The general solution $c_1y_1 + c_2y_2$ needs two independent solutions. For vectors you tested independence with a determinant (Determinants by Cofactor Expansion). For solutions of a differential equation, the right determinant is the Wronskian.
Key idea
The Wronskian of $y_1, y_2$ is $$W = \begin{vmatrix} y_1 & y_2 \\ y_1' & y_2' \end{vmatrix} = y_1y_2' - y_1'y_2.$$ For two solutions of $y'' + p(t)y' + q(t)y = 0$ on an interval, either $W \equiv 0$ (dependent) or $W$ is never 0 (independent: a fundamental set). Abel's formula: $W(t) = Ce^{-\int p\,dt}$.
Why it works
To match initial values $y(t_0) = a$, $y'(t_0) = b$ with $c_1y_1 + c_2y_2$ we must solve $$c_1y_1(t_0) + c_2y_2(t_0) = a,$$ $$c_1y_1'(t_0) + c_2y_2'(t_0) = b.$$ This system has a solution for every $a, b$ exactly when its determinant, $W(t_0)$, is not 0. Abel's formula explains why $W$ cannot vanish at just one point: using $y_i'' = -py_i' - qy_i$, $$W' = y_1y_2'' - y_1''y_2 = -p(y_1y_2' - y_1'y_2) = -pW,$$ so $W = Ce^{-\int p\,dt}$, which is either always 0 ($C = 0$) or never 0.
Worked example: Two exponentials
Find $W(e^{2t}, e^{-3t})$.
- $W$ $= e^{2t}\cdot(-3e^{-3t}) - 2e^{2t}\cdot e^{-3t}$.
- $= -3e^{-t} - 2e^{-t} = -5e^{-t}$.
Answer: $W = -5e^{-t}$, never 0: independent
Worked example: Cosine and sine
Find $W(\cos 3t, \sin 3t)$.
- $W$ $= \cos 3t\cdot 3\cos 3t$ $- (-3\sin 3t)\sin 3t$.
- $= 3(\cos^{2} 3t + \sin^{2} 3t) = 3$.
Answer: $W = 3$
Worked example: Abel's formula without solving
Two solutions of $t^{2}y'' + ty' - 4y = 0$ ($t > 0$) have $W(1) = -4$. Find $W(t)$, and check with $y_1 = t^{2}$, $y_2 = t^{-2}$.
- Standard form: $y'' + \frac{1}{t}y' - \frac{4}{t^{2}}y = 0$, so $p = \frac{1}{t}$.
- $W = Ce^{-\ln t} = \frac{C}{t}$; $W(1) = -4$ gives $C = -4$.
- Check: $W(t^{2}, t^{-2})$ $= t^{2}(-2t^{-3}) - 2t\cdot t^{-2}$ $= -\frac{4}{t}$. ✓
Answer: $W(t) = -\frac{4}{t}$
| $y_1, y_2$ | $W$ |
|---|---|
| $e^{at}, e^{bt}$ | $(b - a)e^{(a+b)t}$ |
| $e^{rt}, te^{rt}$ | $e^{2rt}$ |
| $e^{\alpha t}\cos\beta t, e^{\alpha t}\sin\beta t$ | $\beta e^{2\alpha t}$ |
| $t^{m}, t^{n}$ | $(n - m)t^{m+n-1}$ |
Common mistake
Concluding “dependent” because $W = 0$ at one point, or “independent” after checking a single point of functions that are not solutions of one equation. For two solutions of the same equation, one value of $W$ decides everything (Abel). Also, a constant multiple is the most common hidden dependence: $e^{3t}$ and $5e^{3t}$ have $W \equiv 0$.
Where does the Wronskian come from for two functions? Dependence means one is a constant multiple of the other, so $\frac{y_2}{y_1}$ is constant. By the quotient rule, $\left(\frac{y_2}{y_1}\right)' = \frac{y_1y_2' - y_1'y_2}{y_1^{2}} = \frac{W}{y_1^{2}}$. So on an interval where $y_1 \ne 0$, $W \equiv 0$ says exactly that the ratio never changes. For $n$ solutions of an $n$th-order equation, $W$ is the $n\times n$ determinant of the functions and their derivatives up to order $n - 1$, and Abel's formula still holds.