It is closely related to the annihilator method , but instead of using a particular kind of differential operator the annihilator in order to find the best possible form of the particular solution, a "guess" is made as to the appropriate form, which is then tested by differentiating the resulting 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.
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.
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. In this section we will take a look at the first method that can be used to find a particular solution to a nonhomogeneous differential equation. One of the main advantages of this method is that it reduces the problem down to an algebra problem. The algebra can get messy on occasion, but for most of the problems it will not be terribly difficult.
There are two disadvantages to this method. Second, it is generally only useful for constant coefficient differential equations. The method is quite simple. Plug the guess into the differential equation and see if we can determine values of the coefficients. Recall that the complementary solution comes from solving,. At this point the reason for doing this first will not be apparent, however we want you in the habit of finding it before we start the work to find a particular solution.
As mentioned prior to the start of this example we need to make a guess as to the form of a particular solution to this differential equation. Okay, we found a value for the coefficient.
This means that we guessed correctly. A particular solution to the differential equation is then,. At this point do not worry about why it is a good habit. Now, back to the work at hand. Any of them will work when it comes to writing down the general solution to the differential equation. Speaking of which… This section is devoted to finding particular solutions and most of the examples will be finding only the particular solution.
This however, is incorrect. The complementary solution is only the solution to the homogeneous differential equation and we are after a solution to the nonhomogeneous differential equation and the initial conditions must satisfy that solution instead of the complementary solution.
So, we need the general solution to the nonhomogeneous differential equation. Taking the complementary solution and the particular solution that we found in the previous example we get the following for a general solution and its derivative. This means that the coefficients of the sines and cosines must be equal.
First, since there is no cosine on the right hand side this means that the coefficient must be zero on that side. More importantly we have a serious problem here. What this means is that our initial guess was wrong.
If we get multiple values of the same constant or are unable to find the value of a constant then we have guessed wrong. One of the nicer aspects of this method is that when we guess wrong our work will often suggest a fix. In this case the problem was the cosine that cropped up.
Our new guess is. We found constants and this time we guessed correctly. Notice that if we had had a cosine instead of a sine in the last example then our guess would have been the same.
In fact, if both a sine and a cosine had shown up we will see that the same guess will also work. For this we will need the following guess for the particular solution. So, differentiate and plug into the differential equation. Notice that in this case it was very easy to solve for the constants. A particular solution for this differential equation is then. Notice that there are really only three kinds of functions given above. If you think about it the single cosine and single sine functions are really special cases of the case where both the sine and cosine are present.
Also, we have not yet justified the guess for the case where both a sine and a cosine show up. We will justify this later. We now need move on to some more complicated functions. The more complicated functions arise by taking products and sums of the basic kinds of functions.
Doing this would give. However, we will have problems with this. As we will see, when we plug our guess into the differential equation we will only get two equations out of this. So, we will use the following for our guess.
Following this rule we will get two terms when we collect like terms. Now, set coefficients equal. This last example illustrated the general rule that we will follow when products involve an exponential. When a product involves an exponential we will first strip out the exponential and write down the guess for the portion of the function without the exponential, then we will go back and tack on the exponential without any leading coefficient.
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. The method of undetermined coefficients is used to solve a class of nonhomogeneous second order differential equations. This method makes use of.
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.