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The Method of Undetermined Coefficients

The Method of Undetermined Coefficients

❶Example 7 Find a particular solution for the following differential equation.

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This will arise because we have two different arguments in them. The main point of this problem is dealing with the constant. We just wanted to make sure that an example of that is somewhere in the notes. If you recall that a constant is nothing more than a zeroth degree polynomial the guess becomes clear.

All we did was move the 9. However, upon doing that we see that the function is really a sum of a quadratic polynomial and a sine. The guess for this is then. Notice however that if we were to multiply the exponential in the second term through we would end up with two terms that are essentially the same and would need to be combined. This is a case where the guess for one term is completely contained in the guess for a different term.

Therefore, we will take the one with the largest degree polynomial in front of it and write down the guess for that one and ignore the other term. So, the guess for the function is. This last part is designed to make sure you understand the general rule that we used in the last two parts. This time there really are three terms and we will need a guess for each term. The guess here is. We can only combine guesses if they are identical up to the constant.

So, when dealing with sums of functions make sure that you look for identical guesses that may or may not be contained in other guesses and combine them. This will simplify your work later on. We have one last topic in this section that needs to be dealt with. In the first few examples we were constantly harping on the usefulness of having the complementary solution in hand before making the guess for a particular solution.

It is now time to see why having the complementary solution in hand first is useful. This problem seems almost too simple to be given this late in the section. There is not much to the guess here. From our previous work we know that the guess for the particular solution should be,. Something seems wrong here. So, what went wrong? We finally need the complementary solution.

Notice that the second term in the complementary solution listed above is exactly our guess for the form of the particular solution and now recall that both portions of the complementary solution are solutions to the homogeneous differential equation,.

In other words, we had better have gotten zero by plugging our guess into the differential equation, it is a solution to the homogeneous differential equation! So, what did we learn from this last example. This work is avoidable if we first find the complementary solution and comparing our guess to the complementary solution and seeing if any portion of your guess shows up in the complementary solution.

The following set of examples will show you how to do this. Notice that the last term in the guess is the last term in the complementary solution. Notice that if we multiplied the exponential term through the parenthesis that we would end up getting part of the complementary solution showing up.

The second and third terms are okay as they are. In this case both the second and third terms contain portions of the complementary solution. We only need to worry about terms showing up in the complementary solution if the only difference between the complementary solution term and the particular guess term is the constant in front of them. The complementary solution this time is. This time however it is the first term that causes problems and not the second or third.

Recall that we will only have a problem with a term in our guess if it only differs from the complementary solution by a constant. So this means that we only need to look at the term with the highest degree polynomial in front of it. A first guess for the particular solution is. Notice that if we multiplied the exponential term through the parenthesis the last two terms would be the complementary solution. This still causes problems however. Upon multiplying this out none of the terms are in the complementary solution and so it will be okay.

As this last set of examples has shown, we really should have the complementary solution in hand before even writing down the first guess for the particular solution. Notes Quick Nav Download. Go To Notes Practice and Assignment problems are not yet written. As time permits I am working on them, however I don't have the amount of free time that I used to so it will take a while before anything shows up here. Assignment Problems Downloads Problems not yet written. You appear to be on a device with a "narrow" screen width i.

Now, combinbing like terms yields. The first equation immediately gives. Therefore, a particular solution of the given differential equation is. Find a particular solution and the complete solution of the differential equation.

Now, combining like terms and simplifying yields. A particular solution of the given differential equation is therefore. Find the solution of the IVP. The first step is to obtain the general solution of the corresponding homogeneous equation. Since the auxiliary polynomial equation has distinct real roots,. Combining like terms and simplifying yields. Therefore, the desired solution of the IVP is. Now that the basic process of the method of undetermined coefficients has been illustrated, it is time to mention that is isn't always this straightforward.

A problem arises if a member of a family of the nonhomogeneous term happens to be a solution of the corresponding homogeneous equation. For complex equations, the annihilator method or variation of parameters is less time consuming to perform. Undetermined coefficients is not as general a method as variation of parameters , since it only works for differential equations that follow certain forms.

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. 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.


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In this section we introduce the method of undetermined coefficients to find particular solutions to nonhomogeneous differential equation. We work a wide variety of examples illustrating the many guidelines for making the initial guess of the form of the particular solution that is needed for the method.

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The Method of Undetermined Coefficients In order to give the complete solution of a nonhomogeneous linear differential equation, Theorem B says that a particular solution must be added to the general solution of the corresponding homogeneous equation.

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The method of undetermined coefficients is a technique for determining the particular solution to linear constant-coefficient differential equations. The method of undetermined coefficients is used to solve a class of nonhomogeneous second order differential equations. This method makes use of.

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Method of Undetermined Coefficients Problem. Find a particular solution y p of the constant coefficients linear equation a ny (n) +···+a 2y 00 +a 1y 0 +a 0y = g(x). We assume that g(x) = [polynomial]×[exponential]×[sinusoid]. Introduction to the method of undetermined coefficients for obtaining the particular solutions of ordinary differential equations, a list of trial functions, and a brief discussion of pors and cons of this method.