Wronskians in Second Order ODE
In these notes we discuss the Wronskian of a second order, linear differential equation. We
start with the equation
y 00 + a(x)y 0 + b(x)y = 0. (1)
We can be slightly more general, and also consider equations of the form
c(x)y 00 + a(x)y 0 + b(x)y = 0,
but we recover (1) when we divide by c(x), and so (1) covers all the interesting cases we want to
examine.
Let y1 and y2 solve (1), and define
y1 (x) y2 (x)
W (x) = y1 (x)y20 (x) − y2 (x)y10 (x) = det .
y10 (x) y20 (x)
Observe that W 6= 0 precisely when the 2 × 2 matrix above is nonsingular, that is, precisely when
y1 and y2 are linearly independent solutions. We state this as a theorem.
Theorem 1 The Wronskian of two solutions of a differential equation is nonzero precisely when
those two functions are linearly independent.
Next, we find an equation for the Wronskian itself. Take a derivative:
W0 = (y1 y20 − y2 y10 )0 = y10 y20 + y1 y200 − y20 y10 − y2 y100 (2)
= y1 y200 − y2 y100 = y1 (−ay20 − by2 ) − y2 (−ay10 − by1 )
= −a(y1 y20 − y2 y10 ) = −aW,
or
W 0 + aW = 0.
This is a separable equation (or one we can solve using integrating factors, either way you get
the same answer), and we see Rx
W (x) = ce− 0 a(s)ds .
Observe that W is an exponential, so the only way it can vanish is if c = 0. We summarize this
in a theorem.
Theorem 2 The Wronskian W of y 00 + a(x)y 0 + b(x)y = 0 satisfies the equation
W 0 + a(x)W = 0,
and so Z x
W (x) = ce−p(x) , p(x) = a(s)ds
0
for some constant c determined by initial conditions. In particular, either W is never zero or it
is identically zero.
In fact, if we’re given one solution y1 (x) we can use the Wronskian to find a second, linearly
independent solution y2 (x) (and thus all possible solutions, by taking linear combinations). We
illustrate this with an example.
Consider the equation
y 00 − 3x2 y 0 − 6xy = 0, (3)
and verify that
3
y1 = ex
1
is a solution. We’d like to fine a second, linearly independent solution y2 (x). The Wronskian
W (x) = y1 y20 − y2 y10
satisfies the differential equation
d W0
W 0 − 3x2 W = 0 ⇒ (ln(W )) = = 3x2 . (4)
dx W
3
We can directly integrate this equation to get W (x) = cex ; in fact, we can rescale (multiplying
y2 by the appropriate constant) to make c = 1, and so
3 3
W = ex = y1 y20 − y2 y10 = ex (y20 − 3x2 y2 ) ⇒ y20 − 3x2 y2 = 1.
This last equation has the solution
Z x
3 3
y2 (x) = ex e−s ds + k ,
0
where k is a constant. It remains to find k, which we do by matching the initial conditions. We
have, using the fundamental theorem of calculus,
1 = W (0) = y1 (0)y20 (0) − y2 (0)y10 (0) = y20 (0) = e0 (e−0 + k) = 1 + k.
Thus k = 0 and Z x
x3 3
y2 (x) = e e−s ds.
0
I’ll end these notes with advice/remarks. In the last week or so, I’ve seen several students pull
out formulas for solutions to differential equations using several determinants. These formulas
involving determinants are essentially Cramer’s rule for the solution to a linear system of equation,
so they’re correct. However, let’s think about what you’re doing. You’re using a complicated
formula to solve a system of two linear equations with two unknows. Why? Why
are you making your life more complicated than you need to? Doing this goes against the most
fundamental thing I want to teach you, which is to
Think before you compute.
