Differential Equations - Second Order: Undetermined Coefficients - iii

Example 9 Find a particular solution for the following differential equation.$$y^{\prime \prime }-4y^{\prime }-12y=e^{6t}$$

This problem seems almost too simple to be given this late in the section. This is especially true given the ease of finding a particular solution for $g$($t$)’s that are just exponential functions. Also, because the point of this example is to illustrate why it is generally a good idea to have the complementary solution in hand first we’ll let’s go ahead and recall the complementary solution first. Here it is,
$$y_c\left(t\right)=c_1{e}^{-2t}+c_2{e}^{6t}$$
Now, without worrying about the complementary solution for a couple more seconds let’s go ahead and get to work on the particular solution. There is not much to the guess here. From our previous work we know that the guess for the particular solution should be,
$$Y_P\left(t\right)=A{e}^{6t}$$Plugging this into the differential equation gives,
$$\begin{array}{cc}36A{e}^{6t}-24A{e}^{6t}-12A{e}^{6t}& =e^{6t}\\ 0& =e^{6t}\end{array}$$Hmmmm…. Something seems wrong here. Clearly an exponential can’t be zero. 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,
$$y^{\prime \prime }-4y^{\prime }-12y=0$$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, how do we fix this? The way that we fix this is to add a $t$ to our guess as follows.
$$Y_P\left(t\right)=At{e}^{6t}$$Plugging this into our differential equation gives,
$$\begin{array}{cc}\left(12A{e}^{6t}+36At{e}^{6t}\right)-4\left(A{e}^{6t}+6At{e}^{6t}\right)-12At{e}^{6t}& =e^{6t}\\ \left(36A-24A-12A\right)t{e}^{6t}+\left(12A-4A\right)e^{6t}& =e^{6t}\\ 8A{e}^{6t}& =e^{6t}\end{array}$$Now, we can set coefficients equal.
$$8A=1\Rightarrow A=\frac{1}{8}$$So, the particular solution in this case is,
$$Y_P\left(t\right)=\frac{t}{8}{e}^{6t}$$

So, what did we learn from this last example. While technically we don’t need the complementary solution to do undetermined coefficients, you can go through a lot of work only to figure out at the end that you needed to add in a $t$ to the guess because it appeared in the complementary solution. 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.
If a portion of your guess does show up in the complementary solution then we’ll need to modify that portion of the guess by adding in a $t$ to the portion of the guess that is causing the problems. We do need to be a little careful and make sure that we add the $t$ in the correct place however. The following set of examples will show you how to do this.

Example 10 Write down the guess for the particular solution to the given differential equation. Do not find the coefficients.

1. $y^{\prime \prime }+3y^{\prime }-28y=7t+e^{-7t}-1$
2. $y^{\prime \prime }-100y=9t^2{e}^{10t}+\cos t-t\sin t$
3. $4y^{\prime \prime }+y=e^{-2t}\sin \left(\frac{t}{2}\right)+6t\cos \left(\frac{t}{2}\right)$
4. $4y^{\prime \prime }+16y^{\prime }+17y=e^{-2t}\sin \left(\frac{t}{2}\right)+6t\cos \left(\frac{t}{2}\right)$
5. $y^{\prime \prime }+8y^{\prime }+16y=e^{-4t}+\left(t^2+5\right)e^{-4t}$

In these solutions we’ll leave the details of checking the complementary solution to you.

a) $y^{\prime \prime }+3y^{\prime }-28y=7t+e^{-7t}-1$ 

The complementary solution is
$$y_c\left(t\right)=c_1{e}^{4t}+c_2{e}^{-7t}$$Remembering to put the “-1” with the 7$t$ gives a first guess for the particular solution.
$$Y_P\left(t\right)=At+B+C{e}^{-7t}$$Notice that the last term in the guess is the last term in the complementary solution. The first two terms however aren’t a problem and don’t appear in the complementary solution. Therefore, we will only add a $t$ onto the last term.
The correct guess for the form of the particular solution is.
$$Y_P\left(t\right)=At+B+Ct{e}^{-7t}$$

b) $y^{\prime \prime }-100y=9t^2{e}^{10t}+\cos t-t\sin t$ 

The complementary solution is
$$y_c\left(t\right)=c_1{e}^{10t}+c_2{e}^{-10t}$$A first guess for the particular solution is
$$Y_P\left(t\right)=\left(A{t}^2+Bt+C\right)e^{10t}+\left(Et+F\right)\cos t+\left(Gt+H\right)\sin t$$Notice that if we multiplied the exponential term through the parenthesis that we would end up getting part of the complementary solution showing up. Since the problem part arises from the first term the whole first term will get multiplied by $t$. The second and third terms are okay as they are.
The correct guess for the form of the particular solution in this case is.
$$Y_P\left(t\right)=t\left(A{t}^2+Bt+C\right)e^{10t}+\left(Et+F\right)\cos t+\left(Gt+H\right)\sin t$$So, in general, if you were to multiply out a guess and if any term in the result shows up in the complementary solution, then the whole term will get a $t$ not just the problem portion of the term.

c) $4y^{\prime \prime }+y=e^{-2t}\sin \left(\frac{t}{2}\right)+6t\cos \left(\frac{t}{2}\right)$ 

The complementary solution is
$$y_c\left(t\right)=c_1\cos \left(\frac{t}{2}\right)+c_2\sin \left(\frac{t}{2}\right)$$A first guess for the particular solution is
$$Y_P\left(t\right)=e^{-2t}\left(A\cos \left(\frac{t}{2}\right)+B\sin \left(\frac{t}{2}\right)\right)+\left(Ct+D\right)\cos \left(\frac{t}{2}\right)+\left(Et+F\right)\sin \left(\frac{t}{2}\right)$$In this case both the second and third terms contain portions of the complementary solution. The first term doesn’t however, since upon multiplying out, both the sine and the cosine would have an exponential with them and that isn’t part 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.
So, in this case the second and third terms will get a $t$ while the first won’t
The correct guess for the form of the particular solution is.
$$Y_P\left(t\right)=e^{-2t}\left(A\cos \left(\frac{t}{2}\right)+B\sin \left(\frac{t}{2}\right)\right)+t\left(Ct+D\right)\cos \left(\frac{t}{2}\right)+t\left(Et+F\right)\sin \left(\frac{t}{2}\right)$$

d) $4y^{\prime \prime }+16y^{\prime }+17y=e^{-2t}\sin \left(\frac{t}{2}\right)+6t\cos \left(\frac{t}{2}\right)$ 

To get this problem we changed the differential equation from the last example and left the $g\left(t\right)$ alone. The complementary solution this time is
$$y_c\left(t\right)=c_1{e}^{-2t}\cos \left(\frac{t}{2}\right)+c_2{e}^{-2t}\sin \left(\frac{t}{2}\right)$$As with the last part, a first guess for the particular solution is
$$Y_P\left(t\right)=e^{-2t}\left(A\cos \left(\frac{t}{2}\right)+B\sin \left(\frac{t}{2}\right)\right)+\left(Ct+D\right)\cos \left(\frac{t}{2}\right)+\left(Et+F\right)\sin \left(\frac{t}{2}\right)$$This time however it is the first term that causes problems and not the second or third. In fact, the first term is exactly the complementary solution and so it will need a $t$. 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. The second and third terms in our guess don’t have the exponential in them and so they don’t differ from the complementary solution by only a constant.
The correct guess for the form of the particular solution is.
$$Y_P\left(t\right)=t{e}^{-2t}\left(A\cos \left(\frac{t}{2}\right)+B\sin \left(\frac{t}{2}\right)\right)+\left(Ct+D\right)\cos \left(\frac{t}{2}\right)+\left(Et+F\right)\sin \left(\frac{t}{2}\right)$$

e) $y^{\prime \prime }+8y^{\prime }+16y=e^{-4t}+\left(t^2+5\right)e^{-4t}$ 

The complementary solution is
$$y_c\left(t\right)=c_1{e}^{-4t}+c_{2}t{e}^{-4t}$$The two terms in $g\left(t\right)$ are identical with the exception of a polynomial in front of them. 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
$$Y_P\left(t\right)=\left(A{t}^2+Bt+C\right)e^{-4t}$$Notice that if we multiplied the exponential term through the parenthesis the last two terms would be the complementary solution. Therefore, we will need to multiply this whole thing by a $t$.
The next guess for the particular solution is then.
$$Y_P\left(t\right)=t\left(A{t}^2+Bt+C\right)e^{-4t}$$This still causes problems however. If we multiplied the $t$ and the exponential through, the last term will still be in the complementary solution. In this case, unlike the previous ones, a $t$ wasn’t sufficient to fix the problem. So, we will add in another $t$ to our guess.
The correct guess for the form of the particular solution is.
$$Y_P\left(t\right)=t^2\left(A{t}^2+Bt+C\right)e^{-4t}$$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. By doing this we can compare our guess to the complementary solution and if any of the terms from your particular solution show up we will know that we’ll have problems. Once the problem is identified we can add a $t$ to the problem term(s) and compare our new guess to the complementary solution. If there are no problems we can proceed with the problem, if there are problems add in another $t$ and compare again.



Can you see a general rule as to when a $t$ will be needed and when a t² will be needed for second order differential equations?