For arbitrary functions the Wronskian test runs in one direction only. For solutions of a common linear equation it runs both ways, and the reason is a striking fact discovered by Abel: the Wronskian of such solutions satisfies its own first-order differential equation, whose solution is an exponential. Because an exponential is never zero, the Wronskian is either zero everywhere or nowhere.

Quick Reference

Item	Statement
Setting	$y_{1}, y_{2}$ solve y'' + P(x)y' + Q(x)y = 0 on an interval $I$
Abel's differential equation	$\frac{d W}{d x} = - P (x) W$
Abel's formula	$W (x) = c e^{- \int P (x) d x}$
Point form	$W (y_{1}, y_{2}) (x) = W (y_{1}, y_{2}) (x_{0}) e^{- \int_{x_{0}}^{x} P (t) d t}$
Dichotomy	$W$ is either never zero on $I$ or zero everywhere on $I$
Independence test	Solutions are independent $⟺ W (x_{0}) \neq 0$ at any one point $x_{0} \in I$
Fundamental set	$n$ linearly independent solutions of an $n$ th order homogeneous equation
General solution	$y_{h} = c_{1} ϕ_{1} + c_{2} ϕ_{2} + \dots + c_{n} ϕ_{n}$

Abel's Formula

Suppose $y_{1} (x)$ and $y_{2} (x)$ are two solutions of a second-order linear differential equation

y'' + P(x)y' + Q(x)y = 0

on an interval $I$ . Their Wronskian is

W\left(y_{1}, y_{2}\right) = \begin{vmatrix} y_{1}(x) & y_{2}(x) \\ y_{1}'(x) & y_{2}'(x) \end{vmatrix} = y_{1}(x)y_{2}'(x) - y_{2}(x)y_{1}'(x) .

We will show that this function is either never zero on $I$ or zero everywhere on $I$ .

Let us find $\frac{d W}{d x}$ . Differentiating the product form,

\begin{aligned} \frac{dW}{dx} &= y_{1}'(x)y_{2}'(x) + y_{1}(x)y_{2}''(x) - y_{2}'(x)y_{1}'(x) - y_{2}(x)y_{1}''(x) \\ &= y_{1}(x)y_{2}''(x) - y_{2}(x)y_{1}''(x) \end{aligned}

The two middle terms cancel, which is the first small miracle: the derivative of the Wronskian involves only second derivatives, not a mixture.

Because $y_{1}$ and $y_{2}$ satisfy the differential equation,

\begin{aligned} y_{1}'' + P(x)y_{1}' + Q(x)y_{1} &= 0 & \Longrightarrow \qquad y_{1}'' &= -P(x)y_{1}' - Q(x)y_{1} \\ y_{2}'' + P(x)y_{2}' + Q(x)y_{2} &= 0 & \Longrightarrow \qquad y_{2}'' &= -P(x)y_{2}' - Q(x)y_{2} \end{aligned}

Substituting these expressions,

\begin{aligned} \frac{dW}{dx} &= y_{1}\left[-P(x)y_{2}' - Q(x)y_{2}\right] - y_{2}\left[-P(x)y_{1}' - Q(x)y_{1}\right] \\ &= -P(x)\left[y_{1}y_{2}' - y_{2}y_{1}'\right] - Q(x)y_{1}y_{2} + Q(x)y_{1}y_{2} \\ &= -P(x)\left[y_{1}y_{2}' - y_{2}y_{1}'\right] \\ &= -P(x)\,W \end{aligned}

The terms involving $Q$ cancel exactly, which is the second small miracle: the Wronskian is blind to the coefficient $Q$ and responds only to $P$ .

Note that $W$ is a function of $x$ , so we have obtained a first-order linear differential equation:

\begin{matrix} \frac{d W}{d x} = - P (x) W \\ ⟹ \frac{d W}{W} = - P (x) d x \\ ⟹ \ln | W | = - \int P (x) d x \\ ⟹ W = c e^{- \int P (x) d x} x \in I \end{matrix}

Abel's Formula. If $y_{1}$ and $y_{2}$ are solutions of y'' + P(x)y' + Q(x)y = 0 on an interval $I$ , then

W (y_{1}, y_{2}) (x) = c e^{- \int P (x) d x} .

Equivalently, evaluating at a point $x_{0} \in I$ ,

W (y_{1}, y_{2}) (x) = W (y_{1}, y_{2}) (x_{0}) e^{- \int_{x_{0}}^{x} P (t) d t} .

The All-or-Nothing Dichotomy

Because an exponential function is never zero, we conclude that if $c = 0$ then $W (x) \equiv 0$ , and if $c \neq 0$ then the Wronskian $W (x)$ is never zero.

If $W (y_{1}, y_{2}) (x_{0}) = 0$ at one point, then $W (y_{1}, y_{2}) (x) = 0$ for every $x \in I$ . If $W (y_{1}, y_{2}) (x_{0}) \neq 0$ at one point, then $W (y_{1}, y_{2}) (x) \neq 0$ for every $x \in I$ . There is no in-between: the Wronskian of two solutions cannot vanish at some points and not others.

This is precisely the extra ingredient that was missing in Section 3.2. For arbitrary functions a zero Wronskian proves nothing; for solutions of a common linear equation it proves dependence.

The same argument extends to order $n$ , where the corresponding statement is sometimes called Liouville's formula.

If $ϕ_{1} (x), \dots, ϕ_{n} (x)$ are solutions of a homogeneous linear differential equation of order $n$ ,

L [y] = 0,

on an interval $I$ , then they are linearly independent if and only if

W (ϕ_{1}, \dots, ϕ_{n}) (x_{0}) \neq 0

for any arbitrary point $x_{0}$ in $I$ .

The phrase "for any arbitrary point" is doing real work. Because of the dichotomy, checking one convenient point settles the question for the entire interval. In practice you choose whichever $x_{0}$ makes the determinant easiest, usually $x_{0} = 0$ .

Worked Examples

Two solutions of y'' - \dfrac{2}{x}y' + \dfrac{2}{x^{2}}y = 0 on $x > 0$ are $y_{1} = x$ and $y_{2} = x^{2}$ . Verify Abel's formula for them.

Solution

Direct computation gives

W = | \begin{matrix} x & x^{2} \\ 1 & 2 x \end{matrix} | = 2 x^{2} - x^{2} = x^{2} .

Now check against Abel's formula. Here $P (x) = - \frac{2}{x}$ , so

- \int P (x) d x = \int \frac{2}{x} d x = 2 \ln x,

and

c e^{2 \ln x} = c x^{2} .

The two agree with $c = 1$ . Note that $W = x^{2}$ is never zero on $x > 0$ , so $y_{1}$ and $y_{2}$ are independent there, and $y = c_{1} x + c_{2} x^{2}$ is the general solution.

Find the Wronskian of any two solutions of y'' + 3y' + 2y = 0 without solving the equation.

Solution

Abel's formula applies immediately, with $P (x) = 3$ :

W = c e^{- \int 3 d x} = c e^{- 3 x} .

No knowledge of the solutions themselves was needed. As a check, the actual solutions are $e^{- x}$ and $e^{- 2 x}$ , whose Wronskian is

| \begin{matrix} e^{- x} & e^{- 2 x} \\ - e^{- x} & - 2 e^{- 2 x} \end{matrix} | = - 2 e^{- 3 x} + e^{- 3 x} = - e^{- 3 x},

which is the predicted form with $c = - 1$ .

Given that $y_{1} = x$ solves x^{2}y'' - x(x+2)y' + (x+2)y = 0 on $x > 0$ , use Abel's formula to find a second independent solution.

Solution

This technique is called reduction of order via the Wronskian, and it is a standard use of Abel's formula.

First put the equation in standard form by dividing by $x^{2}$ :

y'' - \frac{x+2}{x}y' + \frac{x+2}{x^{2}}y = 0 ,

so $P (x) = - \frac{x + 2}{x} = - 1 - \frac{2}{x}$ .

By Abel's formula, taking $c = 1$ ,

W = e^{- \int (- 1 - 2 / x) d x} = e^{x + 2 \ln x} = x^{2} e^{x} .

But by definition W = y_1 y_2' - y_2 y_1' = x y_2' - y_2. So

x y_2' - y_2 = x^{2}e^{x} .

This is a first-order linear equation for $y_{2}$ . Dividing by $x$ ,

y_2' - \frac{1}{x}y_2 = x e^{x} ,

with integrating factor $\frac{1}{x}$ :

\frac{d}{d x} (\frac{y_{2}}{x}) = e^{x} ⟹ \frac{y_{2}}{x} = e^{x} + k .

Taking $k = 0$ gives the second solution $y_{2} = x e^{x}$ .

Check. With $y_{2} = x e^{x}$ : y_2' = (1+x)e^{x} and y_2'' = (2+x)e^{x}. Substituting,

x^{2} (2 + x) e^{x} - x (x + 2) (1 + x) e^{x} + (x + 2) x e^{x} = x e^{x} (x + 2) [x - (1 + x) + 1] = 0 . ✓

The general solution is $y = c_{1} x + c_{2} x e^{x}$ .

Fundamental Sets of Solutions

A set of $n$ linearly independent solutions ${ϕ_{1}, ϕ_{2}, \dots, ϕ_{n}}$ of an $n$ th order homogeneous linear equation $L [y] = 0$ on an interval $I$ is called a fundamental set of solutions on $I$ .

If ${ϕ_{1}, \dots, ϕ_{n}}$ is a fundamental set of solutions of $L [y] = 0$ on $I$ , then every solution of $L [y] = 0$ on $I$ has the form

y = c_{1} ϕ_{1} (x) + c_{2} ϕ_{2} (x) + \dots + c_{n} ϕ_{n} (x)

for suitable constants $c_{1}, \dots, c_{n}$ . This expression is called the general solution of the homogeneous equation, often written $y_{h}$ .

Sketch of proof

Let $ψ$ be any solution and fix $x_{0} \in I$ . We seek constants making $y = \sum c_{k} ϕ_{k}$ match $ψ$ in value and in the first $n - 1$ derivatives at $x_{0}$ :
$[\begin{matrix} ϕ_{1} (x_{0}) & \dots & ϕ_{n} (x_{0}) ⋮ & ⋱ & ⋮ ϕ_{1}^{(n - 1)} (x_{0}) & \dots & ϕ_{n}^{(n - 1)} (x_{0}) \end{matrix}] [\begin{matrix} c_{1} ⋮ c_{n} \end{matrix}] < / h 1 >< p > [\begin{matrix} ψ (x_{0}) ⋮ ψ^{(n - 1)} (x_{0}) \end{matrix}]$

The coefficient matrix is exactly the Wronskian matrix, whose determinant is nonzero by independence and the theorem above. So the system has a unique solution $c_{1}, \dots, c_{n}$ .

Now $y = \sum c_{k} ϕ_{k}$ and $ψ$ are both solutions of $L [y] = 0$ with identical data at $x_{0}$ . By the existence and uniqueness theorem for linear equations, they must be the same function on $I$ .

The theorem says the solution space has dimension exactly $n$ . This is the single most important structural fact about linear equations, and it tells you when to stop looking: once you have $n$ independent solutions, you have them all. Finding an $(n + 1)$ st solution that seems new means it is secretly a combination of the ones you have.

Exercises

Without solving, find the form of the Wronskian of two solutions of y'' + \dfrac{1}{x}y' + y = 0 on $x > 0$ .

Solution

Here $P (x) = \frac{1}{x}$ , so

W = c e^{- \int d x / x} = c e^{- \ln x} = \frac{c}{x} .

This is Bessel's equation of order zero, whose solutions are not elementary, yet Abel's formula gives their Wronskian exactly.

The functions $y_{1} = e^{x}$ and $y_{2} = 2 e^{x}$ both solve y'' - y' = 0. Compute their Wronskian and explain the result.

Solution

W = | \begin{matrix} e^{x} & 2 e^{x} \\ e^{x} & 2 e^{x} \end{matrix} | = 2 e^{2 x} - 2 e^{2 x} = 0 .

The Wronskian vanishes identically, and by the theorem this means the solutions are dependent, which is obvious here since $y_{2} = 2 y_{1}$ . Abel's formula is consistent: it gives $W = c e^{x}$ , and the constant happens to be $c = 0$ .

Note the contrast with the general-function case. Here the vanishing Wronskian really does prove dependence, because both functions solve the same linear equation.

Given that $y_{1} = e^{x}$ solves xy'' - (x+1)y' + y = 0 on $x > 0$ , use Abel's formula to find a second independent solution.

Solution

In standard form, $P (x) = - \frac{x + 1}{x} = - 1 - \frac{1}{x}$ , so

W = e^{\int (1 + 1 / x) d x} = e^{x + \ln x} = x e^{x} .

Then W = y_1 y_2' - y_2 y_1' = e^{x}y_2' - y_2 e^{x} = xe^{x}, so dividing by $e^{x}$ ,

y_2' - y_2 = x .

With integrating factor $e^{- x}$ ,

\frac{d}{d x} (y_{2} e^{- x}) = x e^{- x} ⟹ y_{2} e^{- x} = - x e^{- x} - e^{- x} + k .

Taking $k = 0$ gives $y_{2} = - x - 1$ , or equivalently $y_{2} = x + 1$ after dropping the sign.

Check. With $y_{2} = x + 1$ : y_2' = 1, y_2'' = 0, and $x \cdot 0 - (x + 1) \cdot 1 + (x + 1) = 0$ . ✓

The general solution is $y = c_{1} e^{x} + c_{2} (x + 1)$ .

Frequently Asked Questions

What is Abel's formula?

It states that the Wronskian of solutions of y'' + P(x)y' + Q(x)y = 0 equals $c e^{- \int P d x}$ . The Wronskian therefore satisfies its own first-order equation and is determined up to a single constant by the coefficient $P$ alone, without any knowledge of the solutions.

Why does

Q (x)

not appear in Abel's formula?

Because the $Q$ terms cancel in the derivation: substituting the two expressions for y_1'' and y_2'' produces $- Q y_{1} y_{2} + Q y_{1} y_{2} = 0$ . Only the $P$ terms survive. This is why two equations with the same $P$ but wildly different $Q$ have Wronskians of the same shape.

Why is the all-or-nothing dichotomy so useful?

Because it converts a statement about an entire interval into a check at a single point. Without it, verifying independence would require confirming $W \neq 0$ everywhere. With it, one convenient evaluation, typically at $x_{0} = 0$ , settles the matter.

What is a fundamental set of solutions?

A set of $n$ linearly independent solutions of an $n$ th order homogeneous linear equation. It is a basis for the solution space, so every solution is a unique linear combination of its members. A second-order equation needs two, a third-order equation three.

Can I use Abel's formula to find a second solution?

Yes, and it is one of the formula's most practical applications. If $y_{1}$ is known, Abel gives $W$ explicitly, and the definition W = y_1y_2' - y_2y_1' then becomes a first-order linear equation for $y_{2}$ . The last two worked examples do exactly this.

Is a fundamental set unique?

No. Any $n$ independent solutions form one. For y'' + y = 0, both ${\sin x, \cos x}$ and ${e^{i x}, e^{- i x}}$ and even ${\sin x, \sin x + \cos x}$ are fundamental sets. They describe the same two-dimensional solution space in different coordinates, exactly as different bases describe the same vector space.

Abel's Formula and Fundamental Sets of Solutions

Quick Reference

Abel's Formula

The All-or-Nothing Dichotomy

Worked Examples

Fundamental Sets of Solutions

Exercises

Frequently Asked Questions

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