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,

\[y'' - 4y' - 12y = 0\]

The characteristic equation for this differential equation and its roots are.

\[{r^2} - 4r - 12 = \left( {r - 6} \right)\left( {r + 2} \right) = 0\hspace{0.25in} \Rightarrow \hspace{0.25in}\,\,{r_1} = - 2,\,\,\,\,{r_2} = 6\]

The complementary solution is then,

\[{y_c}\left( t \right) = {c_1}{{\bf{e}}^{ - 2t}} + {c_2}{{\bf{e}}^{6t}}\]

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 \(g(t)\) 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

\[{Y_P}\left( t \right) = A{{\bf{e}}^{5t}}\]

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

\[\begin{align*}25A{{\bf{e}}^{5t}} - 4\left( {5A{{\bf{e}}^{5t}}} \right) - 12\left( {A{{\bf{e}}^{5t}}} \right) & = 3{{\bf{e}}^{5t}}\\ - 7A{{\bf{e}}^{5t}} & = 3{{\bf{e}}^{5t}}\end{align*}\]

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,

\[ - 7A = 3\hspace{0.25in} \Rightarrow \hspace{0.25in}A = - \frac{3}{7}\]

Okay, we found a value for the coefficient. This means that we guessed correctly. A particular solution to the differential equation is then,

\[{Y_P}\left( t \right) = - \frac{3}{7}{{\bf{e}}^{5t}}\]

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.

\[\begin{align*}y\left( t \right) & = {c_1}{{\bf{e}}^{ - 2t}} + {c_2}{{\bf{e}}^{6t}} - \frac{3}{7}{{\bf{e}}^{5t}}\\ y'\left( t \right) & = - 2{c_1}{{\bf{e}}^{ - 2t}} + 6{c_2}{{\bf{e}}^{6t}} - \frac{{15}}{7}{{\bf{e}}^{5t}}\end{align*}\]

Now, apply the initial conditions to these.

\[\begin{align*}\frac{{18}}{7} = y\left( 0 \right) & = {c_1} + {c_2} - \frac{3}{7}\\ - \frac{1}{7} = y'\left( 0 \right) & = - 2{c_1} + 6{c_2} - \frac{{15}}{7}\end{align*}\]

Solving this system gives \(c_{1} = 2\) and \(c_{2} = 1\). The actual solution is then.

\[y\left( t \right) = 2{{\bf{e}}^{ - 2t}} + {{\bf{e}}^{6t}} - \frac{3}{7}{{\bf{e}}^{5t}}\]

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 \(g(t)\) 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,

\[\begin{align*}& \cos \left( {2t} \right)\,: & - 8A & = 0\hspace{0.25in} \Rightarrow \hspace{0.25in}A = 0\\ & \sin \left( {2t} \right)\,:& - 16A & = 1\hspace{0.25in} \Rightarrow \hspace{0.25in}A = - \frac{1}{{16}}\end{align*}\]

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,

\[\begin{align*} - 4A\cos \left( {2t} \right) - 4B\sin \left( {2t} \right) - 4\left( { - 2A\sin \left( {2t} \right) + 2B\cos \left( {2t} \right)} \right) - \\ 12\left( {A\cos \left( {2t} \right) + B\sin \left( {2t} \right)} \right) & = \sin \left( {2t} \right)\\ \left( { - 4A - 8B - 12A} \right)\cos \left( {2t} \right) + \left( { - 4B + 8A - 12B} \right)\sin \left( {2t} \right) & = \sin \left( {2t} \right)\\ \left( { - 16A - 8B} \right)\cos \left( {2t} \right) + \left( {8A - 16B} \right)\sin \left( {2t} \right) & = \sin \left( {2t} \right)\end{align*}\]

Now, set the coefficients equal

\[\begin{align*} & \cos \left( {2t} \right)\,: &- 16A - 8B & = 0\\ & \sin \left( {2t} \right)\,: & 8A - 16B & = 1\end{align*}\]

Solving this system gives us

\[A = \frac{1}{{40}}\hspace{0.25in}\hspace{0.25in}B = - \frac{1}{{20}}\]

We found constants and this time we guessed correctly. A particular solution to the differential equation is then,

\[{Y_P}\left( t \right) = \frac{1}{{40}}\cos \left( {2t} \right) - \frac{1}{{20}}\sin \left( {2t} \right)\]

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, \(g(t)\) 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 \(g(t)\) doesn’t have a \({t^2}\) in it our guess will still need one! So, differentiate and plug into the differential equation.

\[\begin{align*}6At + 2B - 4\left( {3A{t^2} + 2Bt + C} \right) - 12\left( {A{t^3} + B{t^2} + Ct + D} \right) & = 2{t^3} - t + 3\\ - 12A{t^3} + \left( { - 12A - 12B} \right){t^2} + \left( {6A - 8B - 12C} \right)t + 2B - 4C - 12D & = 2{t^3} - t + 3\end{align*}\]

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.

\[\begin{align*} & {t^3}\,: & - 12A & = 2 & \Rightarrow \hspace{0.25in}A & = - \frac{1}{6}\\ & {t^2}\,: & - 12A - 12B & = 0 & \Rightarrow \hspace{0.25in}B & = \frac{1}{6}\\ & {t^1}\,: & 6A - 8B - 12C & = - 1 & \Rightarrow \hspace{0.25in}C & = - \frac{1}{9}\\ & {t^0}\,: & 2B - 4C - 12D & = 3 & \Rightarrow \hspace{0.25in}D & = - \frac{5}{{27}}\end{align*}\]

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

\[{Y_P}\left( t \right) = - \frac{1}{6}{t^3} + \frac{1}{6}{t^2} - \frac{1}{9}t - \frac{5}{{27}}\]

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

\[C{{\bf{e}}^{4t}}\]

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

\[C{{\bf{e}}^{4t}}\left( {At + B} \right)\]

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

\[C{{\bf{e}}^{4t}}\left( {At + B} \right) = {{\bf{e}}^{4t}}\left( {ACt + BC} \right)\]

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.

\[{Y_P}\left( t \right) = {{\bf{e}}^{4t}}\left( {At + B} \right)\]

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.

\[\begin{align*}{{\bf{e}}^{4t}}\left( {16At + 16B + 8A} \right) - 4\left( {{{\bf{e}}^{4t}}\left( {4At + 4B + A} \right)} \right) - 12\left( {{{\bf{e}}^{4t}}\left( {At + B} \right)} \right) & = t{{\bf{e}}^{4t}}\\ \left( {16A - 16A - 12A} \right)t{{\bf{e}}^{4t}} + \left( {16B + 8A - 16B - 4A - 12B} \right){{\bf{e}}^{4t}} & = t{{\bf{e}}^{4t}}\\ - 12At{{\bf{e}}^{4t}} + \left( {4A - 12B} \right){{\bf{e}}^{4t}} & = t{{\bf{e}}^{4t}}\end{align*}\]

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.

\[\begin{align*} & t{{\bf{e}}^{4t}}\,: & - 12A & = 1 & \Rightarrow \hspace{0.25in}A & = - \frac{1}{{12}}\\ & {{\bf{e}}^{4t}}\,: & 4A - 12B & = 0 & \Rightarrow \hspace{0.25in}B & = - \frac{1}{{36}}\end{align*}\]

A particular solution for this differential equation is then

\[{Y_P}\left( t \right) = {{\bf{e}}^{4t}}\left( { - \frac{t}{{12}} - \frac{1}{{36}}} \right) = - \frac{1}{{36}}\left( {3t + 1} \right){{\bf{e}}^{4t}}\]

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,

\[{Y_P}\left( t \right) = {{\bf{e}}^{7t}}\left( {A\cos \left( {10t} \right) + B\sin \left( {10t} \right)} \right)\]

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.

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.

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.

\[{Y_P}\left( t \right) = {{\bf{e}}^{ - 2t}}\left( {At + B} \right)\cos \left( {9t} \right) + {{\bf{e}}^{ - 2t}}\left( {Ct + D} \right)\sin \left( {9t} \right)\]

Notice that we put the exponential on both terms.

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

\[{Y_P}\left( t \right) = - \frac{3}{7}{{\bf{e}}^{5t}} + \frac{1}{{40}}\cos \left( {2t} \right) - \frac{1}{{20}}\sin \left( {2t} \right) - \frac{1}{{36}}\left( {3t + 1} \right){{\bf{e}}^{4t}}\]

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.

\[\underbrace {A\cos \left( {6t} \right) + B\sin \left( {6t} \right)}_{{\mbox{guess for the cosine}}} + \underbrace {C\cos \left( {6t} \right) + D\sin \left( {6t} \right)}_{{\mbox{guess for the sine}}}\]

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.

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.

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

\[{Y_P}\left( t \right) = A\cos t + B\sin t + C\cos \left( {14t} \right) + D\sin \left( {14t} \right)\]

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

\[{Y_p}\left( t \right) = A{{\bf{e}}^{7t}} + B\]

This one can be a little tricky if you aren’t paying attention. Let’s first rewrite the function

\[\begin{align*}g\left( t \right) & = 6{t^2} - 7\sin \left( {3t} \right) + 9\hspace{0.25in}{\mbox{as}}\\ g\left( t \right) & = 6{t^2} + 9 - 7\sin \left( {3t} \right)\end{align*}\]

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.

So, this look like we’ve got a sum of three terms here. Let’s write down a guess for that.

\[A{{\bf{e}}^t} + \left( {Bt + C} \right){{\bf{e}}^{ - 8t}} + D{{\bf{e}}^{ - 8t}}\]

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.

\[{Y_P}\left( t \right) = A{{\bf{e}}^t} + \left( {Bt + C} \right){{\bf{e}}^{ - 8t}}\]

Notice that this arose because we had two terms in our \(g(t)\) 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.

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

\[{Y_P}\left( t \right) = \left( {A{t^2} + Bt + C} \right)\cos t + \left( {D{t^2} + Et + F} \right)\sin t\]

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

\[{Y_P}\left( t \right) = A{{\bf{e}}^{ - 3t}} + {{\bf{e}}^{ - 3t}}\left( {B\cos \left( {6t} \right) + C\sin \left( {6t} \right)} \right) + D\cos \left( {6t} \right) + E\sin \left( {6t} \right)\]

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.

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,

\[{y_c}\left( t \right) = {c_1}{{\bf{e}}^{ - 2t}} + {c_2}{{\bf{e}}^{6t}}\]

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,

\[{Y_P}\left( t \right) = A{{\bf{e}}^{6t}}\]

Plugging this into the differential equation gives,

\[\begin{align*}36A{{\bf{e}}^{6t}} - 24A{{\bf{e}}^{6t}} - 12A{{\bf{e}}^{6t}} & = {{\bf{e}}^{6t}}\\ 0 & = {{\bf{e}}^{6t}}\end{align*}\]

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,

\[y'' - 4y' - 12y = 0\]

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.

\[{Y_P}\left( t \right) = At{{\bf{e}}^{6t}}\]

Plugging this into our differential equation gives,

\[\begin{align*}\left( {12A{{\bf{e}}^{6t}} + 36At{{\bf{e}}^{6t}}} \right) - 4\left( {A{{\bf{e}}^{6t}} + 6At{{\bf{e}}^{6t}}} \right) - 12At{{\bf{e}}^{6t}} & = {{\bf{e}}^{6t}}\\ \left( {36A - 24A - 12A} \right)t{{\bf{e}}^{6t}} + \left( {12A - 4A} \right){{\bf{e}}^{6t}} & = {{\bf{e}}^{6t}}\\ 8A{{\bf{e}}^{6t}} & = {{\bf{e}}^{6t}}\end{align*}\]

Now, we can set coefficients equal.

\[8A = 1\hspace{0.25in}\hspace{0.25in} \Rightarrow \hspace{0.25in}\,\,\,\,\,A = \frac{1}{8}\]

So, the particular solution in this case is,

\[{Y_P}\left( t \right) = \frac{t}{8}{{\bf{e}}^{6t}}\]

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

The complementary solution is

\[{y_c}\left( t \right) = {c_1}{{\bf{e}}^{4t}} + {c_2}{{\bf{e}}^{ - 7t}}\]

Remembering to put the “-1” with the 7\(t\) gives a first guess for the particular solution.

\[{Y_P}\left( t \right) = At + B + C{{\bf{e}}^{ - 7t}}\]

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.

\[{Y_P}\left( t \right) = At + B + Ct{{\bf{e}}^{ - 7t}}\]

The complementary solution is

\[{y_c}\left( t \right) = {c_1}{{\bf{e}}^{10t}} + {c_2}{{\bf{e}}^{ - 10t}}\]

A first guess for the particular solution is

\[{Y_P}\left( t \right) = \left( {A{t^2} + Bt + C} \right){{\bf{e}}^{10t}} + \left( {Et + F} \right)\cos t + \left( {Gt + H} \right)\sin t\]

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.

\[{Y_P}\left( t \right) = t\left( {A{t^2} + Bt + C} \right){{\bf{e}}^{10t}} + \left( {Et + F} \right)\cos t + \left( {Gt + H} \right)\sin t\]

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.

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

\[{Y_P}\left( t \right) = {{\bf{e}}^{ - 2t}}\left( {A\cos \left( {\frac{t}{2}} \right) + B\sin \left( {\frac{t}{2}} \right)} \right) + \left( {Ct + D} \right)\cos \left( {\frac{t}{2}} \right) + \left( {Et + F} \right)\sin \left( {\frac{t}{2}} \right)\]

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.

\[{Y_P}\left( t \right) = {{\bf{e}}^{ - 2t}}\left( {A\cos \left( {\frac{t}{2}} \right) + B\sin \left( {\frac{t}{2}} \right)} \right) + t\left( {Ct + D} \right)\cos \left( {\frac{t}{2}} \right) + t\left( {Et + F} \right)\sin \left( {\frac{t}{2}} \right)\]

To get this problem we changed the differential equation from the last example and left the \(g(t)\) alone. The complementary solution this time is

\[{y_c}\left( t \right) = {c_1}{{\bf{e}}^{ - 2t}}\cos \left( {\frac{t}{2}} \right) + {c_2}{{\bf{e}}^{ - 2t}}\sin \left( {\frac{t}{2}} \right)\]

As with the last part, a first guess for the particular solution is

\[{Y_P}\left( t \right) = {{\bf{e}}^{ - 2t}}\left( {A\cos \left( {\frac{t}{2}} \right) + B\sin \left( {\frac{t}{2}} \right)} \right) + \left( {Ct + D} \right)\cos \left( {\frac{t}{2}} \right) + \left( {Et + F} \right)\sin \left( {\frac{t}{2}} \right)\]

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.

\[{Y_P}\left( t \right) = t{{\bf{e}}^{ - 2t}}\left( {A\cos \left( {\frac{t}{2}} \right) + B\sin \left( {\frac{t}{2}} \right)} \right) + \left( {Ct + D} \right)\cos \left( {\frac{t}{2}} \right) + \left( {Et + F} \right)\sin \left( {\frac{t}{2}} \right)\]

The complementary solution is

\[{y_c}\left( t \right) = {c_1}{{\bf{e}}^{ - 4t}} + {c_2}t{{\bf{e}}^{ - 4t}}\]

The two terms in \(g(t)\) 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

\[{Y_P}\left( t \right) = \left( {A{t^2} + Bt + C} \right){{\bf{e}}^{ - 4t}}\]

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.

\[{Y_P}\left( t \right) = t\left( {A{t^2} + Bt + C} \right){{\bf{e}}^{ - 4t}}\]

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.

\[{Y_P}\left( t \right) = {t^2}\left( {A{t^2} + Bt + C} \right){{\bf{e}}^{ - 4t}}\]

Upon multiplying this out none of the terms are in the complementary solution and so it will be okay.