summarize all the important info i need to know
4.5 Connection to Linear Algebra
Recall that Ck(I) is the vector space of all functions with at least k continuous derivatives
on an interval I. Recall also that differentiation is a linear transformation.
Definition 4.5.1. A differential operator is the the linear transformation
D : C1(I) → C0(I)
D(f (x)) = f ′(x)
Dk : Ck → C0(I)
Dk(f (x)) = D(Dk−1(f (x))) = dkf
dxk .
An nth order linear differential operator
L = Dn + a1(x)Dn−1 + · · · + an−1(x)D + an
L(y) = y(n) + a1(x)y(n−1) + · · · + an−1(x)y′ + an(x)y
Namely, we can think of a linear differential equation as a sum of linear transforma-
tions applied to a function y = f (x).
Why would this be useful? Well, think of all we know about vector spaces and linear
transformations. If a linear diff eq is really the image of a linear transformation, then solving
a differential equation can be turned into a question about linear tranformations and their
images.
Example 4.5.2.
Find each of the following
(a) Find L(cos(x)) if L = D3 + 3xD − 2x Solution:
L(cos(x)) = D3(cos(x)) + 3xD(cos(x)) − 2x cos(x)
= sin(x) − 3x sin(x) − 2x cos(x)
(b) Find the kernel of L = D − 2x Solution:
L(y) = D(y) − 2xy = 0
y′ − 2xy = 0
dy
dx = 2xy
∫ dy
y =
∫
2xdx
150
ln |y| = x2 + c1
y = c2ex2
so
ker(L) = {cex2
| c ∈ R} = span{ex2
}.
(c) Find all solutions to Ly = 1
x where L = D + 1
x ln(x) . Solution:
1
x = Ly
=⇒ =
(
D + 1
x ln(x)
)
y
= y′ + 1
x ln(x)y
∫
p(x)dx =
∫
+ 1
x ln(x)dx =
∫ 1
udu u = ln(x), du = 1
x
= ln |u| = ln | ln(x)|
I(x) = e∫ p(x)dx = eln | ln(x)| = ln(x)
∫
I(x)q(x)dx =
∫ 1
x ln(x)dx =
∫
udu u = ln(x), du = 1
x
= u2 + c = (ln(x))2 + c
So a general solution is
y = 1
ln(x)((ln(x))2 + c) = ln(x) + c
ln(x).
Definition 4.5.3. A homogeneous differential equation is
y(n) + a1(x)y(n−1) + · · · + an(x)y = 0
⇐⇒ L(y) = Dn + a1(x)Dn−1 + · · · + an−1(x)D + an = 0
L(y) = 0
Theorem 4.5.4.
The set of all solution to L(y) = 0 on an interval I is an n-dimensional
vector space called the solution space.
151
Proof 4.5.5. We know the set S of all y ∈ Cn(I) so L(y) = 0 is the kernel of a linear
transformation so it must be a subspace of Cn(I), so specifically its a vector space.
Thus, we just need to show it has dimension n.
We will use Theorem 4.1.10 which says that an initial value problem has a unique
solution.
Fix some x0 ∈ I. Consider the following initial conditions.
y1 : y1(x0) = 1, y′
1(x0) = 0, · · · , y(n−1)
1 (x0) = 0
y2 : y2(x0) = 0, y′
2(x0) = 1, · · · , y(n−1)
2 (x0) = 0
...
yn : yn(x0) = 1, y′
n(x0) = 0, · · · , y(n−1)
n (x0) = 0
Then, the Wronskian
W (x0) =
∣
∣
∣
∣
∣
∣
∣
∣
∣
y1(x0) y2(x0) · · · yn(x0)
y′
1(x0) y2(x0) · · · y′
n(x0)
... ... . . . ...
y(n−1)
1 (x0) y(n−1)
2 (x0) · · · y(n−1)
n (x0)
∣
∣
∣
∣
∣
∣
∣
∣
∣
=
∣
∣
∣
∣
∣
∣
∣
∣
∣
1 0 · · · 0
0 1 · · · 0
... ... . . . ...
0 0 · · · 1
∣
∣
∣
∣
∣
∣
∣
∣
∣
= 1
since the Wronskian is nonzero at a point in the interval, the solutions are all li so
dim(S) ≥ n.
Now, let c0, c1, ..., cn−1 but any real numbers and let y = u(x) be the unique solution
coming from those initial conditions so
u(x0) = c0, u′(x0) = c1, · · · , u(n−1)(x0) = cn−1.
Let
w(x) = c0y1(x) + c1y2(x) + · · · + cn−1yn(x).
Certainly w(x) is a solution since it is a sum of solutions and
w(x0) = c0y1(x0) + c1y2(x0) + · · · + cn−1yn(x0) = c0
w′(x0) = c0y′
1(x0) + c1y′
2(x0) + · · · + cn−1y′
n(x0) = c1
...
w(n−1)(x0) = c0y(n−1)
1 (x0) + c1y(n−1)
2 (x0) + · · · + cn−1y(n−1)
n (x0) = cn−1
namely, w(x) has the same initial conditions as u(x) and both are solutions so by
uniquness, u(x) = w(x) = a linear combination of y1, ..., yn. So {y1(x), ..., yn(x)} form
a basis for S so dim(S) = n.
152
Theorem 4.5.6.
If y1, ..., yn are solutions to an nth order homogeneous diff eq L(y) = 0 on
some interval I, then
{y1, .., yn} li on I ⇐⇒ W [y1, ..., yn](x) ̸ = 0 for any x ∈ I.
This says, a set of solutions is li if and only if the Wronskian of those solutions
is nonzero everywhere.
Proof 4.5.7. First, suppose the Wronskian is nonzero at every point. Then because
it’s nonzero at at least one point x0 ∈ I the solutions are li.
Now, suppose the Wronskian is zero at some x0 ∈ I.
W (x0) =
∣
∣
∣
∣
∣
∣
∣
∣
∣
y1(x0) y2(x0) · · · yn(x0)
y′
1(x0) y2(x0) · · · y′
n(x0)
... ... . . . ...
y(n−1)
1 (x0) y(n−1)
2 (x0) · · · y(n−1)
n (x0)
∣
∣
∣
∣
∣
∣
∣
∣
∣
= 0
Recall the Wronskian is the determinant of an n × n matrix W (x0) so the Wronskian
being 0 at x0 means the matrix W (x0) is not invertible and so nullity(W (x0)) > 0. Let
⃗
c =





c1
c2
...
cn




 ∈ Null(W (x0)) =⇒ W (x0)⃗c =





c1y1(x0) + · · · + cnyn(x0)
c1y′
1(x0) + · · · + cny′
n(x0)
...
c1y(n−1)
1 (x0) + · · · + cny(n−1)
n (x0)




 =





0
0
...
0





Then let u(x) = c1y1(x) + · · · + cnyn(x) and consider the initial conditions of u(x) and
compare them to the Wonskian computation we just did
u(x0) = c1y1(x0) + · · · + cnyn(x0) = 0
u′(x0) = c1y′
1(x0) + · · · + cny′
n(x0) = 0
u(n−1)(x0) = c1y(n−1)
1 (x0) + · · · + cny(n−1)
n (x0) = 0
so u(x) is the unique solution with the initial conditions u(x0) = 0, u′(x0) =
0, ..., u(n−1)(x0) = 0 but these initial conditions are satisfied by the solution y = 0
so u(x) = 0. Namely, we’ve found a linear dependence
c1y1(x) + · · · + cnyn(x) = 0 =⇒ {y1, ..., yn} is ld.
Importantly, we review with the following notes.
153
Fact 4.5.8.
(1) If we are given a set of arbitrary functions {f1, ..., fn} then the Wronksian being
nonzero for some point x0 ∈ I means the set is li
(2) If we are given a set of arbitrary functions {f1, ..., fn} then the Wronksian being
zero for some point x0 ∈ I or zero everywhere on I means NOTHING.
(3) If we are given a set of solutions {y1, ..., yn} to a linear diff eq L(y) = 0, then the
Wronskian can only be entirely nonzero on I or entirely 0 on I and these two
states determine the linear Independence of the set.
Example 4.5.9.
Show that {e−3x, e5x} are two linearly independent solutions to L(y) = 0 where
L = D2 − 2D − 15.
Solution: First, we’ll show that they are solutions.
L(e−3x) = 9e−3x + 6e−3x − 15e−3x = 0
L(e5x) = 25e5x − 10e5x − 15e5x = 0
now we’ll verify they are li. Importantly, now that we know they are solutions to a
linear diff eq, we can plug in any real number before we take the derivate! Plugging
in x0 = 0 is generally the easiest!
W (x) =
∣
∣
∣
∣ e−3x e5x
−3e−3x 5e5x
∣
∣
∣
∣
W (0) =
∣
∣
∣
∣ 1 1
−3 5
∣
∣
∣
∣ = 5 + 3 ̸ = 0.
so they are li solutions.
Theorem 4.5.10.
If {y1, ..., yn} are li solutions to L(y) = 0 on an interval I, and yp is a
particular solution (any solution) to L(y) = F (x), then all solutions to L(y) =
F (x) are of the form
y(x) = c1y1(x) + · · · + cnyn(x) + yp(x)
for some constants c1, ..., cn.
154
Example 4.5.11.
Verify that yp(x) = 2e6x is a particular solution of L(y) = 45e6x where
L = D2 − 2D − 15.
Then write down a general solution for L(y) = 45e6x.
Solution: We’ve already seen that {e−3x, e5x} form a basis for the homogeneious solu-
tions space. Now let’s verify the particular solution
L(2e6x) = 72e6x − 12e6x − 15e6x = 45e6x
and so a general solution is
y(x) = c1e−3x + c2e5x + 2e6x.
155