All that we need to do it go back to the appropriate examples above and get the particular solution from that example and add them all together. This is in the table of the basic functions. However, we wanted to justify the guess that we put down there. Using the fact on sums of function we would be tempted to write down a guess for the cosine and a guess for the sine.
So, we would get a cosine from each guess and a sine from each guess. The problem with this as a guess is that we are only going to get two equations to solve after plugging into the differential equation and yet we have 4 unknowns. We will never be able to solve for each of the constants. The general rule of thumb for writing down guesses for functions that involve sums is to always combine like terms into single terms with single coefficients.
This will greatly simplify the work required to find the coefficients. For this one we will get two sets of sines and cosines. 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.
Due to the nature of the mathematics on this site it is best views in landscape mode. If your device is not in landscape mode many of the equations will run off the side of your device should be able to scroll to see them and some of the menu items will be cut off due to the narrow screen width. The actual solution is then. Example 3 Find a particular solution for the following differential equation.
Example 4 Find a particular solution for the following differential equation. Example 5 Find a particular solution for the following differential equation. This is easy to fix however. First multiply the polynomial through as follows. Example 7 Find a particular solution for the following differential equation. To fix this notice that we can combine some terms as follows. Look for problems where rearranging the function can simplify the initial guess. So, the guess here is actually.
Example 9 Find a particular solution for the following differential equation. So, how do we fix this? Example 10 Write down the guess for the particular solution to the given 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.
In this case, that family must be modified before the general linear combination can be substituted into the original nonhomogeneous differential equation to solve for the undetermined coefficients. The specific modification procedure will be introduced through the following alteration of Example 6.
Find the complete solution of the differential equation. The general solution of the corresponding homogeneous equation was obtained in Example Multiply each member of the family by x and try again. Since the modified family no longer contains a solution of the corresponding homogeneous equation, the method of undetermined coefficients can now proceed.
First, obtain the general solution of the corresponding homogeneous equation. These equations determine the values of the coefficients: Find the complete solution of the equation. Since the auxiliary polynomial equation has distinct conjugate complex roots,.
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.
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.
The method of undetermined coefficients is a technique for determining the particular solution to linear constant-coefficient differential equations for certain types of nonhomogeneous terms f(t).