Differential Equations - Second Order: Undetermined Coefficients - i

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.
$$y^{\prime \prime }+p\left(t\right)y^{\prime }+q\left(t\right)y=g\left(t\right)$$

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. Another nice thing about this method is that the complementary solution will not be explicitly required, although as we will see knowledge of the complementary solution will be needed in some cases and so we’ll generally find that as well.

There are two disadvantages to this method. First, it will only work for a fairly small class of $g\left(t\right)$’s. The class of $g\left(t\right)$’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 quite simple. All that we need to do is look at $g\left(t\right)$ and make a guess as to the form of $Y_P\left(t\right)$ leaving the coefficient(s) undetermined (and hence the name of the method). Plug the guess into the differential equation and see if we can determine values of the coefficients. If we can determine values for the coefficients then we guessed correctly, if we can’t find values for the coefficients then we guessed incorrectly.

It’s usually easier to see this method in action rather than to try and describe it, so let’s jump into some examples.

Example 1 Determine a particular solution to$$y^{\prime \prime }-4y^{\prime }-12y=3e^{5t}$$

The point here is to find a particular solution, however the first thing that we’re going to do is find the complementary solution to this differential equation. Recall that the complementary solution comes from solving,
$$y^{\prime \prime }-4y^{\prime }-12y=0$$The characteristic equation for this differential equation and its roots are.
$$r^2-4r-12=\left(r-6\right)\left(r+2\right)=0\Rightarrow r_1=-2,r_2=6$$The complementary solution is then,
$$y_c\left(t\right)=c_1{e}^{-2t}+c_2{e}^{6t}$$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. Eventually, as we’ll see, having the complementary solution in hand will be helpful and so it’s best to be in the habit of finding it first prior to doing the work for undetermined coefficients.
Now, let’s proceed with finding 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. Since $g\left(t\right)$ is an exponential and we know that exponentials never just appear or disappear in the differentiation process it seems that a likely form of the particular solution would be
$$Y_P\left(t\right)=A{e}^{5t}$$Now, all that we need to do is do a couple of derivatives, plug this into the differential equation and see if we can determine what $A$ needs to be.
Plugging into the differential equation gives
$$\begin{array}{cc}25A{e}^{5t}-4\left(5A{e}^{5t}\right)-12\left(A{e}^{5t}\right)& =3e^{5t}\\ -7A{e}^{5t}& =3e^{5t}\end{array}$$So, in order for our guess to be a solution we will need to choose $A$ so that the coefficients of the exponentials on either side of the equal sign are the same. In other words we need to choose $A$ so that,
$$-7A=3\Rightarrow A=-\frac{3}{7}$$Okay, we found a value for the coefficient. This means that we guessed correctly. A particular solution to the differential equation is then,
$$Y_P\left(t\right)=-\frac{3}{7}{e}^{5t}$$

Before proceeding any further let’s again note that we started off the solution above by finding the complementary solution. This is not technically part the method of Undetermined Coefficients however, as we’ll eventually see, having this in hand before we make our guess for the particular solution can save us a lot of work and/or headache. Finding the complementary solution first is simply a good habit to have so we’ll try to get you in the habit over the course of the next few examples. At this point do not worry about why it is a good habit. We’ll eventually see why it is a good habit.

Now, back to the work at hand. Notice in the last example that we kept saying “a” particular solution, not “the” particular solution. This is because there are other possibilities out there for the particular solution we’ve just managed to find one of them. 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. However, we should do at least one full blown IVP to make sure that we can say that we’ve done one.

Example 2 Solve the following IVP$$y^{\prime \prime }-4y^{\prime }-12y=3e^{5t}y\left(0\right)=\frac{18}{7}{y}^{\prime }\left(0\right)=-\frac{1}{7}$$

We know that the general solution will be of the form,
$$y\left(t\right)=y_c\left(t\right)+Y_P\left(t\right)$$and we already have both the complementary and particular solution from the first example so we don’t really need to do any extra work for this problem.
One of the more common mistakes in these problems is to find the complementary solution and then, because we’re probably in the habit of doing it, apply the initial conditions to the complementary solution to find the constants. 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.
$$\begin{array}{cc}y\left(t\right)& =c_1{e}^{-2t}+c_2{e}^{6t}-\frac{3}{7}{e}^{5t}\\ y^{\prime }\left(t\right)& =-2c_1{e}^{-2t}+6c_2{e}^{6t}-\frac{15}{7}{e}^{5t}\end{array}$$Now, apply the initial conditions to these.
$$\begin{array}{cc}\frac{18}{7}=y\left(0\right)& =c_1+c_2-\frac{3}{7}\\ -\frac{1}{7}=y^{\prime }\left(0\right)& =-2c_1+6c_2-\frac{15}{7}\end{array}$$Solving this system gives $c_1=2$ and $c_2=1$. The actual solution is then.
$$y\left(t\right)=2e^{-2t}+e^{6t}-\frac{3}{7}{e}^{5t}$$

This will be the only IVP in this section so don’t forget how these are done for nonhomogeneous differential equations!

Let’s take a look at another example that will give the second type of $g\left(t\right)$ for which undetermined coefficients will work.

Example 3 Find a particular solution for the following differential equation.$$y^{\prime \prime }-4y^{\prime }-12y=\sin \left(2t\right)$$

Again, let’s note that we should probably find the complementary solution before we proceed onto the guess for a particular solution. However, because the homogeneous differential equation for this example is the same as that for the first example we won’t bother with that here.
Now, let’s take our experience from the first example and apply that here. The first example had an exponential function in the $g\left(t\right)$ and our guess was an exponential. This differential equation has a sine so let’s try the following guess for the particular solution.
$$Y_P\left(t\right)=A\sin \left(2t\right)$$Differentiating and plugging into the differential equation gives,
$$-4A\sin \left(2t\right)-4\left(2A\cos \left(2t\right)\right)-12\left(A\sin \left(2t\right)\right)=\sin \left(2t\right)$$Collecting like terms yields
$$-16A\sin \left(2t\right)-8A\cos \left(2t\right)=\sin \left(2t\right)$$We need to pick $A$ so that we get the same function on both sides of the equal sign. This means that the coefficients of the sines and cosines must be equal. Or,
$$\begin{array}{cccc}& \cos \left(2t\right):& -8A& =0\Rightarrow A=0\\ & \sin \left(2t\right):& -16A& =1\Rightarrow A=-\frac{1}{16}\end{array}$$Notice two things. 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. In order for the cosine to drop out, as it must in order for the guess to satisfy the differential equation, we need to set $A=0$, but if $A=0$, the sine will also drop out and that can’t happen. Likewise, choosing $A$ to keep the sine around will also keep the cosine around.
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. So, to counter this let’s add a cosine to our guess. Our new guess is
$$Y_P\left(t\right)=A\cos \left(2t\right)+B\sin \left(2t\right)$$Plugging this into the differential equation and collecting like terms gives,
$$\begin{array}{cc}-4A\cos \left(2t\right)-4B\sin \left(2t\right)-4\left(-2A\sin \left(2t\right)+2B\cos \left(2t\right)\right)-& \\ 12\left(A\cos \left(2t\right)+B\sin \left(2t\right)\right)& =\sin \left(2t\right)\\ \left(-4A-8B-12A\right)\cos \left(2t\right)+\left(-4B+8A-12B\right)\sin \left(2t\right)& =\sin \left(2t\right)\\ \left(-16A-8B\right)\cos \left(2t\right)+\left(8A-16B\right)\sin \left(2t\right)& =\sin \left(2t\right)\end{array}$$Now, set the coefficients equal
$$\begin{array}{cccc}& \cos \left(2t\right):& -16A-8B& =0\\ & \sin \left(2t\right):& 8A-16B& =1\end{array}$$Solving this system gives us
$$A=\frac{1}{40}B=-\frac{1}{20}$$We found constants and this time we guessed correctly. A particular solution to the differential equation is then,
$$Y_P\left(t\right)=\frac{1}{40}\cos \left(2t\right)-\frac{1}{20}\sin \left(2t\right)$$

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.

Let’s take a look at the third and final type of basic $g\left(t\right)$ that we can have. There are other types of $g\left(t\right)$ that we can have, but as we will see they will all come back to two types that we’ve already done as well as the next one.

Example 4 Find a particular solution for the following differential equation.$$y^{\prime \prime }-4y^{\prime }-12y=2t^3-t+3$$

Once, again we will generally want the complementary solution in hand first, but again we’re working with the same homogeneous differential equation (you’ll eventually see why we keep working with the same homogeneous problem) so we’ll again just refer to the first example.
For this example, $g\left(t\right)$ is a cubic polynomial. For this we will need the following guess for the particular solution.
$$Y_P\left(t\right)=A{t}^3+B{t}^2+Ct+D$$Notice that even though $g\left(t\right)$ doesn’t have a $t^2$ in it our guess will still need one! So, differentiate and plug into the differential equation.
$$\begin{array}{cc}6At+2B-4\left(3A{t}^2+2Bt+C\right)-12\left(A{t}^3+B{t}^2+Ct+D\right)& =2t^3-t+3\\ -12A{t}^3+\left(-12A-12B\right)t^2+\left(6A-8B-12C\right)t+2B-4C-12D& =2t^3-t+3\end{array}$$Now, as we’ve done in the previous examples we will need the coefficients of the terms on both sides of the equal sign to be the same so set coefficients equal and solve.
$$\begin{array}{cccccc}& t^3:& -12A& =2& \Rightarrow A& =-\frac{1}{6}\\ & t^2:& -12A-12B& =0& \Rightarrow B& =\frac{1}{6}\\ & t^1:& 6A-8B-12C& =-1& \Rightarrow C& =-\frac{1}{9}\\ & t^0:& 2B-4C-12D& =3& \Rightarrow D& =-\frac{5}{27}\end{array}$$Notice that in this case it was very easy to solve for the constants. The first equation gave $A$. Then once we knew $A$ the second equation gave $B$, etc. A particular solution for this differential equation is then
$$Y_P\left(t\right)=-\frac{1}{6}{t}^3+\frac{1}{6}{t}^2-\frac{1}{9}t-\frac{5}{27}$$

Now that we’ve gone over the three basic kinds of functions that we can use undetermined coefficients on let’s summarize.

$g\left(t\right)$	$Y_P\left(t\right)$ guess
$a{e}^{\beta t}$	$A{e}^{\beta t}$
$a\cos \left(\beta t\right)$	$A\cos \left(\beta t\right)+B\sin \left(\beta t\right)$
$b\sin \left(\beta t\right)$	$A\cos \left(\beta t\right)+B\sin \left(\beta t\right)$
$a\cos \left(\beta t\right)+b\sin \left(\beta t\right)$	$A\cos \left(\beta t\right)+B\sin \left(\beta t\right)$
$n^{\text{th}}$ degree polynomial	$A_n{t}^n+A_{n-1}{t}^{n-1}+\cdots A_{1}t+A_0$

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. Let’s first look at products.

Example 5 Find a particular solution for the following differential equation.$$y^{\prime \prime }-4y^{\prime }-12y=t{e}^{4t}$$

You’re probably getting tired of the opening comment, but again finding the complementary solution first really a good idea but again we’ve already done the work in the first example so we won’t do it again here. We promise that eventually you’ll see why we keep using the same homogeneous problem and why we say it’s a good idea to have the complementary solution in hand first. At this point all we’re trying to do is reinforce the habit of finding the complementary solution first.
Okay, let’s start off by writing down the guesses for the individual pieces of the function. The guess for the $t$ would be
$$At+B$$while the guess for the exponential would be
$$C{e}^{4t}$$Now, since we’ve got a product of two functions it seems like taking a product of the guesses for the individual pieces might work. Doing this would give
$$C{e}^{4t}\left(At+B\right)$$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. The problem is that with this guess we’ve got three unknown constants. With only two equations we won’t be able to solve for all the constants.
This is easy to fix however. Let’s notice that we could do the following
$$C{e}^{4t}\left(At+B\right)=e^{4t}\left(ACt+BC\right)$$If we multiply the $C$ through, we can see that the guess can be written in such a way that there are really only two constants. So, we will use the following for our guess.
$$Y_P\left(t\right)=e^{4t}\left(At+B\right)$$Notice that this is nothing more than the guess for the $t$ with an exponential tacked on for good measure.
Now that we’ve got our guess, let’s differentiate, plug into the differential equation and collect like terms.
$$\begin{array}{cc}{e}^{4t}\left(16At+16B+8A\right)-4\left(e^{4t}\left(4At+4B+A\right)\right)-12\left(e^{4t}\left(At+B\right)\right)& =t{e}^{4t}\\ \left(16A-16A-12A\right)t{e}^{4t}+\left(16B+8A-16B-4A-12B\right)e^{4t}& =t{e}^{4t}\\ -12At{e}^{4t}+\left(4A-12B\right)e^{4t}& =t{e}^{4t}\end{array}$$Note that when we’re collecting like terms we want the coefficient of each term to have only constants in it. Following this rule we will get two terms when we collect like terms. Now, set coefficients equal.
$$\begin{array}{cccccc}& t{e}^{4t}:& -12A& =1& \Rightarrow A& =-\frac{1}{12}\\ & e^{4t}:& 4A-12B& =0& \Rightarrow B& =-\frac{1}{36}\end{array}$$A particular solution for this differential equation is then
$$Y_P\left(t\right)=e^{4t}\left(-\frac{t}{12}-\frac{1}{36}\right)=-\frac{1}{36}\left(3t+1\right)e^{4t}$$

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.

Let’s take a look at some more products. 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. Also, because we aren’t going to give an actual differential equation we can’t deal with finding the complementary solution first.