r - 2 = 0$ . Quadratic formula: $r = \frac{1 \pm 1 + 24}{6} = \frac{1 \pm 5}{6}$ . $r_{1} = 1, r_{2} = - \frac{2}{3}$ . $y = C_{1} e^{x} + C_{2} e^{- 2 x /3}$ , $y^{'} = C_{1} e^{x} - \frac{2}{3} C_{2} e^{- 2 x /3}$ . At $x = 0$ : $C_{1} + C_{2} = 1$ , $C_{1} - \frac{2}{3} C_{2} = 0 \Rightarrow C_{1} = \frac{2}{3} C_{2}$ . Then $\frac{2}{3} C_{2} + C_{2} = 1 \Rightarrow \frac{5}{3} C_{2} = 1 \Rightarrow C_{2} = \frac{3}{5}, C_{1} = \frac{2}{5}$ . $y = \frac{2}{5} e^{x} + \frac{3}{5} e^{- 2 x /3} .$

Recall Self-test recap (cloze)

A distinct root $r = 0$ contributes the solution ==the constant $1$ ==. Both roots negative $\Leftrightarrow$ $b > 0$ and == $c > 0$ == (monic). The Wronskian of $e^{r_{1} x}, e^{r_{2} x}$ equals == $(r_{2} - r_{1}) e^{(r_{1} + r_{2}) x}$ ==, nonzero iff $r_{1} \neq = r_{2}$ . From roots $r_{1}, r_{2}$ the monic characteristic polynomial is == $(r - r_{1}) (r - r_{2})$ ==.

Connections

Parent: Case 1 — the theory these exercises drill.
Characteristic equation — the two-way bridge used in L4/L5.
Wronskian and linear independence — the tool of Exercise 4.3.
Superposition principle — why every general solution is $C_{1} y_{1} + C_{2} y_{2}$ .
Exponential function and its derivative — why $e^{r x}$ is the ansatz.
Case 2 repeated real root / Case 3 complex conjugate roots — the discriminant siblings from L1.