Fundamental Sets and the Wronskian

Main idea

Two solutions are enough to build every solution when you can match any starting position and velocity with them. That matching is a 2×2 linear system, and its determinant, the Wronskian $W = y_1y_2' - y_1'y_2$ , tells you whether it always works.

{\htmlData{sym=0}{W}}(y_1, y_2) = \begin{vmatrix} y_1 & y_2\\ y_1' & y_2'\end{vmatrix} = y_1y_2' - y_1'y_2

What the symbols mean

$W$
Wronskian
If it's nonzero at a point, any initial conditions there can be matched by c₁y₁ + c₂y₂.

Explanation

A fundamental set is simply a pair of solutions you can mix to get every possible solution, and the Wronskian is a single number you compute to check that a pair qualifies. (It's named after the Polish mathematician Józef Hoene-Wroński.) All we're really asking is: are your two solutions genuinely different, or is one secretly just a copy of the other? The Wronskian $W = y_1y_2' - y_1'y_2$ looks like a mouthful, but it's just four numbers: $y_1$ and $y_2$ (the two solutions) and $y_1'$ and $y_2'$ (their slopes), cross-multiplied and subtracted, the same way you'd take a 2×2 determinant. If that number isn't zero, the two solutions are different enough to build anything. That's all there is to it.

We've been saying "two independent solutions give the general solution" for a while now. This lesson makes that precise. Two solutions $y_1, y_2$ form a fundamental set when every solution can be written as $c_1y_1 + c_2y_2$ .

Why it matters. Every time you solved an IVP in this unit, you did the same thing: matched $y(t_0) = y_0$ and $y'(t_0) = y_0'$ by solving

c_1y_1(t_0) + c_2y_2(t_0) = y_0, \qquad c_1y_1'(t_0) + c_2y_2'(t_0) = y_0'.

That's a 2×2 linear system for $c_1$ and $c_2$ . From algebra, it has a solution for every right side exactly when its determinant is nonzero, and that determinant is the Wronskian. So a nonzero Wronskian means $y_1$ and $y_2$ can hit any starting position and velocity, which is exactly what "every solution" needs.

It makes sense that a zero Wronskian means trouble. If $y_2$ is just a multiple of $y_1$ , the two columns of the determinant are proportional, $W = 0$ , and you really only have one solution.

Abel's theorem. For solutions of $y'' + p(t)y' + q(t)y = 0$ , the Wronskian is $Ce^{-\int p\,dt}$ , so it's either never zero or always zero. Checking one point is enough.

A common mistake. It's tempting to use the Wronskian on any two functions and conclude that $W = 0$ means they're multiples of each other. For solutions of the same linear equation that's right, but not in general: $t^2$ and $t|t|$ have $W = 0$ everywhere, yet on $(-1, 1)$ neither is a multiple of the other. Abel's theorem is what makes the test airtight for solutions.

See it

Two solutions of y″ + 0.5y′ + 4y = 0 and their Wronskian. Notice that the solutions themselves cross zero again and again, but W = 1.98e^(−0.5t) never does. It starts at about 1.98 and shrinks toward zero without ever touching it, just as Abel's theorem promises. Never zero means y₁ and y₂ can match any starting position and velocity.

Worked examples

Example 1

Is $\{e^{2t}, e^{-3t}\}$ a fundamental set?

1
$W = y_1y_2' - y_1'y_2 = e^{2t}(-3e^{-3t}) - 2e^{2t}e^{-3t} = -5e^{-t}$ , which is never zero.

Yes: $W \ne 0$ .

Practice

Type your answer and check it. Solve them all to complete the lesson.

Problem 1+5 XP

Find $W(\cos t, \sin t)$ .

Problem 2+5 XP

Find $W(e^{2t}, e^{-3t})$ at $t = 0$ .

Problem 3+5 XP

Do $e^t$ and $2e^t$ form a fundamental set?

Problem 4+5 XP

Find $W(e^{rt}, te^{rt})$ at $t = 0$ .

Problem 5+5 XP

Find $W(t, t^2)$ at $t = 1$ .

Problem 6+5 XP

$x_1 = e^{-2t}\cos 3t + 4e^{-2t}\sin 3t$ and $x_2 = e^{-2t}\sin 3t$ both solve the same homogeneous second order linear equation. Which of these is its general solution?

Problem 7+5 XP

Two modes of a building's sway are $y_1 = e^{-t}\cos 2t$ and $y_2 = e^{-t}\sin 2t$ . To be sure every starting displacement and velocity can be produced by $c_1y_1 + c_2y_2$ , an engineer computes $W(y_1, y_2)$ at $t = 0$ . What is it?

Challenge problem+25 XP

The hardest one here. Get it, and you've got this lesson.

Two solutions of $ty'' + 2y' + ty = 0$ (for $t > 0$ ) have Wronskian $W(1) = 3$ . Find $W(2)$ without solving the equation.

Where it's used

Engineers rely on this guarantee: any starting position and velocity of a linear system can be reproduced by combining two basic modes.

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