484 Higher Engineering Mathematics
Table 51.1Form of particular integral for different functions
Type Straightforward cases
Try as particular integral:‘Snag’ cases
Try as particular integral:See
problem(a) f(x)=a constant v=k v=kx(usedwhenC.F.
contains a constant)1, 2(b) f(x)=polynomial (i.e. v=a+bx+cx^2 + ··· 3
f(x)=L+Mx+Nx^2 + ···
where any of the coefficients
may be zero)
(c) f(x)=an exponential function v=keax (i)v=kxeax(usedwheneax 4, 5
(i.e.f(x)=Aeax) appears in the C.F.)
(ii)v=kx^2 eax(used when eax 6
andxeaxboth appear in
the C.F.)(d) f(x)=a sine or cosine function v=Asinpx+Bcospx v=x(Asinpx+Bcospx) 7, 8
(i.e.f(x)=asinpx+bcospx, (usedwhensinpxand/or
whereaorbmay be zero) cospxappears in the C.F.)
(e) f(x)=a sum e.g. 9
(i) f(x)= 4 x^2 −3sin2x (i)v=ax^2 +bx+c
+dsin2x+ecos2x
(ii)f(x)= 2 −x+e^3 x (ii)v=ax+b+ce^3 x(f) f(x)=a product e.g. v=ex(Asin 2x+Bcos2x) 10
f(x)=2excos2x(iv) To determine the particular integral,v, firstly
assume a particular integral which is sug-
gested by f(x), but which contains unde-
termined coefficients. Table 51.1 gives some
suggested substitutions for different functions
f(x).(v) Substitute the suggested P.I. into the dif-
ferential equation(aD^2 +bD+c)v=f(x)and
equate relevant coefficients to find the constants
introduced.(vi) The general solution is given by
y=C.F.+P.I.,i.e.y=u+v.(vii) Given boundary conditions, arbitrary constants
in the C.F. may be determined and the particular
solution of the differential equation obtained.51.3 Worked problemson
differential equations of the
forma
d^2 y
dx^2
+b
dy
dx
+cy=f(x)
wheref(x)is a constant or
polynomial
Problem 1. Solve the differential equation
d^2 y
dx^2+dy
dx− 2 y=4.Using the procedure of Section 51.2:(i)d^2 y
dx^2+dy
dx− 2 y=4 in D-operator form is
(D^2 +D− 2 )y=4.