Differential Equations—Final Exam Review
Format:
One-third of the final will cover chapters 1-4. For this part, study the midterms, review sheets,
handouts, and (if you have time) the homework. Section 4.1 will not be covered.
Two-thirds of the final will cover chapters 5-6. For this part, study this review sheet, the handouts,
and the homework. Section 6.5 will not be covered.
At the end of this review sheet is a three-page formula sheet for chapters 5-6 which will be given to
you for use during the final exam. I will also give you any partial fraction formulas you need to
compute inverse Laplace transforms. I will not give you Euler’s formula:
𝑒
𝑎+𝑖𝑏 𝑡
= 𝑒 𝑎𝑡 𝑒 𝑖(𝑏𝑡 ) = 𝑒 𝑎𝑡 cos 𝑏𝑡 + 𝑖 sin 𝑏𝑡 .
If there is one formula you choose to memorize for the rest of your lives, let it be Euler’s formula.
(Better yet, know how to prove it using Taylor series.) After all, this is where 𝑒 𝑖𝜋 + 1 = 0 comes
from, the formula relating the five most important numbers in mathematics!
Chapters 5-6
1) Compute the Laplace transforms of 𝑓 𝑡 = 𝑡 and of 𝑓 𝑡 = 𝑒 2𝑡 directly from the definition of
the Laplace transform as an integral.
2) (5.1 #33) Solve the IVP 𝑦 ′ + 𝑦 = 3𝑡 using the Laplace transform method. You may use the
following partial fractions formula when computing the inverse transform.
1
1
1
1
𝑏−𝑎
=
−
+
2
2
𝑠−𝑎 𝑠−𝑏
𝑏−𝑎 𝑠−𝑎 𝑠−𝑏
𝑠−𝑏 2
3) (6.3 #1,7) Find the general solution of the system
𝑥1′ = 5𝑥1 + 3𝑥2
𝑥2′ = −𝑥1 + 𝑥2
and then solve the IVP with 𝑥1 0 = 1, 𝑥2 0 = 1. What happens as 𝑡 → ∞?
4) (6.3 #31) Find the general solution of 𝒙′ = 𝐴𝒙 for the matrix
1 1
𝐴=
.
−1 3
5) (6.4 #7) Find the general real-valued solution of 𝒙′ = 𝐴𝒙 for the matrix
6 −1
𝐴=
5 2
−1
and then solve the IVP with 𝒙 0 =
.
1
6) (6.4 #22) Find all real-valued solutions of the second-order ODE 𝑦 ′′ + 2𝑦 ′ + 2𝑦 = 0
by converting to a system of linear ODEs and then solving.
7) Find all real-valued solutions of the vector ODE
0 −1
𝒙′ =
𝒙
1 0
by computing 𝑒 𝐴𝑡 using the Taylor series definition. The following two Taylor series may
come in handy:
𝑥2 𝑥4 𝑥6
cos 𝑥 = 1 − + − + ⋯ ,
2! 4! 6!
𝑥3 𝑥5 𝑥7
sin 𝑥 = 𝑥 − + − + ⋯.
3! 5! 7!
3
, compute 𝑒 𝐴𝑡 using a basic solution matrix 𝑋(𝑡). Then solve the
−1
1
IVP 𝒙′ = 𝐴𝒙, 𝒙 0 =
.
2
8) (6.6 #1) For 𝐴 =
1
1
Formula sheet for use on final exam
The Laplace transform:
function of 𝑡
function of 𝑠
𝐿∶ 𝑓
𝐿[𝑓]
∞
𝐿 𝑓 (𝑠) =
𝜏
𝑒 −𝑠𝑡 𝑓(𝑡) 𝑑𝑡 = lim
𝜏→∞ 0
0
𝑒 −𝑠𝑡 𝑓(𝑡) 𝑑𝑡
Polynomial-exponentials:
𝐿 𝑡𝑛 =
𝑛!
𝐿 𝑡 𝑛 𝑒 𝑎𝑡 =
𝑠 𝑛+1
𝑛!
𝑠−𝑎
𝑛+1
Derivatives:
𝐿 𝑦 ′ = 𝑠𝐿 𝑦 − 𝑦 0
𝐿 𝑦 ′′ = 𝑠 2 𝐿 𝑦 − 𝑠𝑦 0 − 𝑦 ′ 0
Linearity:
𝐿 𝛼𝑓 = 𝛼𝐿 𝑓
𝐿 𝑓+𝑔 = 𝐿 𝑓 +𝐿 𝑔
𝐿−1 𝛼𝑓 = 𝛼𝐿−1 𝑓
𝐿−1 𝑓 + 𝑔 = 𝐿−1 𝑓 + 𝐿−1 𝑔
Determinant (𝐝𝐞𝐭 𝑨):
det
𝑎
𝑐
𝑏
= 𝑎𝑑 − 𝑏𝑐
𝑑
Identity matrix (𝑰):
𝐼=
1 0
0 1
𝐼𝐴 = 𝐴
𝐴𝐼 = 𝐴
Inverse matrix (𝑨−𝟏 ):
𝐴=
𝑎
𝑐
𝑏
𝑑
𝐴−1 =
1
𝑑
det 𝐴 −𝑐
−𝑏
𝑎
Definition of eigenvalues 𝝀 and eigenvectors 𝒗: 𝐴𝒗 = 𝜆𝒗
Finding eigenvalues 𝝀: det 𝐴 − 𝜆𝐼 = 0
Finding eigenvectors 𝒗: 𝐴 − 𝜆𝐼 𝒗 = 𝟎
𝐴−1 𝐴 = 𝐼
𝐴𝐴−1 = 𝐼
𝒙′ = 𝐴𝒙
Vector ODE:
Theorem: the general solution 𝒙(𝑡)
Case 1) Two distinct real eigenvalues 𝜆1 , 𝜆2 , with 𝑉𝜆 𝑖 = Span 𝒗𝒊 .
𝒙 𝑡 = 𝐶1 𝑒 𝜆 1 𝑡 𝒗𝟏 + 𝐶2 𝑒 𝜆 2 𝑡 𝒗𝟐
Case 2a) One (real) eigenvalue 𝜆, with 𝑉𝜆 = Span(𝒗), and also 𝐴 − 𝜆𝐼 𝒖 = 𝒗.
Here, 𝜆 has a “deficient eigenspace” and we must use a generalized eigenvector 𝒖 along with the
eigenvector 𝒗.
𝒙 𝑡 = 𝐶1 𝑒 𝜆𝑡 𝒗 + 𝐶2 𝑒 𝜆𝑡 𝒖 + 𝑡𝒗
= 𝑒 𝜆𝑡 𝐶1 + 𝐶2 𝑡 𝒗 + 𝐶2 𝒖
Case 2b) One (real) eigenvalue 𝜆, with 𝑉𝜆 = ℝ2 .
In this case, any 𝒗 in ℝ2 will make 𝒙 𝑡 = 𝑒 𝜆𝑡 𝒗 a solution. We get all solutions by taking linear
1
0
combinations of the solutions with 𝒗 =
and with 𝒗 =
.
0
1
𝒙 𝑡 = 𝐶1 𝑒 𝜆𝑡
1
0
+ 𝐶2 𝑒 𝜆𝑡
𝜆𝑡
0
= 𝐶1 𝑒 +
𝜆𝑡
𝐶
0
2𝑒
𝐶 𝑒 𝜆𝑡
= 1 𝜆𝑡
𝐶2 𝑒
𝐶
= 𝑒 𝜆𝑡 1
𝐶2
0
1
Case 3) Two distinct complex (conjugate) eigenvalues 𝜆, 𝜆, with 𝑉𝜆 = Span(𝒗), 𝑉𝜆 = Span(𝒗).

The general complex-valued general solution of 𝒛′ = 𝐴𝒛 is as follows, where 𝐾1 , 𝐾2 , are any
complex scalars.
𝒛 𝑡 = 𝐾1 𝑒 𝜆𝑡 𝒗 + 𝐾2 𝑒 𝜆 𝑡 𝒗

The general real-valued general solution of 𝒙′ = 𝐴𝒙 is as follows, where 𝐶1 , 𝐶2 , are any real
scalars.
𝒙 𝑡 = 𝐶1 Re 𝑒 𝜆𝑡 𝒗 + 𝐶2 Im[𝑒 𝜆𝑡 𝒗]
𝒙′ = 𝐴𝒙
Vector ODE:
Theorem: the general solution 𝒙(𝑡)
𝒙 𝑡 = 𝑒 𝐴𝑡 𝑪,
𝑪=𝒙 0
1
1
𝐴𝑡 2 +
𝐴𝑡 3 + ⋯
2
3!
1
1
= 𝐼 + 𝐴𝑡 + 𝐴2 𝑡 2 + 𝐴3 𝑡 3 + ⋯
2
3!
𝑒 𝐴𝑡 = 𝐼 + 𝐴𝑡 +
Definition: A basic solution matrix 𝑋(𝑡) is a matrix (of functions of 𝑡) whose columns are
“independent” solutions to 𝒙′ = 𝐴𝒙.
𝐴
power
series
𝑋 𝑡 𝑋 −1 (0)
𝑒 𝐴𝑡
eigenvalues
eigenvectors
general solution
of 𝒙′ = 𝐴𝒙
𝑒 𝐴𝑡 𝑪
solution of IVP
𝒙 = 𝐴𝒙, 𝒙 0 = 𝑪
′