Homogeneous with constant coefficients — characteristic equation: Exercises

Quick reference for every case (we will use these labels constantly):

WHY $Δ$ decides everything — the picture. The characteristic polynomial $a r^{2} + b r + c$ is a parabola in the variable $r$ . The roots are where that parabola crosses the horizontal axis. The next figure shows all three cases side by side: how many times the parabola touches the axis is exactly $Δ$ in disguise.

Figure — Homogeneous with constant coefficients — characteristic equation

Two crossings (left, red): the parabola dips below the axis, so it cuts it in two places → two distinct real roots → two exponentials.
One touch (middle, red): the parabola just kisses the axis at its vertex → the two roots have merged into one → $Δ = 0$ , and the merging is why we are short a solution and must slip in the $x$ -factor $x e^{r x}$ .
No crossing (right, red): the parabola floats entirely above the axis → no real root. The roots have left the real line and become the complex pair $α \pm i β$ .

And why do complex roots mean waves? The next figure shows the complex root $α \pm i β$ as a point in the plane. Multiplying by $e^{i β x}$ rotates that point around a circle as $x$ grows — and the shadow of a point going round a circle is exactly a $cos$ / $sin$ wave. The distance from the vertical axis, controlled by $α$ , is what makes the circle spiral in ( $α < 0$ ) or out ( $α > 0$ ).

Level 1 — Recognition

Here the characteristic equation is handed to you already factored, or trivial to factor. The skill: read the roots and pick the right shape.

Problem 1.1

The characteristic equation of some ODE is $(r - 4) (r + 1) = 0$ . Write the general solution $y (x)$ .

Recall Solution 1.1

WHAT: find the roots. Setting each factor to zero: $r - 4 = 0 \Rightarrow r = 4$ and $r + 1 = 0 \Rightarrow r = - 1$ . WHY the shape: two different real numbers → $Δ > 0$ → Case 1 (two crossings in the left panel of the parabola figure). $y = c_{1} e^{4 x} + c_{2} e^{- x} .$

Problem 1.2

The characteristic equation is $(r - 7)^{2} = 0$ . Write the general solution.

Recall Solution 1.2

WHAT: $(r - 7)^{2} = 0$ means $r = 7$ counted twice — a repeated root ( $Δ = 0$ , the "just touches" middle panel). WHY the extra $x$ : one root gives only one independent solution $e^{7 x}$ ; the second independent solution is $x e^{7 x}$ (proven in Second Order Linear ODE — Reduction of Order). $y = (c_{1} + c_{2} x) e^{7 x} .$

Problem 1.3

The characteristic equation is $r^{2} + 9 = 0$ . Write the general solution.

Recall Solution 1.3

WHAT: $r^{2} = - 9$ , so $r = \pm 3 i$ . In the form $α \pm i β$ we have $α = 0$ , $β = 3$ (the "no crossing" right panel — the parabola $r^{2} + 9$ never reaches the axis). WHY sin and cos: $α = 0$ means the exponential envelope $e^{0 \cdot x} = 1$ (no growth/decay), and $β = 3$ is the oscillation frequency. Euler turns $e^{\pm 3 i x}$ into real waves. $y = c_{1} cos 3 x + c_{2} sin 3 x .$

Problem 1.4 — the zero-root edge case

The characteristic equation is $r (r - 5) = 0$ . Write the general solution.

Recall Solution 1.4

WHAT: the two roots are $r = 0$ and $r = 5$ . The root $r = 0$ is easy to overlook because it makes $e^{0 \cdot x} = e^{0} = 1$ — a plain constant. WHY a constant counts: $y = c_{1} e^{0 \cdot x} = c_{1}$ genuinely solves the ODE (this comes from an ODE like $y^{''} - 5 y^{'} = 0$ , which never contains $y$ itself, so any constant survives). It is a real, independent solution. $y = c_{1} e^{0 \cdot x} + c_{2} e^{5 x} = c_{1} + c_{2} e^{5 x} .$ If instead the equation were $r^{2} = 0$ (a repeated zero root), the same $x$ -factor rule applies: $y = (c_{1} + c_{2} x) e^{0 \cdot x} = c_{1} + c_{2} x$ — a straight line.

Level 2 — Application

Now you form the characteristic equation yourself and solve it. Watch the leading coefficient $a$ .

Problem 2.1

Solve $y^{''} + 7 y^{'} + 10 y = 0$ .

Recall Solution 2.1

Char. eq. (replace $y^{''} \to r^{2}$ , $y^{'} \to r$ , $y \to 1$ ): $r^{2} + 7 r + 10 = 0$ . Factor: $(r + 2) (r + 5) = 0 \Rightarrow r = - 2, - 5$ . Both real and distinct. $y = c_{1} e^{- 2 x} + c_{2} e^{- 5 x} .$

Problem 2.2

Solve $y^{''} - 4 y^{'} + 4 y = 0$ .

Recall Solution 2.2

Char. eq.: $r^{2} - 4 r + 4 = 0$ . Check the discriminant: $Δ = (- 4)^{2} - 4 (1) (4) = 16 - 16 = 0$ → repeated root. $(r - 2)^{2} = 0 \Rightarrow r = 2$ (double). $y = (c_{1} + c_{2} x) e^{2 x} .$

Problem 2.3

Solve $y^{''} + 6 y^{'} + 13 y = 0$ .

Recall Solution 2.3

Char. eq.: $r^{2} + 6 r + 13 = 0$ . Discriminant $Δ = 36 - 52 = - 16 < 0$ → complex. Quadratic formula: $r = \frac{- 6 \pm - 16}{2} = \frac{- 6 \pm 4 i}{2} = - 3 \pm 2 i$ . So $α = - 3$ (decay), $β = 2$ (frequency). $y = e^{- 3 x} (c_{1} cos 2 x + c_{2} sin 2 x) .$

Problem 2.4

Solve $2 y^{''} - 3 y^{'} - 2 y = 0$ — note $a = 2$ , not $1$ .

Recall Solution 2.4

Char. eq.: $2 r^{2} - 3 r - 2 = 0$ . Do not drop the $2$ . Quadratic formula: $r = \frac{3 \pm 9 + 16}{4} = \frac{3 \pm 5}{4}$ , giving $r = 2$ and $r = - \frac{1}{2}$ . $y = c_{1} e^{2 x} + c_{2} e^{- x /2} .$

Level 3 — Analysis (initial conditions + interpretation)

Problem 3.1

Solve $y^{''} - y^{'} - 6 y = 0$ with $y (0) = 1$ , $y^{'} (0) = 8$ .

Recall Solution 3.1

Roots: $r^{2} - r - 6 = (r - 3) (r + 2) = 0 \Rightarrow r = 3, - 2$ . General: $y = c_{1} e^{3 x} + c_{2} e^{- 2 x}$ . Apply $y (0) = 1$ : at $x = 0$ , $e^{0} = 1$ , so $c_{1} + c_{2} = 1$ . Differentiate: $y^{'} = 3 c_{1} e^{3 x} - 2 c_{2} e^{- 2 x}$ . At $x = 0$ : $3 c_{1} - 2 c_{2} = 8$ . Solve the pair: from the first, $c_{2} = 1 - c_{1}$ ; substitute: $3 c_{1} - 2 (1 - c_{1}) = 8 \Rightarrow 5 c_{1} = 10 \Rightarrow c_{1} = 2$ , hence $c_{2} = - 1$ . $y = 2 e^{3 x} - e^{- 2 x} .$

Problem 3.2

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

Recall Solution 3.2

Roots: $r^{2} + 2 r + 1 = (r + 1)^{2} = 0 \Rightarrow r = - 1$ (double). General: $y = (c_{1} + c_{2} x) e^{- x}$ . Apply $y (0) = 3$ : $x = 0$ gives $(c_{1} + 0) \cdot 1 = c_{1} = 3$ . Differentiate (product rule!): $y^{'} = c_{2} e^{- x} - (c_{1} + c_{2} x) e^{- x} = (c_{2} - c_{1} - c_{2} x) e^{- x}$ . At $x = 0$ : $c_{2} - c_{1} = - 5 \Rightarrow c_{2} = - 5 + 3 = - 2$ . $y = (3 - 2 x) e^{- x} .$

Problem 3.3

Solve $y^{''} + 4 y^{'} + 5 y = 0$ with $y (0) = 2$ , $y^{'} (0) = - 1$ , and describe the motion.

Recall Solution 3.3

Roots: $r^{2} + 4 r + 5 = 0$ , $Δ = 16 - 20 = - 4$ , so $r = \frac{- 4 \pm 2 i}{2} = - 2 \pm i$ . Thus $α = - 2, β = 1$ . General: $y = e^{- 2 x} (c_{1} cos x + c_{2} sin x)$ . Apply $y (0) = 2$ : $e^{0} (c_{1} \cdot 1 + c_{2} \cdot 0) = c_{1} = 2$ . Differentiate (product rule): $y^{'} = - 2 e^{- 2 x} (c_{1} cos x + c_{2} sin x) + e^{- 2 x} (- c_{1} sin x + c_{2} cos x) .$ At $x = 0$ : $- 2 c_{1} + c_{2} = - 1 \Rightarrow c_{2} = - 1 + 2 c_{1} = - 1 + 4 = 3$ . $y = e^{- 2 x} (2 cos x + 3 sin x) .$ Interpretation: $α = - 2 < 0$ → the amplitude envelope $e^{- 2 x}$ shrinks, so this is a damped oscillation — a wiggle fading to zero (see Damped Harmonic Oscillator). The figure below plots this exact solution (red) trapped inside its shrinking envelope $\pm 13 e^{- 2 x}$ (dashed black): the labelled axes are $x$ (horizontal) and $y (x)$ (vertical), and the arrows point out the oscillation and the decaying envelope.

Level 4 — Synthesis (higher order, mixed roots)

Everything scales up: an $n$ -th order equation gives a degree- $n$ characteristic polynomial and needs $== n == "$ independent solutions (from Wronskian and Linear Independence of Solutions).

Problem 4.1

Solve the third-order ODE $y^{'''} - 6 y^{''} + 11 y^{'} - 6 y = 0$ .

Recall Solution 4.1

Char. eq.: $r^{3} - 6 r^{2} + 11 r - 6 = 0$ . Try small integer roots (factors of the constant $6$ ): $r = 1$ gives $1 - 6 + 11 - 6 = 0$ ✓. So by the factor theorem $(r - 1)$ divides the cubic exactly. HOW to divide it out — synthetic division. The figure below lays out the synthetic-division table for dividing by $(r - 1)$ : bring down the leading $1$ , multiply by the root $1$ , add into the next coefficient, repeat. The bottom row reads off the quotient coefficients $1, - 5, 6$ and a final remainder $0$ (confirming $r = 1$ really is a root):

So $r^{3} - 6 r^{2} + 11 r - 6 = (r - 1) (r^{2} - 5 r + 6) = (r - 1) (r - 2) (r - 3)$ . Roots $r = 1, 2, 3$ — three distinct real roots, three exponentials. $y = c_{1} e^{x} + c_{2} e^{2 x} + c_{3} e^{3 x} .$

Problem 4.2

Solve $y^{'''} - 3 y^{''} + 3 y^{'} - y = 0$ .

Recall Solution 4.2

Char. eq.: $r^{3} - 3 r^{2} + 3 r - 1 = 0$ . Recognise the binomial: this is $(r - 1)^{3} = 0$ , so $r = 1$ with multiplicity three. WHY three terms $1, x, x^{2}$ : a root repeated $m$ times contributes $e^{r x}$ multiplied by $1, x, \dots, x^{m - 1}$ — the same "sneak in an $x$ " idea, extended. $y = (c_{1} + c_{2} x + c_{3} x^{2}) e^{x} .$

Problem 4.3

Solve the fourth-order ODE $y^{(4)} + 8 y^{''} + 16 y = 0$ .

Recall Solution 4.3

Char. eq.: $r^{4} + 8 r^{2} + 16 = 0$ . Let $u = r^{2}$ : $u^{2} + 8 u + 16 = (u + 4)^{2} = 0 \Rightarrow u = - 4$ (double). So $r^{2} = - 4 \Rightarrow r = \pm 2 i$ , and because $u = - 4$ was a double root, the pair $\pm 2 i$ is repeated. Here $α = 0, β = 2$ . WHY the extra $x$ on the trig too: a repeated complex pair multiplies each of $cos, sin$ by $1$ and $x$ — exactly the repeated-root rule applied to oscillations. $y = (c_{1} + c_{2} x) cos 2 x + (c_{3} + c_{4} x) sin 2 x .$

Problem 4.4 — repeated zero root among others

Solve $y^{(4)} - y^{'''} = 0$ .

Recall Solution 4.4

Char. eq.: $r^{4} - r^{3} = 0 \Rightarrow r^{3} (r - 1) = 0$ . Roots: $r = 0$ with multiplicity three and $r = 1$ once. WHY the powers of $x$ : the triple root $r = 0$ gives $e^{0 \cdot x}$ times $1, x, x^{2}$ — i.e. just $1, x, x^{2}$ — and $r = 1$ gives $e^{x}$ . $y = c_{1} + c_{2} x + c_{3} x^{2} + c_{4} e^{x} .$

Level 5 — Mastery (build it / prove it)

Problem 5.1 (reverse engineering)

Find a second-order homogeneous ODE with constant coefficients whose general solution is $y = e^{- x} (c_{1} cos 3 x + c_{2} sin 3 x)$ .

Recall Solution 5.1

WHAT the solution tells us: $α = - 1, β = 3$ , so the roots are $r = - 1 \pm 3 i$ . Build the polynomial from its roots: $(r - (- 1 + 3 i)) (r - (- 1 - 3 i)) = (r + 1 - 3 i) (r + 1 + 3 i) = (r + 1)^{2} - (3 i)^{2} = (r + 1)^{2} + 9.$ Expand: $(r + 1)^{2} + 9 = r^{2} + 2 r + 1 + 9 = r^{2} + 2 r + 10$ . Read off the ODE (coefficients of $r^{2}, r, 1$ become $y^{''}, y^{'}, y$ ): $y^{''} + 2 y^{'} + 10 y = 0.$ Check: $Δ = 4 - 40 = - 36 < 0$ ✓ complex, and $r = \frac{- 2 \pm 6 i}{2} = - 1 \pm 3 i$ ✓.

Problem 5.2 (build with a mixed spectrum)

Construct the lowest-order constant-coefficient ODE having $x e^{2 x}$ and $cos x$ among its solutions.

Recall Solution 5.2

Decode each solution into required roots:

$x e^{2 x}$ appearing means $r = 2$ is a double root → factor $(r - 2)^{2}$ .
$cos x$ (i.e. $e^{0 \cdot x} cos 1 x$ ) means the pair $r = \pm i$ → factor $(r^{2} + 1)$ . Lowest degree = multiply exactly these: $(r - 2)^{2} (r^{2} + 1)$ , a degree-4 polynomial → 4th order ODE. Expand: $(r - 2)^{2} = r^{2} - 4 r + 4$ , then $(r^{2} - 4 r + 4) (r^{2} + 1) = r^{4} - 4 r^{3} + 4 r^{2} + r^{2} - 4 r + 4 = r^{4} - 4 r^{3} + 5 r^{2} - 4 r + 4.$ The ODE: $y^{(4)} - 4 y^{'''} + 5 y^{''} - 4 y^{'} + 4 y = 0.$

Problem 5.3 (prove the reduction fact)

For $y^{''} - 2 r y^{'} + r^{2} y = 0$ (a deliberate double root $r$ ), verify directly that $y = x e^{r x}$ is a solution.

Recall Solution 5.3

Compute the derivatives (product rule): $y = x e^{r x}$ , $y^{'} = e^{r x} + r x e^{r x} = (1 + r x) e^{r x}$ , $y^{''} = r e^{r x} + (1 + r x) r e^{r x} = (2 r + r^{2} x) e^{r x}$ . Substitute into $y^{''} - 2 r y^{'} + r^{2} y$ : $(2 r + r^{2} x) e^{r x} - 2 r (1 + r x) e^{r x} + r^{2} x e^{r x} .$ Collect the bracket: $2 r + r^{2} x - 2 r - 2 r^{2} x + r^{2} x = (2 r - 2 r) + (r^{2} x - 2 r^{2} x + r^{2} x) = 0$ . The whole thing is $0 \cdot e^{r x} = 0$ ✓. So $x e^{r x}$ genuinely solves the double-root ODE — this is why the $x$ -factor is not a trick but a theorem.

Connections

Homogeneous with constant coefficients — characteristic equation (parent)
Linear Differential Operators and Superposition
Euler's Formula and Complex Exponentials
Wronskian and Linear Independence of Solutions
Method of Undetermined Coefficients
Damped Harmonic Oscillator
Second Order Linear ODE — Reduction of Order