Mathematical Methods for Physics and Engineering : A Comprehensive Guide

(lu) #1

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.
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