LINEAR-ALGEBRA · Interactive Practice | Unit 11 · Video 2

Your Diff EQ Course Was Linear Algebra All Along

IKey Formulas

Formula	Name	Description
$L[c_1 y_1 + c_2 y_2] = c_1 L[y_1] + c_2 L[y_2]$	Linearity of $L$	Linear combinations of solutions are solutions
$\dim(\ker L) = n$	Dimension Theorem	$n$ th-order homogeneous linear ODE has $n$ -dimensional solution space
$W(y_1, y_2) = y_1 y_2' - y_2 y_1'$	Wronskian	Nonzero $\Rightarrow$ linearly independent
$e^{ix} = \cos x + i\sin x$	Euler's Formula	Change of basis between $\{\cos x, \sin x\}$ and $\{e^{ix}, e^{-ix}\}$

IIVisualization 1 — Superposition

Every linear combination $y = c_1\cos x + c_2\sin x$ of the two basis solutions still solves $y'' + y = 0$ — the superposition principle.

c₁ 1 c₂ 0.5

IIIVisualization 2 — Initial Conditions as Coordinates

For $y'' + y = 0$ , the initial data $\big(y(0),\,y'(0)\big) = (a,b)$ are exactly the coordinates $(c_1,c_2)$ of the solution in the basis $\{\cos x,\sin x\}$ .

y(0) = a 1 y′(0) = b 0

💡 Two conditions fix two coordinates, so the solution space is 2-dimensional — the pattern the next visualization generalizes.

IVVisualization 3 — Order = Dimension

An $n$ th-order homogeneous linear ODE has an $n$ -dimensional solution space — the order sets the number of independent basis functions.

Equation

VQuiz Questions

Question 1

What is the dimension of the solution space of $y'' + y = 0$ ?

✅ Correct! A 2nd-order homogeneous linear ODE has a 2-dimensional solution space.

❌ Not quite. Remember: the dimension of the solution space equals the order of the ODE.

Show solution

Solution:

The equation $y'' + y = 0$ is a 2nd-order homogeneous linear ODE. By the dimension theorem, the solution space has dimension equal to the order:

$\text{Order} = 2 \;\Longrightarrow\; \dim(\text{solution space}) = 2$

A basis is $\{\cos x, \sin x\}$ — exactly two linearly independent functions. The general solution $y = c_1\cos x + c_2\sin x$ has two free constants, one per basis vector.

Question 2

If $y_1$ and $y_2$ are both solutions to a homogeneous linear ODE, which of the following is guaranteed to also be a solution?

✅ Correct! Linear combinations of solutions are always solutions — that's the superposition principle, the hallmark of a vector space.

❌ Not quite. Only linear operations (addition and scalar multiplication) preserve the solution property. Products, quotients, and powers are nonlinear.

Show solution

Solution:

The solution set is a vector space, so it is closed under linear combinations:

$L[c_1 y_1 + c_2 y_2] = c_1 L[y_1] + c_2 L[y_2] = c_1 \cdot 0 + c_2 \cdot 0 = 0$

So $3y_1 - 2y_2$ (with $c_1 = 3$ , $c_2 = -2$ ) is always a solution.

Products, quotients, and powers are nonlinear operations and do NOT preserve solutions in general.

Counterexample: $\cos x$ and $\sin x$ both solve $y'' + y = 0$ , but their product $\cos x \cdot \sin x = \tfrac{1}{2}\sin(2x)$ does not — check: $(\tfrac{1}{2}\sin 2x)'' + \tfrac{1}{2}\sin 2x = -2\sin 2x + \tfrac{1}{2}\sin 2x \neq 0$ .

Question 3

True or False: The set $\{e^{ix},\; e^{-ix}\}$ is a valid basis for the solution space of $y'' + y = 0$ .

✅ Correct! Both functions solve the ODE and are linearly independent — they form a valid (complex-valued) basis for the same 2D space.

❌ Not quite. Try substituting e^(ix) into y'' + y = 0: since i² = −1, the second derivative is −e^(ix), so the sum is zero.

Show solution

Solution: True

Both $e^{ix}$ and $e^{-ix}$ are solutions to $y'' + y = 0$ :

$y = e^{ix}$ : $\;y'' = i^2 e^{ix} = -e^{ix}$ , so $y'' + y = -e^{ix} + e^{ix} = 0$ ✓
$y = e^{-ix}$ : $\;y'' = (-i)^2 e^{-ix} = -e^{-ix}$ , so $y'' + y = -e^{-ix} + e^{-ix} = 0$ ✓

They are linearly independent (neither is a scalar multiple of the other), so they form a basis for the same 2-dimensional solution space that $\{\cos x, \sin x\}$ spans.

Euler's formula provides the change-of-basis matrix between these two bases:

$\cos x = \tfrac{1}{2}e^{ix} + \tfrac{1}{2}e^{-ix}, \qquad \sin x = \tfrac{1}{2i}e^{ix} - \tfrac{1}{2i}e^{-ix}$

Question 4

The ODE $y''' - y' = 0$ is 3rd order. Which of the following is a basis for its solution space?

✅ Correct! The characteristic roots are 0, 1, −1, giving the basis {1, eˣ, e⁻ˣ} — three functions for a 3rd-order equation.

❌ Not quite. Try solving the characteristic equation r³ − r = 0 by factoring out r, then check which functions actually satisfy the ODE.

Show solution

Solution:

Factor the characteristic equation:

$r^3 - r = 0 \;\Longrightarrow\; r(r-1)(r+1) = 0 \;\Longrightarrow\; r = 0,\; 1,\; -1$

Each root gives a basis function:

Root	Basis function	Verify
$r = 0$	$y = e^{0 \cdot x} = 1$	$y''' - y' = 0 - 0 = 0$ ✓
$r = 1$	$y = e^{x}$	$y''' - y' = e^x - e^x = 0$ ✓
$r = -1$	$y = e^{-x}$	$y''' - y' = -e^{-x} - (-e^{-x}) = 0$ ✓

So the basis is $\{1,\; e^x,\; e^{-x}\}$ — three linearly independent solutions for a 3rd-order equation.

Why not the others?

$\{1, x, x^2\}$ : $y = x^2 \Rightarrow y''' - y' = 0 - 2x \neq 0$ ✗
$\{e^x, e^{-x}, e^{2x}\}$ : $y = e^{2x} \Rightarrow y''' - y' = 8e^{2x} - 2e^{2x} = 6e^{2x} \neq 0$ ✗
$\{\cos x, \sin x, e^x\}$ : $y = \cos x \Rightarrow y''' - y' = \sin x + \sin x = 2\sin x \neq 0$ ✗

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