Differential Equations - Second Order: Undetermined Coefficients - ii

Example 6 Write down the form of the particular solution to$$y^{\prime \prime }+p\left(t\right)y^{\prime }+q\left(t\right)y=g\left(t\right)$$

1. $g\left(t\right)=16e^{7t}\sin \left(10t\right)$
2. $g\left(t\right)=\left(9t^2-103t\right)\cos t$
3. $g\left(t\right)=-e^{-2t}\left(3-5t\right)\cos \left(9t\right)$

a) $g\left(t\right)=16e^{7t}\sin \left(10t\right)$ 

So, we have an exponential in the function. Remember the rule. We will ignore the exponential and write down a guess for $16\sin \left(10t\right)$ then put the exponential back in.
The guess for the sine is
$$A\cos \left(10t\right)+B\sin \left(10t\right)$$Now, for the actual guess for the particular solution we’ll take the above guess and tack an exponential onto it. This gives,
$$Y_P\left(t\right)=e^{7t}\left(A\cos \left(10t\right)+B\sin \left(10t\right)\right)$$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.

b) $g\left(t\right)=\left(9t^2-103t\right)\cos t$ 

We will start this one the same way that we initially started the previous example. The guess for the polynomial is
$$A{t}^2+Bt+C$$and the guess for the cosine is
$$D\cos t+E\sin t$$If we multiply the two guesses we get.
$$\left(A{t}^2+Bt+C\right)\left(D\cos t+E\sin t\right)$$Let’s simplify things up a little. First multiply the polynomial through as follows.
$$\begin{array}{c}\left(A{t}^2+Bt+C\right)\left(D\cos t\right)+\left(A{t}^2+Bt+C\right)\left(E\sin t\right)\\ \left(AD{t}^2+BDt+CD\right)\cos t+\left(AE{t}^2+BEt+CE\right)\sin t\end{array}$$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. The guess that we’ll use for this function will be.
$$Y_P\left(t\right)=\left(A{t}^2+Bt+C\right)\cos t+\left(D{t}^2+Et+F\right)\sin t$$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.

c) $g\left(t\right)=-e^{-2t}\left(3-5t\right)\cos \left(9t\right)$ 

This final part has all three parts to it. First, we will ignore the exponential and write down a guess for.
$$-\left(3-5t\right)\cos \left(9t\right)$$The minus sign can also be ignored. The guess for this is
$$\left(At+B\right)\cos \left(9t\right)+\left(Ct+D\right)\sin \left(9t\right)$$Now, tack an exponential back on and we’re done.
$$Y_P\left(t\right)=e^{-2t}\left(At+B\right)\cos \left(9t\right)+e^{-2t}\left(Ct+D\right)\sin \left(9t\right)$$Notice that we put the exponential on both terms.

There a couple of general rules that you need to remember for products.

1. If $g\left(t\right)$ contains an exponential, ignore it and write down the guess for the remainder. Then tack the exponential back on without any leading coefficient.
2. For products of polynomials and trig functions you first write down the guess for just the polynomial and multiply that by the appropriate cosine. Then add on a new guess for the polynomial with different coefficients and multiply that by the appropriate sine.

If you can remember these two rules you can’t go wrong with products. Writing down the guesses for products is usually not that difficult. The difficulty arises when you need to actually find the constants.

Now, let’s take a look at sums of the basic components and/or products of the basic components. To do this we’ll need the following fact.

Fact

If $Y_{P1}\left(t\right)$ is a particular solution for
$$y^{\prime \prime }+p\left(t\right)y^{\prime }+q\left(t\right)y=g_1\left(t\right)$$and if $Y_{P2}\left(t\right)$ is a particular solution for
$$y^{\prime \prime }+p\left(t\right)y^{\prime }+q\left(t\right)y=g_2\left(t\right)$$then $Y_{P1}\left(t\right)$ + $Y_{P2}\left(t\right)$ is a particular solution for
$$y^{\prime \prime }+p\left(t\right)y^{\prime }+q\left(t\right)y=g_1\left(t\right)+g_2\left(t\right)$$

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.

Example 7 Find a particular solution for the following differential equation.$$y^{\prime \prime }-4y^{\prime }-12y=3e^{5t}+\sin \left(2t\right)+t{e}^{4t}$$

This example is the reason that we’ve been using the same homogeneous differential equation for all the previous examples. 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.
Doing this gives
$$Y_P\left(t\right)=-\frac{3}{7}{e}^{5t}+\frac{1}{40}\cos \left(2t\right)-\frac{1}{20}\sin \left(2t\right)-\frac{1}{36}\left(3t+1\right)e^{4t}$$

Let’s take a look at a couple of other examples. As with the products we’ll just get guesses here and not worry about actually finding the coefficients.

Example 8 Write down the form of the particular solution to$$y^{\prime \prime }+p\left(t\right)y^{\prime }+q\left(t\right)y=g\left(t\right)$$

for the following $g\left(t\right)$’s.

1. $g\left(t\right)=4\cos \left(6t\right)-9\sin \left(6t\right)$
2. $g\left(t\right)=-2\sin t+\sin \left(14t\right)-5\cos \left(14t\right)$
3. $g\left(t\right)=e^{7t}+6$
4. $g\left(t\right)=6t^2-7\sin \left(3t\right)+9$
5. $g\left(t\right)=10e^t-5t{e}^{-8t}+2e^{-8t}$
6. $g\left(t\right)=t^2\cos t-5t\sin t$
7. $g\left(t\right)=5e^{-3t}+e^{-3t}\cos \left(6t\right)-\sin \left(6t\right)$

a) $g\left(t\right)=4\cos \left(6t\right)-9\sin \left(6t\right)$ 

This first one we’ve actually already told you how to do. 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. This would give.
$$\underset{\text{guess for the cosine}}{\underset{}{A\cos \left(6t\right)+B\sin \left(6t\right)}}+\underset{\text{guess for the sine}}{\underset{}{C\cos \left(6t\right)+D\sin \left(6t\right)}}$$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.
To fix this notice that we can combine some terms as follows.
$$\left(A+C\right)\cos \left(6t\right)+\left(B+D\right)\sin \left(6t\right)$$Upon doing this we can see that we’ve really got a single cosine with a coefficient and a single sine with a coefficient and so we may as well just use
$$Y_P\left(t\right)=A\cos \left(6t\right)+B\sin \left(6t\right)$$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.

b) $g\left(t\right)=-2\sin t+\sin \left(14t\right)-5\cos \left(14t\right)$ 

For this one we will get two sets of sines and cosines. This will arise because we have two different arguments in them. We will get one set for the sine with just a $t$ as its argument and we’ll get another set for the sine and cosine with the 14$t$ as their arguments.
The guess for this function is
$$Y_P\left(t\right)=A\cos t+B\sin t+C\cos \left(14t\right)+D\sin \left(14t\right)$$

c) $g\left(t\right)=e^{7t}+6$ 

The main point of this problem is dealing with the constant. But that isn’t too bad. 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.
The guess for this function is
$$Y_p\left(t\right)=A{e}^{7t}+B$$

d) $g\left(t\right)=6t^2-7\sin \left(3t\right)+9$ 

This one can be a little tricky if you aren’t paying attention. Let’s first rewrite the function
$$\begin{array}{cc}g\left(t\right)& =6t^2-7\sin \left(3t\right)+9\text{as}\\ g\left(t\right)& =6t^2+9-7\sin \left(3t\right)\end{array}$$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
$$Y_P\left(t\right)=A{t}^2+Bt+C+D\cos \left(3t\right)+E\sin \left(3t\right)$$If we don’t do this and treat the function as the sum of three terms we would get
$$A{t}^2+Bt+C+D\cos \left(3t\right)+E\sin \left(3t\right)+G$$and as with the first part in this example we would end up with two terms that are essentially the same (the $C$ and the $G$) and so would need to be combined. An added step that isn’t really necessary if we first rewrite the function.

Look for problems where rearranging the function can simplify the initial guess.

e) $g\left(t\right)=10e^t-5t{e}^{-8t}+2e^{-8t}$

So, this look like we’ve got a sum of three terms here. Let’s write down a guess for that.
$$A{e}^t+\left(Bt+C\right)e^{-8t}+D{e}^{-8t}$$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. When this happens we just drop the guess that’s already included in the other term.


So, the guess here is actually.
$$Y_P\left(t\right)=A{e}^t+\left(Bt+C\right)e^{-8t}$$Notice that this arose because we had two terms in our $g\left(t\right)$ whose only difference was the polynomial that sat in front of them. When this happens we look at the term that contains the largest degree polynomial, write down the guess for that and don’t bother writing down the guess for the other term as that guess will be completely contained in the first guess.

f) $g\left(t\right)=t^2\cos t-5t\sin t$ 

In this case we’ve got two terms whose guess without the polynomials in front of them would be the same. 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
$$Y_P\left(t\right)=\left(A{t}^2+Bt+C\right)\cos t+\left(D{t}^2+Et+F\right)\sin t$$

g) $g\left(t\right)=5e^{-3t}+e^{-3t}\cos \left(6t\right)-\sin \left(6t\right)$  

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
$$Y_P\left(t\right)=A{e}^{-3t}+e^{-3t}\left(B\cos \left(6t\right)+C\sin \left(6t\right)\right)+D\cos \left(6t\right)+E\sin \left(6t\right)$$We can only combine guesses if they are identical up to the constant. So, we can’t combine the first exponential with the second because the second is really multiplied by a cosine and a sine and so the two exponentials are in fact different functions. Likewise, the last sine and cosine can’t be combined with those in the middle term because the sine and cosine in the middle term are in fact multiplied by an exponential and so are different.

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. We never gave any reason for this other that “trust us”. It is now time to see why having the complementary solution in hand first is useful. This is best shown with an example so let’s jump into one.