HIGHER-ORDER ORDINARY DIFFERENTIAL EQUATIONS
Find the complementary function of the equation
d^2 y
dx^2
− 2
dy
dx
+y=ex. (15.15)
Setting the RHS to zero, substitutingy=Aeλxand dividing through byAeλxwe obtain
the auxiliary equation
λ^2 − 2 λ+1=0.
The rootλ= 1 occurs twice and so, althoughexis a solution to (15.15), we must find
a further solution to the equation that is linearly independent ofex.Fromtheabove
discussion, we deduce thatxexis such a solution, so that the full complementary function
is given by the linear superposition
yc(x)=(c 1 +c 2 x)ex.
Solution method. Set the RHS of the ODE to zero (if it is not already so), and
substitutey=Aeλx. After dividing through the resulting equation byAeλx, obtain
annth-order polynomial equation inλ(the auxiliary equation, see (15.10)). Solve
the auxiliary equation to find thenroots,λ 1 ,λ 2 ,...,λn, say. If all these roots are
real and distinct thenyc(x)is given by (15.11). If, however, some of the roots are
complex or repeated thenyc(x)is given by (15.12) or (15.13), or the extension
(15.14) of the latter, respectively.
15.1.2 Finding the particular integralyp(x)
There is no generally applicable method for finding the particular integralyp(x)
but, for linear ODEs with constant coefficients and a simple RHS,yp(x) can often
be found by inspection or by assuming a parameterised form similar tof(x). The
latter method is sometimes called themethod of undetermined coefficients.Iff(x)
contains only polynomial, exponential, or sine and cosine terms then, by assuming
a trial function foryp(x) of similar form but one which contains a number of
undetermined parameters and substituting this trial function into (15.9), the
parameters can be found andyp(x) deduced. Standard trial functions are as
follows.
(i) Iff(x)=aerxthen try
yp(x)=berx.
(ii) Iff(x)=a 1 sinrx+a 2 cosrx(a 1 ora 2 may be zero) then try
yp(x)=b 1 sinrx+b 2 cosrx.
(iii) Iff(x)=a 0 +a 1 x+···+aNxN(someammay be zero) then try
yp(x)=b 0 +b 1 x+···+bNxN.