Differential-Equations · Unit 11 · Video 2 · Interactive Practice

The Wronskian and Abel's Theorem: Fitting Any Initial Values

IKey Formulas

Formula	Name	What it says
$c_1y_1(x_0) + c_2y_2(x_0) = A,\quad c_1y_1'(x_0) + c_2y_2'(x_0) = B$	Initial value system	Two linear equations in the unknowns $c_1, c_2$
$W(y_1, y_2) = \begin{vmatrix} y_1 & y_2 \\ y_1' & y_2' \end{vmatrix} = y_1y_2' - y_2y_1'$	Wronskian	A function of $x$ ; the system is solvable when $W(x_0) \neq 0$
$W \equiv 0$ or $W(x) \neq 0$ for every $x$	Abel's theorem	For two solutions of $y'' + p\,y' + q\,y = 0$ with $p, q$ continuous
$W(e^{r_1x}, e^{r_2x}) = (r_2 - r_1)\,e^{(r_1 + r_2)x}$	Distinct real roots	Never zero when $r_1 \neq r_2$

Key Insight: For independent solutions the Wronskian is never zero, so $c_1y_1 + c_2y_2$ can match any value and slope at any point.

IIVisualization 1 — Fit a Value and a Slope

For $y'' - 3y' + 2y = 0$ with $y_1 = e^x$ , $y_2 = e^{2x}$ , every value $A$ and slope $B$ at $x_0$ is reached by exactly one $c_1, c_2$ .

x₀ 0 A = y(x₀) 1 B = y′(x₀) 0

IIIVisualization 2 — Identically Zero or Never Zero

Abel's theorem leaves the Wronskian of two solutions only two options: zero everywhere, or zero nowhere.

Pair x 0.5

💡 For $x$ and $x^2$ , the standard form $y'' - \frac{2}{x}y' + \frac{2}{x^2}y = 0$ has $p$ and $q$ discontinuous at $x = 0$ — exactly where $W = x^2$ vanishes. Abel's theorem needs continuous coefficients.

IVQuiz Questions

Problem 1 · A Wronskian of Exponentials

Given: $y_1 = e^{2x}$ and $y_2 = e^{-3x}$ — find $W(y_1, y_2) = y_1y_2' - y_2y_1'$ .

✅ Correct!

(r_2 - r_1)\,e^{(r_1 + r_2)x} = (-3 - 2)\,e^{-x}

, never zero.

❌ Check the order.

W = y_1y_2' - y_2y_1'

, which gives the factor

r_2 - r_1 = -5

; you computed

y_2y_1' - y_1y_2'

❌ Not quite. The derivatives bring down the exponents:

y_2' = -3e^{-3x}

and

y_1' = 2e^{2x}

, so the factor is

-3 - 2 = -5

❌ Check the exponent.

e^{2x}e^{-3x} = e^{(2 - 3)x} = e^{-x}

Show solution

W = e^{2x}\cdot(-3e^{-3x}) - e^{-3x}\cdot 2e^{2x} = -3e^{-x} - 2e^{-x} = -5e^{-x}.

This matches $W(e^{r_1x}, e^{r_2x}) = (r_2 - r_1)\,e^{(r_1 + r_2)x}$ with $r_1 = 2$ , $r_2 = -3$ . Since $e^{-x} > 0$ , $W$ is never zero.

Problem 2 · One Zero of the Wronskian

Given: $y_1$ , $y_2$ solve $y'' + p(x)\,y' + q(x)\,y = 0$ with $p, q$ continuous for all $x$ , and $W(y_1, y_2)(1) = 0$ — what follows?

✅ Correct! Abel's theorem rules out a single zero, so

W

is identically zero; by the converse quoted in the video, the pair is dependent.

❌ Not quite. Abel's theorem makes the Wronskian of two solutions zero everywhere once it is zero somewhere, so the system fails at every

x_0

❌ Not quite. For two solutions with continuous

p, q

an isolated zero is impossible — that is exactly what Abel's theorem says.

❌ Not quite. A sign change would mean

W

is zero at one point and nonzero nearby, which Abel's theorem forbids.

Show solution

Abel's theorem: for two solutions of $y'' + p\,y' + q\,y = 0$ with $p, q$ continuous, either $W \equiv 0$ or $W(x) \neq 0$ for every $x$ . Since $W(1) = 0$ , the second case is impossible, so $W \equiv 0$ .

The converse quoted in the video says an identically zero Wronskian of two solutions means they are dependent. Such a pair cannot fit most initial values at any point.

Problem 3 · Solve the Initial Value Problem

Given: $y'' - 4y = 0$ with $y_1 = e^{2x}$ , $y_2 = e^{-2x}$ , and $y(0) = 3$ , $y'(0) = 2$ — find $W(0)$ and the constants.

What is $W(y_1, y_2)(0)$ ?

What are $c_1$ and $c_2$ ?

✅ Correct!

W(0) = -4 \neq 0

, and

y = 2e^{2x} + e^{-2x}

has

y(0) = 3

y'(0) = 4 - 2 = 2

❌ Check the Wronskian. At

0

y_1 = y_2 = 1

y_1' = 2

y_2' = -2

, so

W = 1\cdot(-2) - 1\cdot 2

❌ Check the constants. The system is

c_1 + c_2 = 3

and

2c_1 - 2c_2 = 2

; the initial values are not the constants themselves.

Show solution

At $x_0 = 0$ : $y_1(0) = 1$ , $y_2(0) = 1$ , $y_1'(0) = 2$ , $y_2'(0) = -2$ , so

W(0) = \begin{vmatrix} 1 & 1 \\ 2 & -2 \end{vmatrix} = -2 - 2 = -4 \neq 0.

The system $c_1 + c_2 = 3$ , $2c_1 - 2c_2 = 2$ gives $c_1 - c_2 = 1$ ; adding, $2c_1 = 4$ , so $c_1 = 2$ and $c_2 = 1$ :

y = 2e^{2x} + e^{-2x}.

Problem 4 · A Wronskian That Vanishes Once

Given: $y_1 = x$ and $y_2 = x^2$ solve $x^2y'' - 2xy' + 2y = 0$ , and $W(x, x^2) = x^2$ is zero at $x = 0$ only — why does this not contradict Abel's theorem?

✅ Correct! Abel's theorem assumes

p

and

q

continuous; dividing by

x^2

breaks that at

x = 0

, the one place

W

vanishes.

❌ Not quite.

x^2/x = x

is not constant, so the pair is independent — and a dependent pair would have

W \equiv 0

❌ Not quite. The theorem covers variable coefficients, as long as

p(x)

and

q(x)

are continuous.

❌ Not quite.

W = x\cdot 2x - x^2\cdot 1 = x^2

, which is

0

x = 0

Show solution

$W(x, x^2) = x\cdot 2x - x^2\cdot 1 = x^2$ : zero at $x = 0$ , nonzero elsewhere. Abel's theorem would forbid this for an equation $y'' + p\,y' + q\,y = 0$ with continuous coefficients.

Dividing by $x^2$ gives the standard form $y'' - \frac{2}{x}y' + \frac{2}{x^2}y = 0$ , with $p = -2/x$ and $q = 2/x^2$ undefined at $x = 0$ . The hypothesis fails exactly there, so there is no contradiction. On $x > 0$ (or $x < 0$ ) the coefficients are continuous and $W$ is never zero, as the theorem says.

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