# 3.4: Method of Undetermined Coefficients - Mathematics

Up to now, we have considered homogeneous second order differential equations. In this discussion, we will investigate nonhomogeneous second order linear differential equations.

Theorem: Solutions of Nonhomogeneous Second Order Linear Differential Equations

Let

[ L(y) = y'' + p(t)y' + q(t)y = g(t) ]

be a second order linear differential equation with p, q, and g continuous and let

[ L(y_1) = L(y_2) = 0 ;;; ext{and} ;;; L(y_p) = g(t)]

and let

[ y_h = c_1y_1 + c_2y_2. ]

Then the general solution is given by

[ y = y_h + y_p .]

Proof

Since (L) is a linear transformation,

[egin{align*} L(y_h + y_p) &= C_1L(y_1) + C_2L(y_2) + L(y_h)[4pt] &= C_1(0) + C_2(0) + g(t) = g(t). end{align*}]

This establishes that (y_h + y_p) is a solution. Next we need to show that all solutions are of this form. Suppose that (y_3) is a solution to the nonhomogeneous differential equation. Then we need to show that

[ y_3 = y_h + y_p ]

for some constants (c_1) and (c_2) with

[ y_h = c_1y_1 + c_2y_2. ]

This is equivalent to

[ y_3 - y_p = y_h .]

We have

[ L(y_3 - y_p) = L(y_3) - L(y_p) = g(t) - g(t) = 0. ]

Therefore (y_3 - y_p) is a solution to the homogeneous solution. We can conclude that

[y_3 - y_p = c_1y_1 + c_2y_2 = y_h.]

(square)

This theorem provides us with a practical way of finding the general solution to a nonhomogeneous differential equation.

• Step 1: Find the general solution (y_h) to the homogeneous differential equation.
• Step 2: Find a particular solution (y_p) to the nonhomogeneous differential equation.
• Step 3: Add (y_h + y_p) .

We have already learned how to do Step 1 for constant coefficients. We will now embark on a discussion of Step 2 for some special functions ( g(t) ).

Definition

A function (g(t)) generates a UC-set if the vector space of functions generated by (g(t)) and all the derivatives of (g(t)) is finite dimensional.

Example (PageIndex{1})

Let ( g(t) = t sin(3t) ).

Then

[egin{align*} g'(t) &= sin(3t) + 3t cos(3t) & g''(t) &= 6 cos(3t) - 9t sin(3t) g^{(3)} (t) &= -27 sin(3t) - 27t cos(3t) & g^{(4)}(t) &= 81 cos(3t) - 108t sin(3t) g^{(4)} (t) &= 405 sin(3t) - 243t cos(3t) & g^{(5)}(t) &= 1458 cos(3t) - 729t cos(3t) end{align*}]

We can see that (g(t)) and all of its derivative can be written in the form

[ g^{(n)} (t) = A sin(3t) + B cos(3t) + Ct sin(3t) + Dt cos(3t). ]

We can say that ( left { sin(3t), cos(3t), t sin(3t), t cos(3t) ight } ) is a basis for the UC-Set.

We now state without proof the following theorem tells us how to find the particular solution of a nonhomogeneous second order linear differential equation.

Theorem:

Let

[ L(y) = ay'' + by' + cy = g(t) ]

be a nonhomogeneous linear second order differential equation with constant coefficients such that g(t) generates a UC-Set

[ {f_1(t), f_2(t), ...f_n(t)} ]

Then there exists a whole number s such that

[ y_p = t^s[c_1f_1(t) + c_2f_2(t) + ... + c_nf_n(t)] ]

is a particular solution of the differential equation.

Remark: The "s" will come into play when the homogeneous solution is also in the UC-Set.

Example (PageIndex{2})

Find the general solution of the differential equation

[ y'' + y' - 2y = e^{-t} ext{sin}, t .]

Solution

First find the solution to the homogeneous differential equation

[ y'' + y' - 2y = 0 .]

We have

[ r^2 + r - 2 = (r - 1)(r + 2) = 0 ]

[ r = -2 ;;; ext{or} ;;; r = 1.]

Thus

[ y_h = c_1 e^{-2t} + c_2 e^t .]

Next notice that ( e^{-t} sin t ) and all of its derivatives are of the form

[ A e^{-t} sin t + B e^{-t} cos t .]

We set

[y_p = A e^{-t} sin t + B e^{-t} cos t ]

and find

[ egin{align*} y'_p &= A ( -e^{-t} sin t + e^{-t} cos t) + B (-e^{-t} cos t - e^{-t} sin t ) [4pt] &= -(A + B)e^{-t} sin t + (A - B)e^{-t} cos t end{align*}]

and

[egin{align*} y''_p &= -(A + B)(-e^{-t} sin t + e^{-t} cos t ) + (A - B)(-e^{-t} cos t - e^{-t} sin t ) &= [(A + B) - (A - B)] e^{-t} sin t + [-(A + B) - (A - B) ] e^{-t} cos t &= 2B e^{-t} sin t - 2A e^{-t} cos t . end{align*}]

Now put these into the original differential equation to get

[ 2B e^{-t} sin t - 2A e^{-t} cos t + -(A + B)e^{-t} sin t + (A - B) e^{-t} cos t - 2(A e^{-t} sin t + B e^{-t} cos t) = e^{-t} sin t. ]

Combine like terms to get

[ (2B - A - B - 2A) e^{-t} sin t + ( -2A + A - B - 2B) e^{-t} cos t = e^{-t} sin t ]

or

[ (-3A + B) e^{-t} sin t + (-A - 3B) e^{-t} cos t = e^{-t} sin t. ]

Equating coefficients, we get

[-3A + B = 1 ;;; ext{and} ;;; -A - 3B = 0.]

This system has solution

[ A = - frac {3}{10}, ;;; B = frac{1}{10}. ]

The particular solution is

[ y_p = - frac {3}{10} e^{-t} sin t + frac {1}{10} e^{-t} cos t. ]

Adding the particular solution to the homogeneous solution gives

[ y = y_h + y_p = c_1 e^{-2t} + c_2 e^{t} + - frac {3}{10} e^{-t} sin t + frac {1}{10} e^{-t} cos t. ]

Example (PageIndex{3})

Solve

[ y'' + y = 5 , sin t. label{ex3.1}]

Solution

The characteristic equation is

[ r^2 + 1 = 0. onumber]

Which has the complex roots

[ r = i ;;; ext{or} ;;; r = -i . onumber]

The homogeneous solution is

[ y_h = c_1 sin t + c_2 cos t. onumber ]

The UC-Set for (sin t) is ( left { sin t , cos t ight } ). Derivatives are all ( sin ) and ( cos ) functions

Notice that both of the functions in the UC-Set are solutions to the homogeneous differential equation. We need to multiply by (t) to get

[ left { t sin t, t cos t ight }. onumber]

The particular solution is

[egin{align*} y_p &= At , sin t + B cos t [4pt] y_p' &= A sin t + At cos t + B cos t - Bt sin t [4pt] y_p'' &= A cos t + A cos t - At sin t - B, sin t - Bsin t - Bt cos t [4pt]&= 2A cos t - At sin t - 2B sin t - Bt cos t. end{align*}]

Now put these back into the original differential equation (Equation ef{ex3.1}) to get

[egin{align*} 2A cos t - At sin t -2B sin t - Bt cos t + At sin t + Bt cos t &= 5 sin t [4pt] 2A cos t - 2B sin t &= 5 sin t. end{align*}]

Equating coefficients gives

[ 2A = 0 ;;; ext{and} ;;; -2 B = 5. ]

So

[ A = 0 ;;; ext{and} ;;; B = - frac {2}{5}. ]

We have

[ y_p = - frac {2}{5} cos t. ]

Adding ( y_p) to (y_h ) gives

[ y = c_1 sin t + c_2 cos t - frac {2}{5} cos t. ]

## Problem 3.4.33: Solve the initial value problem using the method of undetermined coefficients. d²x + w?x = F, sin wt, x(0) = 0, x'(0) = 0 dt? help_outline

#### Image Transcriptionclose

Problem 3.4.33: Solve the initial value problem using the method of undetermined coefficients. d²x + w?x = F, sin wt, x(0) = 0, x'(0) = 0 dt?

## Practice Test! Learning how to set up particular solution forms when solving second-order ODEs! (Method of Undetermined Coefficients Examples)

If you haven’t already, please review example 1, example 2, example 3, and example 4 for solving second-order ODEs using a homogeneous and particular solution.

The really tricky part is assuming a particular solution form, it’s very easy to mess it up and we’ve seen many students mess up the entire question.

Let’s practice many second-order ODEs, but instead of actually solving them completely, let’s instead just answer what we would assume the particular solution form to be!

This is sometimes asked as an examination question, “Explain the particular solution form you would use for solving the second-order ODE, but don’t actually solve it!”. It really means you need to know your theory well!

Remember to always find the homogeneous solution first on your own! If the particular solution has terms in the homogeneous solution, you need to multiply by x.

Example problem: What particular solution form would you use for (y” + 3y = – 48>) ?

Now we have a product of a polynomial and an exponential term, so that’s the form of our particular solution, remember to have the highest degree term and also include all lower degree terms too. Also be careful and keep the e’s exponent to be correct.

Example problem: What particular solution form would you use for (4y” – 4y’ – 3y = cos (2x)) ?

Remember to always include both sine plus cosine, even if you only have one in the actual ODE! Also remember to match the correct frequency.

[ = Acos left( <2x> ight) + Bsin left( <2x> ight)]

Example problem: What particular solution form would you use for (y” + 2y’ = 2x + 5 – >) ?

If you find the homogeneous solution, you get roots of 0 and -2. () conflicts with the polynomial, so that means you need to need to multiply the whole polynomial by x. As well, (>) conflicts, so you need to multiply the e in the particular solution by x.

Example problem: What particular solution form would you use for (y” – y’ + frac <4>= 3 + <><2>>>) ?

The trick here is that the homogeneous solution has two real repeated roots at 1/2, conflicting with the particular solution, so you need to multiply the e term in the particular solution by x twice.

Example problem: What particular solution form would you use for (y” + 4y = 3sin (2x)) ?

The trick is that the roots of the homogeneous solution are complex, leading to cosine and sine terms of 2x So you need to multiply both the sine and cosine terms in the particular solution by x.

Example problem: What particular solution form would you use for (y” – 4y = left( <– 3> ight)sin left( <2x> ight)) ?

Product of a polynomial and trigonometry, don’t forget we need both sine and cosine.

[ = left( <>+ Bx + C> ight)cos left( <2x> ight) + left( <>+ Ex + F> ight)sin left( <2x> ight)]

Example problem: What particular solution form would you use for (y” + y = 2xsin left( x ight)) ?

The trick is you get cosine and sine in the homogeneous solution due to complex roots. So you need to multiply both the sine and cosine terms by x.

[ = left( <>+ Bx> ight)cos left( x ight) + left( <>+ Dx> ight)sin left( <2x> ight)]

Example problem: What particular solution form would you use for (y” – 5y’ = 2 – 4 – x + 6) ?

The trick is that the homogeneous solution has roots of 5 and 0. The 0 is the problem because () is a constant, and a constant is present in our polynomial for our particular solution. That means we need to multiply the entire polynomial by x.

Example problem: What particular solution form would you use for (y” – 2y’ + 5y = cos left( <2x> ight)) ?

This one, like the others, has a trick. Find the homogeneous solution, you actually have three roots, 0, and two complex roots. The homogeneous solution is ( = left[ <cos left( <2x> ight) + sin left( <2x> ight)> ight])

Now, our particular solution is of the form e to the x times sine and cosine, but since that’s present you need to multiply them by x.

Be careful, we saw many students who said “e is present and cosine/sine is present, I need to multiply by x twice!”, but that’s wrong!

If you had e OR sine/cosine, you wouldn’t need to multiply by x at all because e OR sine/cosine is not the same as e times sine/cosine. Since the homogeneous solution has e times sine/cosine, you multiply by x once.

Last, but not least, Example problem: What particular solution form would you use for (y” – 2y’ – 6y = >cos (4x)) ?

The homogeneous solution has roots of 0, and real numbers, but it’s okay because the e to the 0 is not multiplied by trig or e to the 6, so you don’t actually have particular solution terms appearing in the homogeneous solution because everything is multiplied if the right-hand side was added instead of multiplied together, then the terms would technically be appearing.

## Aktosun, Spring 2018, Math 3318

Textbook: C. H. Edwards, D. E. Penney, and D. T. Calvis, Differential Equations and Boundary Value Problems, 5th ed., Pearson, Boston, 2015.

Coverage: We will not rely on the textbook much. The textbook is for you to read the material from another viewpoint. Roughly, we will cover some materials in Chapters 1, 2, 3, 4, 5, and 7. The details of the coverage of the material is given in the course outline .

Exams: Exam 1: Thursday, February 22 (during class) Exam 2: Thursday, April 12 (during class) Exam 3 (Final Exam): Thursday, May 10 during 8:00-10:30 am in PKH 107.

Information on each exam: Each exam will contain 30 questions (the first 15 are true/false questions and the remaining 15 are multiple-choice questions with 4 options to choose from). No materials are allowed during the exams besides a pencil or a pen blank sheets are provided on each exam an answer sheet to put the answers on will be provided and will be collected at the end of each exam. Separate information related to the coverage will be provided for each exam.

Prerequisites: A grade of C or better in Math 2326 or concurrent enrollment.

Math clinic: Free help is available for this course at the Math Clinic located in PKH 325 (on the third floor of Pickard Hall just across from the elevator). One of the doctoral mathematics students, Ms. Niloofar Ghorbani, is available at the Math Clinic during 8-9 am on Tuesdays and Wednesday, to provide help to students enrolled in Math 3318. Those of you who need some mathematical help for Math 3318, you can get help from Ms. Ghorbani during those times (Tuesday and Wednesday during 8-9 am) in the Math Clinic (PKH 325).

• ODE: general solution, particular solution, explicit solution, implicit solution arbitrary constants, initial conditions
• linear ODEs, nonlinear ODEs, linear homogeneous ODEs
• First-order ODEs: linear, separable, exact, homogeneous, Bernoulli methods to solve such ODEs
• First-order linear ODEs: standard form, an integrating factor
• Differential, exact differential, total differential criteria for exactness
• Substitution for Bernoulli equations, substitution for homogeneous equations
• Linear ODEs: nonhomogeneous term, homogeneous linear ODE, superposition principle, general solution, particular solution
• Linear nth-order ODEs: with constant coefficients, Cauchy-Euler equations functions satisfying such homogeneous linear ODEs
• higher-order linear ODEs: general solution, particular solution, arbitrary constants, initial conditions
• linear ODEs: the corresponding homogeneous ODE, general solution to the corresponding ODE, linearly independent solutions
• linear homogeneous ODEs with constant coefficients, the corresponding auxiliary equation, the operator notation D=d/dx
• linear homogeneous ODEs with constant coefficients: y=e rx , y=xe rx with a repeated root, y=e ax cos(bx) and y=e ax sin(bx) with complex roots r=a±ib
• Cauchy-Euler equations: y=x r , y=(lnx)x r with a repeated root, y=x a cos(b(lnx)) and y=x a sin(b(lnx)) with complex roots r=a±ib
• method of undetermined coefficients to find a particular solution
• given the general solution, find the corresponding ODE
• method of reduction of order
• higher-order linear ODEs: general solution, particular solution, arbitrary constants, initial conditions
• linear homogeneous ODEs with constant coefficients, the corresponding auxiliary equation, the operator notation D=d/dx
• linear homogeneous ODEs with constant coefficients: y=e rx , y=xe rx with a repeated root, y=e ax cos(bx) and y=e ax sin(bx) with complex roots r=a±ib
• Cauchy-Euler equations: y=x r , y=(lnx)x r with a repeated root, y=x a cos(b(lnx)) and y=x a sin(b(lnx)) with complex roots r=a±ib
• method of undetermined coefficients to find a particular solution
• given the general solution, find the corresponding ODE
• Laplace transform, Laplace transform formulas
• inverse Laplace transform, inverse Laplace transform formulas
• unit step function
• solving initial value problems by using Laplace transform

Supplementary problems 6
Brief answers to supplementary problems 6
Solution to supplementary problems 6
Practice problems from the textbook: 1.1 Differential equations: 1-26, 37-42 1.2 General and particular solutions: 1-18 1.4 Separable equations: 1-28 1.5 Linear first-order equations: 1-28 1.6 Exact equations: 1-40, 43-54 3.1 Second-order linear equations: 1-28, 36-42 3.2 General solutions to linear equations: 1-24 3.3 Homogeneous equations with constant coefficients: 1-32 3.5 Method of undetermined coefficients: 1-40 7.1 Laplace transform: 1-32 7.2 Solving initial-value problems: 1-24 7.3 Further properties for Laplace transform: 1-24

## Contents

Consider a linear non-homogeneous ordinary differential equation of the form

The method of undetermined coefficients provides a straightforward method of obtaining the solution to this ODE when two criteria are met: 

In order to find the particular integral, we need to 'guess' its form, with some coefficients left as variables to be solved for. This takes the form of the first derivative of the complementary function. Below is a table of some typical functions and the solution to guess for them.

K cos ⁡ ( a x ) + M sin ⁡ ( a x )

If a term in the above particular integral for y appears in the homogeneous solution, it is necessary to multiply by a sufficiently large power of x in order to make the solution independent. If the function of x is a sum of terms in the above table, the particular integral can be guessed using a sum of the corresponding terms for y. 

### Example 1 Edit

Find a particular integral of the equation

The right side t cos t has the form

with n = 2, α = 0, and β = 1.

Since α + = i is a simple root of the characteristic equation

we should try a particular integral of the form

Substituting yp into the differential equation, we have the identity

Comparing both sides, we have

We then have a particular integral

### Example 2 Edit

Consider the following linear nonhomogeneous differential equation:

By substituting this function and its derivative into the differential equation, one can solve for A:

## Sample Schedule

• Prerequisite Skills
• Some Basic Modelling § 1.1
• Solutions to Differential Equations § 1.2
• Direction Fields § 1.1
• Separable First Order ODE's § 2.2
• Linear First Order ODE's § 2.1
• Modelling with First Order ODE's § 2.3
• Population Dynamics § 2.5
• Euler's Method § 2.7
• Review
• Midterm #1
• Second Order Constant Coefficient ODE's § 3.1
• Modelling -- Damped Harmonic oscillator § 3.7 (just modeling--optional)
• Homogeneous equations with distinct real roots § 3.1
• Complex numbers and Euler's formula -- notes on complex numbers
• Homogeneous equations with complex roots § 3.3
• Homogeneous equations with repeated roots § 3.4
• Harmonic Oscillator § 3.7
• Method of Undetermined Coefficients § 3.5
• Forced Harmonic Oscillator § 3.8
• Forced Undamped harmonic Oscillator Beats and Resonance § 3.8
• Forced Damped Harmonic Oscillator -- Frequency Response and Phase § 3.8
• Midterm #2
• Laplace Transform -- definition as an integral § 6.1
• Tables of Laplace Transforms § 6.2
• Inverse Laplace Transform using tables § 6.2
• Solving IVP with Laplace Transforms § 6.2
• Step functions and and discontinuous forcing (i.e .time delay) § 6.3 and 6.4
• Final Exam

Religious Accommodation Policy: Washington state law requires that UW develop a policy for accommodation of student absences or significant hardship due to reasons of faith or conscience, or for organized religious activities. The UW's policy, including more information about how to request an accommodation, is available at Religious Accommodations Policy. Accommodations must be requested within the first two weeks of the course using the Religious Accommodations Request form.

## 3.4: Method of Undetermined Coefficients - Mathematics

OFFICE HOURS: Monday 3:35 to 4:25 PM, Wednesday 1:25 to 2:15 PM, or by appt.

PHONE: 865-1403 (office), 867-1364 (home)

COURSE DESCRIPTION: Ordinary Differential Equations (3:3:0). First- and second-order equations numerical methods special functions Laplace transform solutions higher order equations.

PREREQUISITE: Math 141, or equivalent courses.

TEXTBOOK: Elementary Differential Equations and Boundary Value Problems, 8th edition, W. E. Boyce and R.C. Diprima, John Wiley and Sons, Inc. If you have the 6th or 7th edition textbook from the previous years, you can also use that.

EXAMINATIONS: Two 75-minute evening examinations will be given on February 24 and March 28, and a comprehensive final examination will be given during the final examination period.

GRADING POLICY: Grades will be assigned on the basis of 450 points distributed as follows

100 points midterm examination I (2-24-2005)
100 points midterm examination II (3-28-2005)
100 points quizzes/homework
150 points final examination

INTRODUCTION
1.1 Direction Fields
1.2 Solution of Some Differential Equations
1.3 Classification of Differential Equations

FIRST ORDER DIFFERENTIAL EQUATIONS
2.1 Linear Equations with Variable Coefficients
2.2 Separable Equations
2.3 Modeling with First Order Equations
2.4 Differences Between Linear and Nonlinear Equations
2.5 Autonomous Equations and Population Dynamics
2.7 Numerical Approximations: Euler's Method

NUMERICAL METHODS
8.3 Improvements on the Euler Method
8.4 The Runge-Kutta Method

SECOND ORDER LINEAR DIFFERENTIAL EQUATIONS
3.1 Homogeneous Equations with Constant Coefficients
3.2 Fundamental Solutions of Linear Homogeneous Equations
3.3 Linear Independence and the Wronskian
3.4 Complex Roots of the Characteristic Equations
3.5 Repeated Roots Reduction of Order
3.6 Nonhomogeneous Equations Method of Undetermined Coefficients
3.8 Mechanical and Electrical Vibrations
3.9 Forced Vibrations (w/o damping)

THE LAPLACE TRANSFORM
6.1 Definition of the Laplace transform
6.2 Solution of Initial Value Problems
6.3 Step Functions
6.4 Differential Equations with Discontinuous Forcing Functions
6.5 Impulse Functions

SYSTEMS OF TWO LINEAR DIFFERENTIAL EQUATIONS
7.1 Intoduction to Systems of Differential Equations
7.2-7.3 Introduction to 2 x 2 Matrices
7.5-7.9 2 x 2 Linear Systems of Differential Equations

NONLINEAR DIFFERENTIAL EQUATIONS AND STABILITY
9.1 Phase Portraits of 2 x 2 Linear Systems
9.2 Autonomous Systems and Stability
9.3 Almost Linear Systems
9.5 Predator-Prey Equations Linearization of Nonlinear systems

## Method Undetermined Coefficients

Let ( exttt = < ext d>/< ext d>x ) be the derivative operator. We consider linear differential operators that are generated by a polynomial of degree n: [ L[lambda ] = a_n lambda^n + a_ lambda^ + cdots + a_1 lambda + a_0 , ] which is referred to as the characteristic polynomial to the corresponding differential operator [ L left[ exttt ight] = a_n exttt^n + a_ exttt^ + cdots + a_1 exttt + a_0 . ] It is always assumed that the coefficients of the above operator are continuous functions in some interval. Moreover, we always suppose that the leading coefficient ( a_n e 0. ) Therefore, it could be factored out or set to be 1.

Our main objective is to show how to find a particular solution to the nonhomogeneous equation

1. the coefficients ( a_n , a_ , ldots , a_1 , a_0 ) are constants and
2. f(x) is a constant, a polynomial functions, an exponential function e ax , a sine or cosine function ( sin eta x quad mboxquad cos eta x, ) or finite sums and products of these functions.

function formula control number
polynomial or constant ( p_n x^n + cdots + p_1 x + p_0 ) 0
polynomial times exponential ( left( p_n x^n + cdots + p_1 x + p_0 ight) e^) &alpha
polynomial times exponential & sine ( left( p_n x^n + cdots + p_1 x + p_0 ight) e^ , sin eta x) &alpha + &beta j or &alpha - &beta j
polynomial times exponential & cosine ( left( p_n x^n + cdots + p_1 x + p_0 ight) e^, cos eta x) &alpha + &beta j or &alpha - &beta j

A control number is just a root of the characteristic polynomial that corresponds to the operator annihilating the function. For instance, if a control number is known to be &alpha, we know that the annihilating polynomial for such function must be ( left( exttt - alpha ight)^m , ) for some positive integer m (called the multiplicity). Since the characteristic polynomial for any constant coefficient differential operator can be factored into simple terms, it is natural to start analyzing with some such simple multiple.

Example: Admissible functions and their control numbers

The following theorem explains why the method undetermined coefficients works.

1. every solution of ( L[y] = f ) belongs to Ker(AL)
2. in every subspace V of Ker(AL) complementary to Ker(L) (i.e., in every V such that ( mbox(L) oplus V = mbox(AL)), ) there is a unique solution to ( L(y) = f. )

Corollary: If L and A commute then Ker(A) is a subspace of Ker(AL). If, in addition, Ker(L) and Ker(A) have only 0 in common, then in Ker(A) there is a unique solution to ( L( y) = f. qquad ) ■

If L and A have constant coefficients, then L and A commute. This motivates the standard practice of undetermined coefficients however, the complement of Ker(L) in Ker(AL) needs have very little similarity to Ker(A), depending on just how large ( mbox(L) cap mbox(A) ) is.

It is not the fact that L and A are differential operators which makes the theorem work, but the fact that for operators of these types, dim(Ker(AL)) = dim(Ker(A)) + dim(Ker(L)).

The theorem explains exactly which nonhomogeneous linear differential equations permit finding a particular solution by the method of undetermined coefficients: the right-hand side must be annihilated by some linear differential operator of positive order. In the constant coefficients case for either topic, it is clear from the theory how to obtain A and all of the basis functions specified by the theorem. In fact, the bases for Ker(L) and Ker(AL) so obtained will always give a basis for Ker(L) as a subset of a basis for Ker(AL). This means that one can immediately identify a basis for a subspace complementary to Ker(L) as those basis elements in Ker(AL) which were not in Ker(L). One then knows the form of a particular solution since the existence of a unique solution in this complementary subspace is guaranteed. These ideas are illustrated by the following examples.

Example: Basis for two operators

Example: Consider the differential operator of degree four ( L = left( exttt - 2 ight)^2 left( exttt +3 ight)left( exttt - 4 ight) . ) Let ( f(x) = 2x^3 e^ <2x>) be a function whose control number is 2. This means that ( A= left( exttt -2 ight)^4 ) annihilates it. Since ( AL=LA ) has the basis of functions

Lemma: Let us consider the linear nonhomogeneous differential equation with constant coefficients of order n:

1. if ( a_0 e 0 , ) then the degrees of the polynomials f and g are the same
2. if ( k = max left < j , : , a_l =0, l le j ight>, ) exists, then we can admit that ( g(x) = x^ h(x) , ) where h and f have the same degree /li>
3. if all coefficients are real, the g and h are real.

Theorem: Let f(x) be a nonzero real polynomial and (&gamma, &delta) be a pair of numbers such that &gamma is a complex number, but &delta is always real number, with the condition that &delta = 0 if &gamma is real. Let ( L[ exttt] ) be a constant coefficient differential operator, with the characteristic polynomial ( L[lambda ] = a_n lambda^n + cdots + a_1 lambda + a_0 . ) Then the nonhomogeneous differential equation

1. If &gamma is a real number, then g(x) is a real valued solution of the above differential equation, and a particular solution of ( L[ exttt] y = g(x), e^ delta> ) becomes ( y_p (x) = g(x), e^ . )
2. If &gamma is complex, then ( z (x) = g(x), e^ delta> ) is a complex-valued solution of ( L[ exttt] y = g(x), e^ delta> . ) If ( gamma = alpha + <f j>eta , ) where &alpha and &beta are real numbers, then the real part ( Y_r (x) = Re left[ z(x) ight] = mbox left[ z(x) ight] ) and the imaginary part ( Y_i (x) = Im left[ z(x) ight] = mbox left[ z(x) ight] ) are solutions of the differential equation

The above theorem shows that we can always reduce the problem of finding a particular solution to a nonhomogeneous equation to the case when the input function does not contain an exponential multiple. We illustrate the method of undetermined coefficients in the series of examples.

Example: Control number is zero which does not match the roots of the characteristic equation

Example: Consider the differential equation of third order

Example: Control number does not match the roots of the characteristic equation

Example: Consider the differential equation with the same differential operator as in the previous example, but having exponential input:

Example: Consider the differential equation with the same differential operator as in the previous example, but having exponential input:

The control number of the driving term is &sigma = 1, which matches one of the roots of the characteristic equation ( lambda^3 -2 lambda^2 -5 lambda + 6 = left( lambda -1 ight) left( lambda -2 ight)left( lambda -3 ight) =0 ) of multiplicity 1. Therefore, we seek its particular solution in the form:

Example: Consider the differential equation with the fourth order differential operator that has a polynomial input:

Example: Consider the differential equation with the fourth order differential operator when the input has an exponential term:

Example: Consider the differential equation whose characteristic polynomial has complex roots that do not match the control numbers of the input function:

Example: Consider the differential equation of second order whose characteristic polynomial has complex roots that do match the control numbers of the input function:

Example: Consider the differential equation of fourth order whose characteristic polynomial has complex roots that do match the control numbers of the input function:

## Ch 4.3: Nonhomogeneous Equations: Method of Undetermined Coefficients - PowerPoint PPT Presentation

### The method of undetermined coefficients can be used to find a particular solution Y of an nth order linear, . order equations, . the differential equation . &ndash PowerPoint PPT presentation

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## For the nonhomogeneous equation, using the method of undetermined coefficients, the solution I got for #(d^2y)/(dx^2) - 5(dy)/(dx) + 6y = xe^x# is #y = (xe^x)/2 + (3e^x)/4#, but along the way I got two solutions for #A_1#?

I was letting #y = x(A_1x + A_0)e^x# , and I got #(dy)/(dx) = [A_1x^2 + (2A_1 + A_0)x + A_0]e^x# and #(d^2y)/(dx^2) = [A_1x^2 + (4A_1 + A_0)x + 2A_1 + 2A_0]e^x# . When I solved for #A_0# and #A_1# however, I got:

#A_1 - 5A_1 + 6A_1 = 0 => A_1 = 0# is one solution to the quadratic (whoops, just realized this)

#4A_1 + A_0 - 10A_1 - 5A_0 + 6A_0 = 1# , and if #A_1 = 0# , then:

#A_0 = 1/2#

#2A_1 + 2A_0 - 5A_0 = 0# , and if #A_1 ne 0# , then #A_1 = 3/2A_0 = 3/4# . So the result is indeed #y = (xe^x)/2 + (3e^x)/4# . #(d^2y)/(dx^2) - 5(dy)/(dx) + 6y = xe^x# is a linear non-homogeneous differential equation.

The solution #y_p = (xe^x)/2 + (3e^x)/4 # is a particular solution for the differential equation.

The complete solution can be composed as the addition of a particular solution plus the homogeneous solution. The homogeneous solution has the structure

as can be easily verified, so the complete solution is

#y = y_h+y_p = C_1e^(2x)+C_2e^(3x)+ (xe^x)/2 + (3e^x)/4# without ambiguity in the #C_1,C_2# determination. I'll use #A=A_0# and #B=A_1# to save typing

The problem is you are trying the wrong solution you have
#g(x)=(ax^2+b)e^x# and you should be trying #g(x)=(ax+b)e^x#

## METHOD OF UNDETERMINED COEFFICIENTS

If x 2 + x + 1 is a factor of the polynomial 3x 3 + 8x 2 + 8x + a, then find the value of a. Hence the value of a is 5.

The equations x 2 −6x+a = 0 and x 2 −bx+6 = 0 have one root in common. The other root of the first and the second equations are integers in the ratio 4 : 3. Find the common root.

Let α, 4β be the roots of x 2 − 6x + a = 0.

Let α, 3β be the roots of x 2 − bx+6 = 0.

Then, 4αβ = a and 3αβ = 6 which give αβ = 2 and a = 8.

The roots of x 2 − 6x+8 = 0 are 2, 4.

2 which is not an integer.

Hence, the common root is 2.

Find the values of p for which the difference between the roots of the equation x 2 + px+8 = 0 is 2.

Let α and β be the roots of the equation x2 + px+8 = 0.

Then, α + β = −p, αβ = 8 and |α − β| = 2.

Now, (α + β) 2 − 4αβ = (α − β) 2 , which gives p 2 − 32 = 4.

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