1. Properties of Differential Operators
MATH 374 Lecture 14
Properties of Differential Operators

2. 4.9: Some Properties of Differential Operators
Theorem 4.7: Let
f(D) = a0Dn + a1Dn an-1D + an
be a polynomial in D. Then
(a) For any constant m, f(D)emx = f(m)emx.
(b) For any constant a, f(D-a)[eaxy] = eaxf(D)y. 2

3. Thm 4.7 (a) For any constant m, f(D)emx = f(m)emx.
Proof of Theorem 4.7 (a)
Claim: For any integer k  1, Dk emx = mk emx.
Proof of Claim: Use induction.
k = 1: D emx = m emx
Induction Hypothesis: Assume the claim is true for k  1, to show it is true for k+1.
Dk+1emx = D Dk emx = D(mk emx) = mk+1 emx.
Therefore the claim holds! It follows that
f(D) emx = (a0Dn + a1Dn an-1D + an)emx
= a0Dnemx + a1Dn-1emx an-1Demx + anemx
= a0mnemx + a1mn-1emx an-1memx + anemx
= f(m)emx  3

4. Thm 4.7 (b) For any constant a, f(D-a)[eaxy] = eaxf(D)y.
Proof of Theorem 4.7 (b)
Claim: For any integer k  1, (D-a)k[eaxy]=eaxDky.
Proof of Claim: Use induction.
k = 1: (D-a)[eaxy] = aeaxy + eaxDy – aeaxy = eaxDy.
Induction Hypothesis: Assume the claim is true for k  1 to show it is true for k+1.
(D-a)k+1[eaxy] = (D-a)(D-a)k[eaxy]
= (D-a)[eaxDky]
= eaxD(Dky)
= eax Dk+1y
Therefore the claim holds!
Think of this as the
function to the right
of eax. 4

5. Proof of Theorem 4.7 (b) (continued)
Thm 4.7 (b) For any constant a, f(D-a)[eaxy] = eaxf(D)y.
Proof of Theorem 4.7 (b) (continued)
It follows that
f(D-a)[eaxy] = (a0(D-a)n + … + an)[eaxy]
= a0(D-a)n[eaxy] + … + an[eaxy]
= a0eaxDny + … +aneaxy
= eax[a0Dny + … +any]
= eaxf(D)y.  5

6. Some Useful Results Corollary 1: If m is a root of polynomial equation
f(m) = a0mn + … + an-1m + an = 0,
then f(D)emx = 0.
Proof: From Theorem 4.7 (a),
f(D)emx = f(m)emx
= 0·emx = 0.  6

7. Some Useful Results
Corollary 2: For n a positive integer, m a constant, and k = 0, 1, 2, … , n-1,
(D-m)n[xkemx] = 0.
Proof: Apply Theorem 4.7 (b) with y = xk and f(D) = Dn.
It follows that (D-m)n[emxxk] = emxDn(xk) = 0,
for k = 0, 1, 2, … , n-1.  7

8. Example 1: Solve (D2-6D+5)y = 0. (1)
Solution: Here, f(D) = D2 – 6D + 5.
It follows from Corollary 1 that if m is a root of
f(m) = m2 – 6m + 5 = 0,
then y = emx is a solution of (1).
Since m2-6m+5 = (m-1)(m-5), roots of f(m) = 0 are m = 1 and m = 5.
Hence, y1 = ex and y2 = e5x are solutions of (1).
Because y1 and y2 are linearly independent (check), the general solution to (1) is:
y = c1ex + c2e5x,
where c1 and c2 are arbitrary constants! (See Theorem 4.4). 8