FLUXIONS. 437 
“ane a eaieni aks Ai 
Carty Which gives A = 2, B=—3, C=—E=j, cea? * we have v’=s; therefore also jew. 
D=—§ it will be better to make the fraction = i —~— 
: | i. al 
A B. & D E Inthe fraction ———, U==1, Vza?-ta*; therefore <— 
er (+1)? + z41 + (c—1)* ma! iar ate “ 
This expression, by substituting the values of A, B, C> = me 953, when, xoca, 1 becomes.a== 14 and te be~. 
Selena we rate 1 comes o’ =—2a, therefore, one of the partial fractions 
2 2 @pip ~*aq1t Gay test . . mec) = Tas The other is mee: : 
reap dig gga RERRER BEET PROBS Sipame hae teats Vous 
pa ae Pe g vagt e a a a 
1 1 2 x 
. = + / Let both members of the equation be multiplied by 
I teh Ee ¢ 
Case IV. wee SE trinomial (x — a)", then, observing that Sis 6) 5,54 pe 
= + pt + q to be imaginary, if V has for a factor ting K for-5, we have - 
(2*-4-p2+q)", we may make : 
U_ Aa*'4Br2-24 .., +T P K=A+B (7—a)+C(x#—a)?+ wee 
v= (--pa+q)" + > Now, make « = a, and put u and:s to denote as be- 
or rather, | fore, and have A=K= -“; and since 
pf nine Cog K—A=B(z—a)4C(c—a)'4... 
vo (x*-4-pz+q)” + (2*--p -+49)"""* Ress the first since sah yee pa e Lee Cl; 
1 : 
: ant — it will therefore have the form K, (c—a) ; t 
Ex. 8 Thusyiwe make (e441) BiB Co e)troes 
A B C.. Derk F G H I now «=a, and that then K, becomes & 
Getatst ar teptes thus we have B=, and so on. ; = A 
ik 1 
F=—G=}, H=—I=3. . K=A+B(z—a) +C (c—a)'+... 
117. ‘The manner of determining the quantities by taking the fluxions, 
A, B, C, &c. as expressed in last article, is that sug. 4K — Bye c(ena)$3D (ea): py, 
gested by the theory of indeterminate coefficients, (AL- dz 
GEBRA, agt. 313.) but itis very tedious, We shall now @K 
pointy AA saben, se Gua *C+6D(z—a)+... 
Joni, amen Lat cn, BEE Pir, bo. fede eek 7 fe, 
Soy Fe a a mappesingalo shah s—iaiesiote’ a. respsctively, in the! particulir dhee'efoxsassthes, 
factor of S, we propose to find A independently of P, making that hypothesis, w. 
} e find 
Reducing to a common denominator, we get A=k, B=’, C=} k", Dai kk”, &e. 
become w ails respantividy tee Reet 8 P2842 C@ply tapi? s? 
w=As, and A=—, Hence, staiisa + B(x 4 1)4 Fw )s 
Oe tte + i SB + be 
——, we have L=/ , _ 
aae we. have 1 A(a+2)+B(a 2). When ing now a=— 1,.we fnd A=—}, B= —{. 
z=a, this becomes 1 =2 Aa; and when 3.). Let the equation be it , 
r= —2, it Becpmes 1=>2Ba; hence A= B= (), o Az+B: ry 
z - Vir Sepang? 37)  yoiletgna 
; : ee vee js then U= (Az + B)S + P(a?+ px+q). - is 
T the flaxions of the identical V= , vd UT a 
—a), considering that -V and pene If ‘we substitute for x, one of the imaginary. roots of 
st sill ¥ ieee of the equation 2? +pa$ q =.0, P will disappear, and. 
dv : 
By we find 5 =~ (c— 4) 4p Si supposing now w=, the OLS imagery, Weeation reine ee 
then, putting w for the value of when a is substic. tities on each side of the equation equal to one another, 
