Example: Find out a particular solution to

y'' - 4y' - 12 y = 3e_5t

Solution

The point here is to get a particular solution, though the first thing that we're going to do is get the complementary solution to this differential equation. Recall that the complementary solution arrives from solving,

y'' - 4y' - 12 y = 0

For this differential equation and its roots, the characteristic equation is,

r₂ - 4r -12 = (r - 6) (r + 2) = 0

r₁ = -2 , r₂ = 6

The complementary solution is after that,

yc(t) = c₁e^-2t + c₂ e^6t

At this point the purpose for doing this first will not be apparent, though we need you in the habit of finding this before we begin the work to find an exact solution. Eventually, when we'll see, comprising the complementary solution in hand will be useful and therefore it's best to be in the habit of finding it first previous to doing the work for undetermined coefficients.

Here, let's proceed with finding an exact solution. As mentioned earlier to the start of this illustration we need to make a guess as to the form of an exact solution to this differential equation. As g(t) is an exponential and we know that exponentials never simply appear or disappear in the differentiation process this seems that a probable form of the exact solution would be

Yp(t) = A e^5t

Here, all that we require to do is do a couple of derivatives, plug this in the differential equation and notice if we can find out what A needs to be.

Plugging in the differential equation provides,

25A e^5t - 4 (5A e^5t) - 12(Ae^5t) = 3 e^5t

- 7(Ae^5t) = 3 e^5t

Therefore, in order for our guess to be a solution we will require to choose A hence the coefficients of the exponentials on either side of the equivalent sign are similar. In other words we require to choose A hence,

-7A = 3             ⇒         A = -(3/7)

Okay, we determined a value for the coefficient. It means that we guessed properly.  A particular solution to the differential equation is after that,

Yp(t) = -(3/7)e^5t

Before proceeding any additional let's again note that we started off the solution above through finding the complementary solution. It is not technically part the method of Undetermined Coefficients conversely, as we'll eventually see; having this in hand before we make our guess for the exacting solution can save us many work or/and headache.  Determining the complementary solution first is easily a good habit to have so we'll attempt to get you in the habit over the course of the next few illustrations. At this point does not worry regarding to why it is a good habit. We'll finally notice why it is a good habit.

Here, back to the work at hand and see in the last illustration that we kept saying "a" particular solution, not "the" particular solution. It is as there are other possibilities out there for the particular solution we've just managed to get one of them. One of them will work while it comes to writing down the general solution to the differential equation.

Speaking of which... This section is devoted to determining particular solutions and most of the illustrations will be determining only the particular solution. Though, we should do at least one full blown IVP to ensure that we can say that we've complete one.