Differential Equations – MCS 2423
Dr. Wisam Bukaita
Differential Equations
Sec 4.4 – 4.5
Section 4.4 Undetermined Coefficients – Superposition Approach
Homogeneous Linear Equations with Constant Coefficient
To solve a nonhomogeneous linear DE
𝒂𝒏 ∙ 𝒚(𝒏) + 𝒂𝒏−𝟏 ∙ 𝒚(𝒏−𝟏) + ⋯ + 𝒂𝟏 ∙ 𝒚′ + 𝒂𝟎 𝒚 = 𝒈(𝒙)
we must do two things:
1. Find the complementary function 𝒚𝒄 (the solution of the homogeneous DE)
2. Find any particular solution 𝒚𝒑 of the nonhomogeneous DE.
Then the general solution of the DE is
𝒚 = 𝒚𝒄 + 𝒚𝒑
Method of Undetermined Coefficient
The general method is limited to linear DEs where
• The coefficients 𝒂𝒊 , 𝒊 = 𝟎, 𝟏, 𝟐, … , 𝒏 are constants and
• 𝒈(𝒙) is a constant k, a polynomial function, an exponential function 𝒆𝜶𝒙 , a sine or cosine
function 𝒔𝒊𝒏𝜷𝒙 𝒐𝒓 𝒄𝒐𝒔𝜷𝒙, or finite sums and products of these functions.
The idea behind this method is a conjecture about the form of 𝒚𝒑 which is the particular
(𝒏)
(𝒏−𝟏)
solution of DE. Because the linear combination of derivatives 𝒂𝒏 𝒚𝒑 + 𝒂𝒏−𝟏 𝒚𝒑
+⋯+
′
𝒂𝟏 𝒚𝒑 + 𝒂𝟎 𝒚𝒑 must be identical to 𝒈(𝒙), it seems reasonable to assume that 𝒚𝒑 has the same
form as 𝒈(𝒙). There are two cases:
Case 1: No function in the assumed particular solution 𝒚𝒑 is a solution of the associated
homogeneous DE.
1
2
3
4
5
6
7
8
9
10
11
12
𝒈(𝒙)
𝒂𝒏𝒚 𝒄𝒐𝒏𝒔𝒕𝒂𝒏𝒕
𝟐𝒙 − 𝟑
𝟐
𝟕𝒙 − 𝟐𝒙 + 𝟓
𝟔𝒙𝟑 − 𝒙 + 𝟗
𝟑𝒔𝒊𝒏𝟒𝒙
𝟕𝒄𝒐𝒔𝟐𝒙
𝟑𝒆𝟓𝒙
(𝟖𝒙 − 𝟓)𝒆𝟓𝒙
𝟑𝒙𝟐 𝒆𝟓𝒙
𝟔𝒆𝟐𝒙 𝒔𝒊𝒏𝟒𝒙
𝟕𝒙𝟐 𝒔𝒊𝒏𝟒𝒙
𝟖𝒙𝒆𝟑𝒙 𝒄𝒐𝒔𝟐𝒙
𝑭𝒐𝒓𝒎 𝒐𝒇 𝒚𝒑
𝑨
𝑨𝒙 + 𝑩
𝟐
𝑨𝒙 + 𝑩𝒙 + 𝑪
𝑨𝒙𝟑 + 𝑩𝒙𝟐 + 𝑪𝒙 + 𝑬
𝑨𝒄𝒐𝒔𝟒𝒙 + 𝑩𝒔𝒊𝒏𝟒𝒙
𝑨𝒄𝒐𝒔𝟐𝒙 + 𝑩𝒔𝒊𝒏𝟐𝒙
𝑨𝒆𝟓𝒙
(𝑨𝒙 + 𝑩)𝒆𝟓𝒙
(𝑨𝒙𝟐 + 𝑩𝒙 + 𝑪)𝒆𝟓𝒙
𝑨𝒆𝟐𝒙 𝒄𝒐𝒔𝟒𝒙 + 𝑩𝒆𝟐𝒙 𝒔𝒊𝒏𝟒𝒙
(𝑨𝒙𝟐 + 𝑩𝒙 + 𝑪)𝒄𝒐𝒔𝟒𝒙 + (𝑬𝒙𝟐 + 𝑭𝒙 + 𝑮)𝒔𝒊𝒏𝟒𝒙
(𝑨𝒙 + 𝑩)𝒆𝟑𝒙 𝒄𝒐𝒔𝟒𝒙 + (𝑪𝒙 + 𝑬)𝒆𝟑𝒙 𝒔𝒊𝒏𝟒𝒙
1
Differential Equations – MCS 2423
Dr. Wisam Bukaita
Differential Equations
Sec 4.4 – 4.5
If 𝒈(𝒙) consists of a sum of, say, m terms of the kind listed in the table, then the assumption
for a particular solution 𝒚𝒑 consists of the sum of the trial forms 𝒚𝒑𝟏 , 𝒚𝒑𝟐 , … , 𝒚𝒑𝒎
corresponding to these terms:
𝒚𝒑 = 𝒚𝒑𝟏 + 𝒚𝒑𝟐 + … + 𝒚𝒑𝒎
Case 2: A function in the assumed particular solution is also a solution of the associated
homogeneous differential equation.
Multiplication Rule for Case 2: If any 𝒚𝒑𝒊 contains terms that duplicate terms in 𝒚𝒄 , then that
𝒚𝒑𝒊 must be multiplied by 𝒙𝒏 , where n is the smallest positive integer that eliminated that
duplication.
Example 1: Solve the given differential equation by undetermined coefficients.
𝒚′′ − 𝟖𝒚′ + 𝟐𝟎𝒚 = 𝟏𝟎𝟎𝒙𝟐 − 𝟐𝟔𝒙𝒆𝒙
Solution:
Step1: We first solve the associated homogeneous equation
𝒚′′ − 𝟖𝒚′ + 𝟐𝟎𝒚 = 𝟎
From the auxiliary equation
𝒎𝟐 − 𝟖𝒎 + 𝟐𝟎 = 𝟎
we find the roots
−(−𝟖) ± √(−𝟖)𝟐 − 𝟒 ∙ 𝟏 ∙ (𝟐𝟎) 𝟖 ± √𝟔𝟒 − 𝟖𝟎 𝟖 ± √−𝟏𝟔 𝟖 ± 𝟒𝒊
𝒎𝟏,𝟐 =
=
=
=
𝟐∙𝟏
𝟐
𝟐
𝟐
𝒎𝟏,𝟐 = 𝟒 ± 𝟐𝒊 𝒄𝒐𝒎𝒑𝒍𝒆𝒙 𝒄𝒐𝒏𝒋𝒖𝒈𝒂𝒕𝒆 𝒓𝒐𝒐𝒕𝒔
Hence the complementary function is
𝒚𝒄 = 𝒆 𝟒𝒙 (𝒄𝟏 𝒄𝒐𝒔𝟐𝒙 + 𝒄𝟐 𝒔𝒊𝒏𝟐𝒙)
Step2: Now, we assume that a particular solution is in the form of 𝒈(𝒙):
𝐂𝐨𝐫𝐫𝐞𝐬𝐩𝐨𝐧𝐝𝐢𝐧𝐠 𝐭𝐨 𝟏𝟎𝟎𝒙𝟐 𝒘𝒆 𝒂𝒔𝒔𝒖𝒎𝒆 𝒚𝒑𝟏 = 𝑨𝒙𝟐 + 𝑩𝒙 + 𝑪
𝐂𝐨𝐫𝐫𝐞𝐬𝐩𝐨𝐧𝐝𝐢𝐧𝐠 𝐭𝐨 − 𝟐𝟔𝒙𝒆𝒙 𝒘𝒆 𝒂𝒔𝒔𝒖𝒎𝒆 𝒚𝒑𝟐 = (𝑬𝒙 + 𝑭)𝒆𝒙
The assumption for the particular solution is
𝒚𝒑 = 𝒚𝒑𝟏 + 𝒚𝒑𝟐 = 𝑨𝒙𝟐 + 𝑩𝒙 + 𝑪 + (𝑬𝒙 + 𝑭)𝒆𝒙
No term in this assumption duplicates a term in 𝒚𝒄 .
We seek to determine specific coefficients 𝑨, 𝑩, 𝑪, 𝑬, 𝒂𝒏𝒅 𝑭 for which 𝒚𝒑 is a solution of
nonhomogeneous DE. Substituting 𝒚𝒑 and the derivatives
𝒚′𝒑 = 𝟐𝑨𝒙 + 𝑩 + 𝑬𝒆𝒙 + (𝑬𝒙 + 𝑭)𝒆𝒙 = 𝟐𝑨𝒙 + 𝑩 + (𝑬𝒙 + 𝑬 + 𝑭)𝒆𝒙
𝒙
𝒙
𝒙
𝒚′′
𝒑 = 𝟐𝑨 + 𝑬𝒆 + (𝑬𝒙 + 𝑬 + 𝑭)𝒆 = 𝟐𝑨 + (𝑬𝒙 + 𝟐𝑬 + 𝑭)𝒆
Into the given nonhomogeneous DE, we get
𝟐𝑨 + (𝑬𝒙 + 𝟐𝑬 + 𝑭)𝒆𝒙 − 𝟖(𝟐𝑨𝒙 + 𝑩 + (𝑬𝒙 + 𝑬 + 𝑭)𝒆𝒙 ) + 𝟐𝟎 (𝑨𝒙𝟐 + 𝑩𝒙 + 𝑪 + (𝑬𝒙 + 𝑭)) 𝒆𝒙 =
= 𝟏𝟎𝟎𝒙𝟐 − 𝟐𝟔𝒙𝒆𝒙
𝟐𝑨 + (𝑬𝒙 + 𝟐𝑬 + 𝑭)𝒆𝒙 − 𝟏𝟔𝑨𝒙 − 𝟖𝑩 − 𝟖(𝑬𝒙 + 𝑬 + 𝑭)𝒆𝒙 + 𝟐𝟎𝑨𝒙𝟐 + 𝟐𝟎𝑩𝒙 + 𝟐𝟎𝑪 + 𝟐𝟎(𝑬𝒙 + 𝑭)𝒆𝒙 =
= 𝟏𝟎𝟎𝒙𝟐 − 𝟐𝟔𝒙𝒆𝒙
Because the last equation is supposed to be an identity, the coefficients of like powers of x
must be equal:
2
Differential Equations – MCS 2423
Dr. Wisam Bukaita
Differential Equations
Sec 4.4 – 4.5
𝟐𝑨 − 𝟖𝑩 + 𝟐𝟎𝑪 = 𝟎
−𝟏𝟔𝑨 + 𝟐𝟎𝑩 = 𝟎
{
𝟐𝟎𝑨 = 𝟏𝟎𝟎
𝑬𝒙 + 𝟐𝑬 + 𝑭 − 𝟖(𝑬𝒙 + 𝑬 + 𝑭) + 𝟐𝟎(𝑬𝒙 + 𝑭) = −𝟐𝟔𝒙
𝑨 = 𝟓 ➔ −𝟖𝟎 + 𝟐𝟎𝑩 = 𝟎
𝟏𝟑𝑬 = −𝟐𝟔
➔
➔
𝑬 = −𝟐 ➔
𝑩=𝟒
⇔
𝟐𝑨 − 𝟖𝑩 + 𝟐𝟎𝑪 = 𝟎
−𝟏𝟔𝑨 + 𝟐𝟎𝑩 = 𝟎
𝑨=𝟓
𝟐𝑬 + 𝑭 − 𝟖(𝑬 + 𝑭) + 𝟐𝟎𝑭 = 𝟎
𝑬 − 𝟖𝑬 + 𝟐𝟎𝑬 = −𝟐𝟔
{
𝟏𝟎 − 𝟑𝟐 + 𝟐𝟎𝑪 = 𝟎
−𝟔𝑬 + 𝟏𝟑𝑭 = 𝟎 ➔
➔
𝟏𝟐 + 𝟏𝟑𝑭 = 𝟎
𝑨=𝟓
𝑩=𝟒
𝑨=𝟓
𝑭 = −𝟏𝟐/𝟏𝟑
{ 𝑬 = −𝟐
Thus, a particular solution is
𝟏𝟏
𝟏𝟐
𝒚𝒑 = 𝟓𝒙𝟐 + 𝟒𝒙 +
+ (−𝟐𝒙 − ) 𝒆𝒙
𝟏𝟎
𝟏𝟑
Step3: The general solution of the given equation is
𝒚 = 𝒚𝒄 + 𝒚𝒑
𝟏𝟏
𝟏𝟐
𝒚 = 𝒆 𝟒𝒙 (𝒄𝟏 𝒄𝒐𝒔𝟐𝒙 + 𝒄𝟐 𝒔𝒊𝒏𝟐𝒙) + 𝟓𝒙𝟐 + 𝟒𝒙 +
+ (−𝟐𝒙 − ) 𝒆𝒙
𝟏𝟎
𝟏𝟑
Example 2: Solve the given initial – value problem.
𝒚′′ + 𝟒𝒚′ + 𝟒𝒚 = (𝟑 + 𝒙)𝒆−𝟐𝒙 , 𝒚(𝟎) = 𝟐, 𝒚′(𝟎) = 𝟓
Step1: We first solve the associated homogeneous equation
𝒚′′ + 𝟒𝒚′ + 𝟒𝒚 = 𝟎
From the auxiliary equation
𝒎𝟐 + 𝟒𝒎 + 𝟒 = 𝟎
we find the roots
(𝒎 + 𝟐)𝟐 = 𝟎
𝒎+𝟐=𝟎
𝒎𝟏,𝟐 = −𝟐 𝒓𝒆𝒂𝒍 𝒓𝒐𝒐𝒕𝒔 𝒘𝒊𝒕𝒉 𝒎𝒖𝒍𝒕𝒊𝒑𝒍𝒊𝒄𝒊𝒕𝒚 𝒌 = 𝟐
Hence the complementary function is
𝒚𝒄 = 𝒄𝟏 𝒆−𝟐𝒙 + 𝒄𝟐 𝒙𝒆−𝟐𝒙
Step2: Now, we assume that a particular solution is in the form of 𝒈(𝒙):
𝐂𝐨𝐫𝐫𝐞𝐬𝐩𝐨𝐧𝐝𝐢𝐧𝐠 𝐭𝐨 (𝟑 + 𝒙)𝒆−𝟐𝒙 𝒘𝒆 𝒂𝒔𝒔𝒖𝒎𝒆 𝒚𝒑 = (𝑨𝒙 + 𝑩)𝒆−𝟐𝒙
The assumption for the particular solution is
𝒚𝒑 = (𝑨𝒙 + 𝑩)𝒆−𝟐𝒙
3
𝑪 = 𝟏𝟏/𝟏𝟎
➔ 𝑭 = −𝟏𝟐/𝟏𝟑
Differential Equations – MCS 2423
Dr. Wisam Bukaita
Differential Equations
Sec 4.4 – 4.5
Comparing 𝒚𝒄 with our normal assumption for a particular solution 𝒚𝒑 we see that the
duplication between 𝒚𝒄 and 𝒚𝒑 are eliminated when 𝒚𝒑 is multiplied by 𝒙𝟐 . Thus the correct
assumption for a particular solution is
𝒚𝒑 = 𝒙𝟐 (𝑨𝒙 + 𝑩)𝒆−𝟐𝒙 = (𝑨𝒙𝟑 + 𝑩𝒙𝟐 )𝒆−𝟐𝒙
No term in this assumption duplicates a term in 𝒚𝒄 .
We seek to determine specific coefficients 𝑨, 𝒂𝒏𝒅 𝑩 for which 𝒚𝒑 is a solution of
nonhomogeneous DE. Substituting 𝒚𝒑 and the derivatives
𝒚′𝒑 = (𝟑𝑨𝒙𝟐 + 𝟐𝑩𝒙)𝒆−𝟐𝒙 − 𝟐(𝑨𝒙𝟑 + 𝑩𝒙𝟐 )𝒆−𝟐𝒙 = (−𝟐𝑨𝒙𝟑 + 𝟑𝑨𝒙𝟐 − 𝟐𝑩𝒙𝟐 + 𝟐𝑩𝒙)𝒆−𝟐𝒙
𝟐
−𝟐𝒙
𝒚′′
− 𝟐(−𝟐𝑨𝒙𝟑 + 𝟑𝑨𝒙𝟐 − 𝟐𝑩𝒙𝟐 + 𝟐𝑩𝒙)𝒆−𝟐𝒙
𝒑 = (−𝟔𝑨𝒙 + 𝟔𝑨𝒙 − 𝟒𝑩𝒙 + 𝟐𝑩)𝒆
𝟐
−𝟐𝒙
𝒚′′
+ (𝟒𝑨𝒙𝟑 − 𝟔𝑨𝒙𝟐 + 𝟒𝑩𝒙𝟐 − 𝟒𝑩𝒙)𝒆−𝟐𝒙
𝒑 = (−𝟔𝑨𝒙 + 𝟔𝑨𝒙 − 𝟒𝑩𝒙 + 𝟐𝑩)𝒆
𝟑
𝟐
𝟐
−𝟐𝒙
𝒚′′
𝒑 = (𝟒𝑨𝒙 − 𝟏𝟐𝑨𝒙 + 𝟒𝑩𝒙 + 𝟔𝑨𝒙 − 𝟖𝑩𝒙 + 𝟐𝑩)𝒆
Into the given nonhomogeneous DE, we get
(𝟒𝑨𝒙𝟑 − 𝟏𝟐𝑨𝒙𝟐 + 𝟒𝑩𝒙𝟐 + 𝟔𝑨𝒙 − 𝟖𝑩𝒙 + 𝟐𝑩)𝒆−𝟐𝒙 + 𝟒(−𝟐𝑨𝒙𝟑 + 𝟑𝑨𝒙𝟐 − 𝟐𝑩𝒙𝟐 + 𝟐𝑩𝒙)𝒆−𝟐𝒙
+𝟒(𝑨𝒙𝟑 + 𝑩𝒙𝟐 )𝒆−𝟐𝒙 = (𝟑 + 𝒙)𝒆−𝟐𝒙
𝟑
𝟐
𝟐
(𝟒𝑨𝒙 − 𝟏𝟐𝑨𝒙 + 𝟒𝑩𝒙 + 𝟔𝑨𝒙 − 𝟖𝑩𝒙 + 𝟐𝑩) + 𝟒(−𝟐𝑨𝒙𝟑 + 𝟑𝑨𝒙𝟐 − 𝟐𝑩𝒙𝟐 + 𝟐𝑩𝒙)𝒆−𝟐𝒙
+𝟒(𝑨𝒙𝟑 + 𝑩𝒙𝟐 ) = (𝟑 + 𝒙)
Because the last equation is supposed to be an identity, the coefficients of like powers of x
must be equal:
𝟎=𝟎
𝟎=𝟎
𝟒𝑨 − 𝟖𝑨 + 𝟒𝑨 = 𝟎
𝟎=𝟎
𝟎
=
𝟎
{−𝟏𝟐𝑨 + 𝟒𝑩 + 𝟏𝟐𝑨 − 𝟖𝑩 + 𝟒𝑩 = 𝟎 ⇔ { 𝟔𝑨 = 𝟏 ⇔ { 𝑨 = 𝟏/𝟔
𝟔𝑨 − 𝟖𝑩 + 𝟖𝑩 = 𝟏
𝑩 = 𝟑/𝟐
𝑩 = 𝟑/𝟐
𝟐𝑩 = 𝟑
Thus, a particular solution is
𝟏
𝟑
𝒚𝒑 = ( 𝒙𝟑 + 𝒙𝟐 ) 𝒆−𝟐𝒙
𝟔
𝟐
Step3: The general solution of the given equation is
𝒚 = 𝒚𝒄 + 𝒚𝒑
𝟏
𝟑
𝒚 = 𝒄𝟏 𝒆−𝟐𝒙 + 𝒄𝟐 𝒙𝒆−𝟐𝒙 + ( 𝒙𝟑 + 𝒙𝟐 ) 𝒆−𝟐𝒙
𝟔
𝟐
In order to use initial conditions we need the derivative of the solution.
𝟏
𝟏
𝟑
𝒚′ = −𝟐𝒄𝟏 𝒆−𝟐𝒙 + 𝒄𝟐 𝒆−𝟐𝒙 − 𝒄𝟐 𝒙𝒆−𝟐𝒙 + ( 𝒙𝟐 + 𝟑𝒙) 𝒆−𝟐𝒙 − 𝟐 ( 𝒙𝟑 + 𝒙𝟐 ) 𝒆−𝟐𝒙
𝟐
𝟔
𝟐
Use the initial conditions 𝒚(𝟎) = 𝟐, 𝒚′(𝟎) = 𝟓 to the general solution of the equation.
𝟏 𝟑 𝟑 𝟐 𝟎
𝟎
𝟎
𝒄
𝒆
+
𝒄
∙
𝟎
∙
𝒆
+
(
∙𝟎 + ∙𝟎 )𝒆 = 𝟐
𝟏
𝟐
𝒚(𝟎) = 𝟐
𝟔
𝟐
{
⇔ {
𝟏
𝟏
𝟑
𝒚′(𝟎) = 𝟓
−𝟐𝒄𝟏 𝒆𝟎 + 𝒄𝟐 𝒆𝟎 − 𝒄𝟐 ∙ 𝟎 ∙ 𝒆𝟎 + ( ∙ 𝟎𝟐 + 𝟑 ∙ 𝟎) 𝒆𝟎 − 𝟐 ( ∙ 𝟎𝟑 + ∙ 𝟎𝟐 ) 𝒆𝟎 = 𝟓
𝟐
𝟔
𝟐
{
𝒄𝟏 = 𝟐
−𝟐𝒄𝟏 + 𝒄𝟐 = 𝟓
𝒄𝟏 = 𝟐
⇔ {
−𝟒 + 𝒄𝟐 = 𝟓
4
⇔ {
𝒄𝟏 = 𝟐
𝒄𝟐 = 𝟗
Differential Equations – MCS 2423
Dr. Wisam Bukaita
Differential Equations
Sec 4.4 – 4.5
The solution of the initial – value problem is
𝟏
𝟑
𝒚 = 𝟐𝒆−𝟐𝒙 + 𝟗𝒙𝒆−𝟐𝒙 + ( 𝒙𝟑 + 𝒙𝟐 ) 𝒆−𝟐𝒙
𝟔
𝟐
Example 3:
Solution:
Sec 4.4
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5
Differential Equations – MCS 2423
Dr. Wisam Bukaita
Differential Equations
Sec 4.4 – 4.5
Section 4.5 Undetermined Coefficients –Annihilator Approach
Introduction
An nth–order differential equation
𝒂𝒏 ∙ 𝒚(𝒏) + 𝒂𝒏−𝟏 ∙ 𝒚(𝒏−𝟏) + ⋯ + 𝒂𝟏 ∙ 𝒚′ + 𝒂𝟎 𝒚 = 𝒈(𝒙)
can be written
𝒂𝒏 𝑫𝒏 𝒚 + 𝒂𝒏−𝟏 𝑫𝒏−𝟏 𝒚 + ⋯ + 𝒂𝟏 𝑫𝒚 + 𝒂𝟎 𝒚 = 𝒈(𝒙)
where
𝒅𝒌 𝒚
𝒌
𝑫 =
, 𝒌 = 𝟎, 𝟏, 𝟐, … , 𝒏
𝒅𝒙𝒌
When it suits our purpose, it is also written as 𝑳(𝒚) = 𝒈(𝒙) where L denotes the linear nth –
order differential operator
(𝟐)
𝒂𝒏 𝑫𝒏 𝒚 + 𝒂𝒏−𝟏 𝑫𝒏−𝟏 𝒚 + ⋯ + 𝒂𝟏 𝑫𝒚 + 𝒂𝟎 𝒚
Factoring Operators
When the coefficients 𝒂𝒊 , 𝒊 = 𝟎, 𝟏, 𝟐, … , 𝒏 are real constants, a linear differential operator
can be factored whenever the characteristic polynomial
𝒂𝒏 𝒎𝒏 + 𝒂𝒏−𝟏 𝒎𝒏−𝟏 + ⋯ + 𝒂𝟏 𝒎 + 𝒂𝟎
In other words, if 𝒎𝟏 is a root of the auxiliary equation
𝒂𝒏 𝒎𝒏 + 𝒂𝒏−𝟏 𝒎𝒏−𝟏 + ⋯ + 𝒂𝟏 𝒎 + 𝒂𝟎 = 𝟎,
then 𝑳 = (𝑫 − 𝒎𝟏 )𝑷(𝑫), where the polynomial expression 𝑷(𝑫) is a linear differential
operator of order 𝒏 − 𝟏. For example, a differential equation such as
𝒚′′ + 𝟒𝒚′ + 𝟒𝒚 = 𝟎
can be written as
(𝑫𝟐 + 𝟒𝑫 + 𝟒)𝒚 = 𝟎 𝒐𝒓 (𝑫 + 𝟐)(𝑫 + 𝟐)𝒚 = 𝟎 𝒐𝒓 (𝑫 + 𝟐)𝟐 𝒚 = 𝟎
Example 1: Write the given Differential equation in the form 𝑳(𝒚) = 𝒈(𝒙), where L is a linear
differential operator with constant coefficients. If possible, factor L.
𝒚′′ − 𝟓𝒚 = 𝒙𝟐 − 𝟐𝒙
Solution:
𝑫𝟐 𝒚 − 𝟓𝒚 = 𝒙𝟐 − 𝟐𝒙
(𝑫𝟐 − 𝟓)𝒚 = 𝒙𝟐 − 𝟐𝒙
(𝑫 − √𝟓)(𝑫 + √𝟓)𝒚 = 𝒙𝟐 − 𝟐𝒙
Example 2:Write the given Differential equation in the form 𝑳(𝒚) = 𝒈(𝒙), where L is a linear
differential operator with constant coefficients. If possible, factor L.
𝒚′′′ + 𝟒𝒚′ = 𝒆𝒙 𝒄𝒐𝒔𝟐𝒙
Solution:
𝑫𝟑 𝒚 + 𝟒𝑫𝒚 = 𝒆𝒙 𝒄𝒐𝒔𝟐𝒙
(𝑫𝟑 + 𝟒𝑫)𝒚 = 𝒆𝒙 𝒄𝒐𝒔𝟐𝒙
𝑫(𝑫𝟐 + 𝟒)𝒚 = 𝒆𝒙 𝒄𝒐𝒔𝟐𝒙
6
Differential Equations – MCS 2423
Sec 4.4 – 4.5
Dr. Wisam Bukaita
Differential Equations
Annihilator Operators
If L is a linear differential operator with constant coefficients and f is a sufficiently
differentiable function such that 𝑳(𝒇(𝒙)) = 𝟎, then L is said to be an annihilator of the
function. For example,
• The differential operator 𝑫𝒏 annihilates each of the functions
𝟏, 𝒙, 𝒙𝟐 , … , 𝒙𝒏−𝟏
• The differential operator (𝑫 − 𝜶)𝒏 annihilates each of the functions
𝒆𝜶𝒙 , 𝒙𝒆𝜶𝒙 , 𝒙𝟐 𝒆𝜶𝒙 , … , 𝒙𝒏−𝟏 𝒆𝜶𝒙
• The differential operator [𝑫𝟐 − 𝟐𝜶𝑫 + (𝜶𝟐 + 𝜷𝟐 )]𝒏 annihilates each of the functions
𝒆𝜶𝒙 𝒄𝒐𝒔𝜷𝒙, 𝒙𝒆𝜶𝒙 𝒄𝒐𝒔𝜷𝒙, 𝒙𝟐 𝒆𝜶𝒙 𝒄𝒐𝒔𝜷𝒙, … , 𝒙𝒏−𝟏 𝒆𝜶𝒙 𝒄𝒐𝒔𝜷𝒙
𝒆𝜶𝒙 𝒔𝒊𝒏𝜷𝒙, 𝒙𝒆𝜶𝒙 𝒔𝒊𝒏𝜷𝒙, 𝒙𝟐 𝒆𝜶𝒙 𝒔𝒊𝒏𝜷𝒙, … , 𝒙𝒏−𝟏 𝒆𝜶𝒙 𝒔𝒊𝒏𝜷𝒙
Example 3: Verify that the given differential operator annihilates the indicated functions.
𝟐𝑫 − 𝟏; 𝒚 = 𝟒𝒆𝒙/𝟐
Solution:
(𝟐𝑫 − 𝟏)𝒚 = 𝟎 ?
(𝟐𝑫 − 𝟏)𝟒𝒆𝒙/𝟐 = 𝟎
𝟐𝑫𝟒𝒆𝒙/𝟐 − 𝟒𝒆𝒙/𝟐 = 𝟎
𝟖𝑫𝒆𝒙/𝟐 − 𝟒𝒆𝒙/𝟐 = 𝟎
𝟏
𝟖 ∙ 𝒆𝒙/𝟐 − 𝟒𝒆𝒙/𝟐 = 𝟎
𝟐
𝟒𝒆𝒙/𝟐 − 𝟒𝒆𝒙/𝟐 = 𝟎
Example 4: Find a linear differential operator that annihilates the given function.
𝒙𝟑 (𝟏 − 𝟓𝒙) = 𝒙𝟑 − 𝟓𝒙𝟒
Solution:
Annihilator: 𝑫𝟒 𝒃𝒆𝒄𝒂𝒖𝒔𝒆 𝒐𝒇 𝒙𝟑 𝒏 − 𝟏 = 𝟑
𝑫𝟓 𝒃𝒆𝒄𝒂𝒖𝒔𝒆 𝒐𝒇 − 𝟓𝒙𝟒 𝒏 − 𝟏 = 𝟒
The highest degree of operator will annihilate the sum:
𝑫𝟓 (𝒙𝟑 − 𝟓𝒙𝟒 ) = 𝟎
Example 5: Find a linear differential operator that annihilates the given function.
𝒙 + 𝟑𝒙𝒆𝟔𝒙
Solution:
Annihilator:𝑫𝟐 𝒃𝒆𝒄𝒂𝒖𝒔𝒆 𝒐𝒇 𝒙, 𝑫𝟐 𝒙 = 𝟎
(𝑫 − 𝟔)𝟐 𝒃𝒆𝒄𝒂𝒖𝒔𝒆 𝒐𝒇 𝒙𝒆𝟔𝒙 : 𝒏 − 𝟏 = 𝟏, 𝒂𝒏𝒅 𝜶 = 𝟔
Then the product will annihilate the sum:
𝑫𝟐 (𝑫 − 𝟔)𝟐 (𝒙 + 𝟑𝒙𝒆𝟔𝒙 ) = 𝟎
7
Differential Equations – MCS 2423
Sec 4.4 – 4.5
Dr. Wisam Bukaita
Differential Equations
Example 6: Find a linear differential operator that annihilates the given function.
𝟖𝒙 − 𝒔𝒊𝒏𝒙 + 𝟏𝟎𝒄𝒐𝒔𝟓𝒙
Solution:
Annihilator:𝑫𝟐 𝒃𝒆𝒄𝒂𝒖𝒔𝒆 𝒐𝒇 𝒙, 𝑫𝟐 𝒙 = 𝟎
(𝑫𝟐 + 𝟏) 𝒃𝒆𝒄𝒂𝒖𝒔𝒆 𝒐𝒇 𝒔𝒊𝒏𝒙: 𝒏 − 𝟏 = 𝟎, 𝜶 = 𝟎, 𝒂𝒏𝒅 𝜷 = 𝟏
(𝑫𝟐 + 𝟐𝟓) 𝒃𝒆𝒄𝒂𝒖𝒔𝒆 𝒐𝒇 𝟏𝟎𝒄𝒐𝒔𝟓𝒙: 𝒏 − 𝟏 = 𝟎, 𝜶 = 𝟎, 𝒂𝒏𝒅 𝜷 = 𝟓
Then the product will annihilate the sum:
𝑫𝟐 (𝑫𝟐 + 𝟏)(𝑫𝟐 + 𝟐𝟓)(𝟖𝒙 − 𝒔𝒊𝒏𝒙 + 𝟏𝟎𝒄𝒐𝒔𝟓𝒙) = 𝟎
Example 7: Find linearly independent functions that are annihilated by the given differential
operator.
𝑫𝟐 + 𝟒𝑫
Solution:
𝑫𝟐 + 𝟒𝑫 = 𝑫(𝑫 + 𝟒)
The functions are 1, and 𝒆−𝟒𝒙
Undetermined Coefficient – Annihilator Approach
Suppose that 𝑳(𝒚) = 𝒈(𝒙) is a linear differential equation with constant coefficients and that
the input 𝒈(𝒙) is a linear combination of functions of the form
𝒌 (𝒄𝒐𝒏𝒔𝒕𝒂𝒏𝒕), 𝒙𝒎 , 𝒙𝒎 𝒆𝜶𝒙 , 𝒙𝒎 𝒆𝜶𝒙 𝒄𝒐𝒔𝜷𝒙, 𝒂𝒏𝒅 𝒙𝒎 𝒆𝜶𝒙 𝒔𝒊𝒏𝜷𝒙
where m is a nonnegative integer and 𝜶 𝒂𝒏𝒅 𝜷 are real numbers. We know that such a function
𝒈(𝒙) can be annihilated by a differential operator 𝑳𝟏 of lowest order, consisting of a product of the
operators 𝑫𝒏 , (𝑫 − 𝜶)𝒏 , 𝒂𝒏𝒅 [𝑫𝟐 − 𝟐𝜶𝑫 + (𝜶𝟐 + 𝜷𝟐 )]𝒏 .
Applying 𝑳𝟏 to both sides of the equation 𝑳(𝒚) = 𝒈(𝒙) yields 𝑳𝟏 𝑳(𝒚) = 𝑳𝟏 (𝒈(𝒙)) = 𝟎.
By solving the homogeneous higher – order equation 𝑳𝟏 𝑳(𝒚) = 𝟎, we can discover the form of a
particular solution 𝒚𝒑 for the original nonhomogeneous equation 𝑳(𝒚) = 𝒈(𝒙). We then substitute
this assumed form into 𝑳(𝒚) = 𝒈(𝒙) to find an explicit particular solution. This procedure for
determining 𝒚𝒑 , called the method of undetermined coefficients – annihilator method.
The procedure of Undetermined Coefficient – Annihilator Approach method
1. Find the complementary solution 𝒚𝒄 for the homogeneous DE 𝑳(𝒚) = 𝟎.
2. Operate on both sides of the nonhomogeneous equation 𝑳(𝒚) = 𝒈(𝒙) with a differential
operator 𝑳𝟏 that annihilates the function 𝒈(𝒙).
3. Find the general solution of the higher – order homogeneous DE 𝑳𝟏 𝑳(𝒚) = 𝟎.
4. Delete from the solution in step 3 all those terms that are duplicated in the complementary
solution 𝒚𝒄 found in step 1. Form a linear combination 𝒚𝒑 of the terms that remain. This
is the form of a particular solution of 𝑳(𝒚) = 𝒈(𝒙).
5. Substitute 𝒚𝒑 found in step 4 into 𝑳(𝒚) = 𝒈(𝒙). Match coefficients of the various functions
on each side of the equality, and solve the resulting system of equations for the unknown
coefficients in 𝒚𝒑 .
6. With the particular solution found in step 5, form the general solution 𝒚 = 𝒚𝒄 + 𝒚𝒑 of the
given differential equation.
8
Differential Equations – MCS 2423
Sec 4.4 – 4.5
Dr. Wisam Bukaita
Differential Equations
Example 8: Solve the given differential equation by undetermined coefficients.
𝒚′′ − 𝟐𝒚′ + 𝒚 = 𝒙𝟑 + 𝟒𝒙
Solution:
The differential equation we can rewrite in the equivalent differential operator form
(𝑫𝟐 − 𝟐𝑫 + 𝟏)𝒚 = 𝒙𝟑 + 𝟒𝒙
Step 1: First, we solve the homogeneous equation
𝒚′′ − 𝟐𝒚′ + 𝒚 = 𝟎
The auxiliary equation
𝒎𝟐 − 𝟐𝒎 + 𝟏 = 𝟎
(𝒎 − 𝟏)𝟐 = 𝟎
𝒎−𝟏=𝟎
𝒎 = 𝟏 𝒓𝒆𝒂𝒍 𝒓𝒐𝒐𝒕 𝒘𝒊𝒕𝒉 𝒎𝒖𝒍𝒕𝒊𝒑𝒍𝒊𝒄𝒊𝒕𝒚 𝒌 = 𝟐
So, the complementary function is
𝒚𝒄 = 𝒄𝟏 𝒆𝒙 + 𝒄𝟐 𝒙𝒆𝒙
Step 2: Now, since 𝒙𝟑 + 𝟒𝒙 is annihilated by the differential operator 𝑳𝟏 = 𝑫𝟒 , we apply 𝑳𝟏 =
𝑫𝟒 to the differential equation
𝑫𝟒 (𝑫𝟐 − 𝟐𝑫 + 𝟏)𝒚 = 𝑫𝟒 (𝒙𝟑 + 𝟒𝒙) = 𝟎
𝑫𝟒 (𝑫 − 𝟏)𝟐 𝒚 = 𝟎
The auxiliary equation of the 6th order equation is
𝒎𝟒 (𝒎 − 𝟏)𝟐 = 𝟎
and has roots 𝒎𝟏 = 𝒎𝟐 = 𝟏, 𝒂𝒏𝒅 𝒎𝟑 = 𝒎𝟒 = 𝒎𝟓 = 𝒎𝟔 = 𝟎. Thus, the general solution is
𝒚 = 𝒄𝟏 𝒆𝒙 + 𝒄𝟐 𝒙𝒆𝒙 + 𝒄𝟑 + 𝒄𝟒 𝒙 + 𝒄𝟓 𝒙𝟐 + 𝒄𝟔 𝒙𝟑
The first two terms form the complementary function of the original equation. The remaining
terms form a particular solution 𝒚𝒑 of the original equation
𝒚𝒑 = 𝑨 + 𝑩𝒙 + 𝑪𝒙𝟐 + 𝑬𝒙𝟑
To find the specific coefficients 𝑨, 𝑩, 𝑪, 𝒂𝒏𝒅 𝑬 substitute 𝒚𝒑 and its derivatives
𝒚′𝒑 = 𝑩 + 𝟐𝑪𝒙 + 𝟑𝑬𝒙𝟐
𝒚′′
𝒑 = 𝟐𝑪 + 𝟔𝑬𝒙
in the original equation.
𝒚′′ − 𝟐𝒚′ + 𝒚 = 𝒙𝟑 + 𝟒𝒙
𝟐𝑪 + 𝟔𝑬𝒙 − 𝟐(𝑩 + 𝟐𝑪𝒙 + 𝟑𝑬𝒙𝟐 ) + 𝑨 + 𝑩𝒙 + 𝑪𝒙𝟐 + 𝑬𝒙𝟑 = 𝒙𝟑 + 𝟒𝒙
𝟐𝑪 − 𝟐𝑩 + 𝑨 + 𝟔𝑬𝒙 − 𝟒𝑪𝒙 + 𝑩𝒙 − 𝟔𝑬𝒙𝟐 + 𝑪𝒙𝟐 + 𝑬𝒙𝟑 = 𝒙𝟑 + 𝟒𝒙
𝟐𝑪 − 𝟐𝑩 + 𝑨 = 𝟎
𝑨 = 𝟑𝟐
{𝟔𝑬 − 𝟒𝑪 + 𝑩 = 𝟒
⇔ {𝑩 = 𝟐𝟐
𝑪=𝟔
−𝟔𝑬 + 𝑪 = 𝟎
𝑬=𝟏
𝑬=𝟏
Thus, the particular solution is
𝒚𝒑 = 𝟑𝟐 + 𝟐𝟐𝒙 + 𝟔𝒙𝟐 + 𝒙𝟑
Step 3: The general solution of the original equation is
𝒚 = 𝒚𝒄 + 𝒚𝒑
𝒚 = 𝒄𝟏 𝒆𝒙 + 𝒄𝟐 𝒙𝒆𝒙 + 𝟑𝟐 + 𝟐𝟐𝒙 + 𝟔𝒙𝟐 + 𝒙𝟑
9
Differential Equations – MCS 2423
Dr. Wisam Bukaita
Differential Equations
Sec 4.4 – 4.5
Example 9: Solve the given differential equation by undetermined coefficients.
𝒚′′ + 𝟒𝒚 = 𝟒𝒄𝒐𝒔𝒙 + 𝟑𝒔𝒊𝒏𝒙 − 𝟖
Solution:
The differential equation we can rewrite in the equivalent differential operator form
(𝑫𝟐 + 𝟒)𝒚 = 𝟒𝒄𝒐𝒔𝒙 + 𝟑𝒔𝒊𝒏𝒙 − 𝟖
Step 1: First, we solve the homogeneous equation
𝒚′′ + 𝟒𝒚 = 𝟎
The auxiliary equation
𝒎𝟐 + 𝟒 = 𝟎
𝒎𝟐 = −𝟒
𝒎 = ±𝟐𝒊 𝒄𝒐𝒎𝒑𝒍𝒆𝒙 𝒄𝒐𝒏𝒋𝒖𝒈𝒂𝒕𝒆 𝒓𝒐𝒐𝒕𝒔
So, the complementary function is
𝒚𝒄 = 𝒄𝟏 𝒄𝒐𝒔𝟐𝒙 + 𝒄𝟐 𝒔𝒊𝒏𝟐𝒙
Step 2: For the functions 𝒄𝒐𝒔𝒙 𝒂𝒏𝒅 𝒔𝒊𝒏𝒙 we have 𝜶 = 𝟎, 𝜷 = 𝟏, 𝒂𝒏𝒅 𝒏 − 𝟏 = 𝟎 thus, the
operator 𝑫𝟐 + 𝟏 annihilates the functions 𝒄𝒐𝒔𝒙 𝒂𝒏𝒅 𝒔𝒊𝒏𝒙.
For the function (−𝟖) we have 𝒏 − 𝟏 = 𝟎 and the operator D annihilates the function (−𝟖).
The sum of these functions 𝟒𝒄𝒐𝒔𝒙 + 𝟑𝒔𝒊𝒏𝒙 − 𝟖 is annihilated by the product of
operators:𝑫( 𝑫𝟐 + 𝟏).
Now, since 𝟒𝒄𝒐𝒔𝒙 + 𝟑𝒔𝒊𝒏𝒙 − 𝟖 is annihilated by the differential operator 𝑳𝟏 = 𝑫( 𝑫𝟐 + 𝟏), we apply
𝑳𝟏 = 𝑫( 𝑫𝟐 + 𝟏) to the differential equation
𝑫( 𝑫𝟐 + 𝟏)(𝑫𝟐 + 𝟒)𝒚 = 𝑫( 𝑫𝟐 + 𝟏)(𝟒𝒄𝒐𝒔𝒙 + 𝟑𝒔𝒊𝒏𝒙 − 𝟖) = 𝟎
𝑫( 𝑫𝟐 + 𝟏)(𝑫𝟐 + 𝟒)𝒚 = 𝟎
The auxiliary equation of the 5th order equation is
𝒎( 𝒎𝟐 + 𝟏)(𝒎𝟐 + 𝟒) = 𝟎
and has roots 𝒎𝟏 = −𝟐𝒊, 𝒎𝟐 = 𝟐𝒊, 𝒎𝟑 = −𝒊, 𝒎𝟒 = 𝒊, 𝒎𝟓 = 𝟎. Thus, the general solution is
𝒚 = 𝒄𝟏 𝒄𝒐𝒔𝟐𝒙 + 𝒄𝟐 𝒔𝒊𝒏𝟐𝒙 + 𝒄𝟑 𝒄𝒐𝒔𝒙 + 𝒄𝟒 𝒔𝒊𝒏𝒙 + 𝒄𝟓
The first two terms form the complementary function of the original equation. The remaining
terms form a particular solution 𝒚𝒑 of the original equation
𝒚𝒑 = 𝑨𝒄𝒐𝒔𝒙 + 𝑩𝒔𝒊𝒏𝒙 + 𝑪
To find the specific coefficients 𝑨, 𝑩, 𝑪, 𝒂𝒏𝒅 𝑬 substitute 𝒚𝒑 and its derivatives
𝒚′𝒑 = −𝑨𝒔𝒊𝒏𝒙 + 𝑩𝒄𝒐𝒔𝒙
𝒚′′
𝒑 = −𝑨𝒄𝒐𝒔𝒙 − 𝑩𝒔𝒊𝒏𝒙
in the original equation.
𝒚′′ + 𝟒𝒚 = 𝟒𝒄𝒐𝒔𝒙 + 𝟑𝒔𝒊𝒏𝒙 − 𝟖
−𝑨𝒄𝒐𝒔𝒙 − 𝑩𝒔𝒊𝒏𝒙 + 𝟒(𝑨𝒄𝒐𝒔𝒙 + 𝑩𝒔𝒊𝒏𝒙 + 𝑪) = 𝟒𝒄𝒐𝒔𝒙 + 𝟑𝒔𝒊𝒏𝒙 − 𝟖
−𝑨𝒄𝒐𝒔𝒙 − 𝑩𝒔𝒊𝒏𝒙 + 𝟒𝑨𝒄𝒐𝒔𝒙 + 𝟒𝑩𝒔𝒊𝒏𝒙 + 𝟒𝑪 = 𝟒𝒄𝒐𝒔𝒙 + 𝟑𝒔𝒊𝒏𝒙 − 𝟖
−𝑨 + 𝟒𝑨 = 𝟒
𝑨 = 𝟒/𝟑
{−𝑩 + 𝟒𝑩 = 𝟑 ⇔ { 𝑩 = 𝟏
𝟒𝑪 = −𝟖
𝑪 = −𝟐
Thus, the particular solution is
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Differential Equations – MCS 2423
Dr. Wisam Bukaita
Differential Equations
Sec 4.4 – 4.5
𝟒
𝒄𝒐𝒔𝒙 + 𝒔𝒊𝒏𝒙 − 𝟐
𝟑
Step 3: The general solution of the original equation is
𝒚 = 𝒚𝒄 + 𝒚𝒑
𝟒
𝒚 = 𝒄𝟏 𝒄𝒐𝒔𝟐𝒙 + 𝒄𝟐 𝒔𝒊𝒏𝟐𝒙 + 𝒄𝒐𝒔𝒙 + 𝒔𝒊𝒏𝒙 − 𝟐
𝟑
𝒚𝒑 =
Sec 4.5
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