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The Method of Undetermined Coefficients

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❶This still causes problems however. This time however it is the first term that causes problems and not the second or third.

The Method of Undetermined Coefficients




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 the interest of brevity we will just write down the guess for a particular solution and not go through all the details of finding the constants. So, we have an exponential in the function. One final note before we move onto the next part. The 16 in front of the function has absolutely no bearing on our guess.

Any constants multiplying the whole function are ignored. We will start this one the same way that we initially started the previous example. The guess for the polynomial is. Notice that everywhere one of the unknown constants occurs it is in a product of unknown constants. This means that if we went through and used this as our guess the system of equations that we would need to solve for the unknown constants would have products of the unknowns in them. These types of systems are generally very difficult to solve.

So, to avoid this we will do the same thing that we did in the previous example. Everywhere we see a product of constants we will rename it and call it a single constant. This is a general rule that we will use when faced with a product of a polynomial and a trig function. We write down the guess for the polynomial and then multiply that by a cosine.

We then write down the guess for the polynomial again, using different coefficients, and multiply this by a sine. This final part has all three parts to it. First, we will ignore the exponential and write down a guess for.

Writing down the guesses for products is usually not that difficult. The difficulty arises when you need to actually find the constants. This fact can be used to both find particular solutions to differential equations that have sums in them and to write down guess for functions that have sums in them.

There is nothing to do with this problem. 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.

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. 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.


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The class of \(g(t)\)’s for which the method works, does include some of the more common functions, however, there are many functions out there for which undetermined coefficients simply won’t work. Second, it is generally only useful for constant coefficient differential equations. The method is .

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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).

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The central idea of the method of undetermined coefficients is this: Form the most general linear combination of the functions in the family of the nonhomogeneous term d(x), substitute this expression into the given nonhomogeneous differential equation, and solve for . 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]. More precisely, we assume that g(x) = p(x)eαx (acos(ωx)+bsin(ωx)), where p(x) = c 0 +c 1x+c 2x2 +···+c mxm. Basic Guess.

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Method of Undetermined Coefficients For a linear non-homogeneous ordinary differential equation with constant coefficients where are all constants and, the non-homogeneous term sometimes contains only linear combinations or multiples of some simple functions whose . Method of Undetermined Coefficients The Method of Undetermined Coefficients (sometimes referred to as the method of Judicious Guessing) is a systematic way (almost, but not quite, like using “educated guesses”) to determine the general form/type of the particular solution Y(t) based on the nonhomogeneous term g(t) in the given equation.