We now require starting looking into finding a particular solution for nth order differential equations. The two ways which we'll be looking at are similar as those which we looked at in the 2nd order section.
In this section we'll see the method of Undetermined Coefficients and it will be a fairly short section. With one little extension that we'll notice in the lone illustration in this section the method is identical to what we looked at back when we were finding undetermined coefficients in the 2nd order differential equations section.
Specified the differential equation,
y(n) + pn-1(t) y(n-1) + ......+ p1(t) y' + p0(t) y = g(t) (1)
If g (t) is an exponential function, cosine, polynomial, sum/difference, sine of one of these and/or a product of one of such then we guess the form of a exact solution similar guidelines which we used in the 2nd order material. We then plug the guess in the differential equation, set and simplify the coefficients equal to find for the constants.
The one thing which we require to recall is that we first require the complementary solution prior to making our guess for an exact solution. If any of term in our guess is into the complementary solution then we require multiplying the portion of our guess which contains this term by a t. It is where the one extension to the method comes in play. With a 2nd order differential equation the most we would ever require to multiply with is t2. With higher order differential equations that may require being more than t2.
The work include here is almost identical to the work we've previously done and actually it isn't even which much more difficult unless the guess is specifically messy and which makes for more mess while we take the derivatives and resolve for the coefficients. Since there isn't much difference in the work now we only going to do a single illustration in this section for the extension. Therefore, let's take a look at the illustration we're going to do now.