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Exam A1
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17calculus > differential equations > undetermined coefficients

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Undetermined Coefficients

\(ay'' + by' + cy = g(t)\)

classify - second-order, linear, constant coefficients, inhomogeneous → \(g(t) \neq 0\)

The idea of this technique is to use \(g(t)\) to guess what the form of the solution looks like.

For certain functions, the solutions will look a lot like the given \(g(t)\). However, this works only for certain standard forms of \(g(t)\). This technique for handling second order inhomogeneous linear differential equations is great when \(g(t)\) is in a standard form. Here is a list of some of the standard forms that this technique will work with.

polynomials (including constants)

exponentials in the form \(e^{at}\)

trig functions \(\sin(bt)\) and \(\cos(ct)\)

If \(g(t)\) is not one of the standard forms above, the method of variation of parameters, which is a bit more involved, needs to be used.

General Inhomogeneous Solutions

The general solution to an inhomogeneous differential equation is made up of two parts, the homogeneous solution plus what many mathematicians call the particular solution. The idea is that you can add zero to anything you don't change the problem. So, you can look at the differential equation
\(\displaystyle{ y'' + p(t)y' + q(t)y = g(t) = 0 + g(t) }\)
So we find the homogeneous solution, \(y_H(t)\) which solves \(y'' + p(t)y' + q(t)y = 0\) and add it to the inhomogeneous solution or particular solution \(y_P(t)\) which solves \(y'' + p(t)y' + q(t)y = g(t)\) to get the general solution \(y(t) = y_H(t) + y_P(t)\).

Before we start discussing how to get solutions to inhomogeneous equations using undetermined coefficients, let's watch a video talking about inhomogeneous equations is general.

MIT OCW - solutions

Method of Undetermined Coefficients

The discussion on this page assumes you know how to get the homogeneous solution or it is given to you. We will work to get the inhomogeneous solution. You need to know what the homogeneous solution is first before using this technique. The reason is that sometimes you will need to adjust your 'guess' based on the form of the homogeneous solution.

Let\'s start out with a video. This video explains very well, this idea of a general solution built from the homogeneous and particular solutions, as well as the method of undetermined coefficients. It is a bit longer than others you will find on this site but this one is especially worth taking the time to watch. He does several related examples that build on one another and will help you understand what to do. At the very end, he also explains why this technique works. So this is one video you don\'t want to skip.

Dr Chris Tisdell - general solution

If you want another quick overview of this method, here is a quick video clip that might help cement this technique in your mind.

PatrickJMT - quick overview

Okay, let's look at each of the standard forms individually and see what your initial 'guesses' look like, as well as variations that are required to avoid duplication with the homogeneous solution. In the following tables, we are talking about the roots to the characteristic polynomial that you used to find the homogeneous solution.

Standard Form 1 - - Polynomials

\(g(t) = a_2t^2 + a_1t + a_0\)

basic form

\(y_P = At^2+Bt+C\)

variations

\(r=0\) is a single root

\(y_P = ( At^2+Bt+C)t \)

\(r=0\) is a double root

\(y_P = ( At^2+Bt+C)t^2 \)

Note - - In the table above, we showed a second order polynomial. In general, your basic form polynomial needs to be of the same order as your \(g(t)\) polynomial. For example, if \(g(t) = t^4\), your basic polynomial is \(y_P = At^4 + Bt^3 + Ct^2 + Dt + E \), then you would adjust based on the roots to the homogeneous solution, if necessary.

Standard Form 2 - - Exponentials

\(g(t) = e^{ct}\)

basic form

\(y_P = Ae^{ct}\)

variations

\(r=c\) is a single root

\(y_P = Ate^{ct} \)

\(r=c\) is a double root

\(y_P = At^2e^{ct} \)

Standard Form 3 - - Trig Functions

\(g(t) = a_0\sin(kt)\) or \(g(t) = a_0\cos(kt)\) or both

basic form

\(y_P = A\sin(kt)+B\cos(kt)\)

In each case, the function that you build based on \(g(t)\) must be different than any factor in the homogeneous solution, i.e. it must not be a constant multiple of any other factor. You may need to multiply by \(t\) or \(t^2\) to accomplish this, but, as you saw in the first video above, doing so will resolve any conflicts.

After working some practice problems, you will be ready for the next topic, variation of parameters, which is a technique that handles a larger range of problems when \(g(t)\) is not one of the above standard forms.

next: variation of parameters →

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Practice Problems

Instructions - - Unless otherwise instructed, find the general solution to each differential equation using the method of undetermined coefficients. If initial conditions are given, find the particular solution also. Give your answers in exact terms and completely factored.

Level A - Basic

Practice A01

\(y''-5y'+6y=4e^t\)

answer

solution

Practice A02

\(\dot{x}+3x=t^2+t\)

answer

solution

Practice A03

\(\ddot{x}+\dot{x}=t^4\)

answer

solution

Practice A04

\(y''+3y'+2y=x^2\)

answer

solution

Practice A05

\(y''-5y'+6y=2x+3\)

answer

solution

Practice A06

\(y''-5y'+6y=12e^{5x}\)

answer

solution

Practice A07

\(y''-3y'-4y=3e^{2x}\)

answer

solution

Practice A08

\(y''+3y'-4y=2e^x\); \(y(0)=1, y'(0)=2\)

answer

solution


Level B - Intermediate

Practice B01

\(y''-5y'+6y=10e^{2x}\)

answer

solution

Practice B02

\(y''-3y'-4y=2\sin(x)\)

answer

solution

Practice B03

\(y''-3y'-4y=4x^2\)

answer

solution

Practice B04

\(y''+2y'=24x+e^{-2x}\)

answer

solution

Practice B05

\(y''+4y'+4y=130\cos(3x)\)

answer

solution

Practice B06

\(y''+7y'+12y=3e^{-3t}\)

answer

solution

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