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then making 4-2, we readily perceive that 음 will be some func
y tion 2 of z. Therefore we shall have c + Zy=0, or zy +ya+ Zy=0, and separating, we shall find
ง For example, (ax +by) z =(mx+ny) y becomes (az+6) 2 , on making < =; an equation easily inte
% y az? +(6—m) 2-1
y grated by what has been already explained.
Examples for Practice. 1. Solve the fluxional equation x.: +yy=2yc. 2. Integrate the equation xy-yr=(x2 +y“). 3. Integrate the equation yy +(x+2y) r=0.
126. Let there now be proposed the equation (ax+by+c) * + (mx +ny+p) y = 0; we must first make ax +by+c=u, and
nu_bz+bp-on Mx + ny + p =%, and we shall find x =
anemo az—nu +mc-ap; substituting, we have (nu-mz) i + (az-bu) ż=0,
; ап — бm of which the fluent may be found by the last article.
bx Let again axy + byi + momo ya (fxy + gyr) == + + 'y
y +8), by dividing each term by xy. If we make yo mb=2, g' * = 1, we shall have ay + b = r , and fy 153 = Again
ข groep op z', and y's =1, and yup-se gbp-1 =zPt4 Let now ap-fq=m, bp-gq=n, and we shall find p= =
i and by substitution
+ t=, or 2-pad ag-bf itty-i=0; integrating qz--+pt-2=C, or (br-am) (yo 30)oz=4f + (ng.m) (y' sor) or W =(ag-bf)C=C'.
127. Let now 9+ Pyr=aqi, P and Q being functions of x: Suppose, according to the method of Bernoulli, that y=Xz, X being some other function of X, then X2+2X + PX ze=aQui. If we nov make 28+PXzi=0, we shall have Å =-Px, and log X=-|Pé,
bn-am and q>
or Xar, and consequently aQi=arriz. Therefore separating the variables and integrating z=as (erri Qi), or yerrizasieri Qi)
For example, the equation ý +yi -armigives y=ac (C+Ser:) acetat
"mam'+m (m-1)=_&c.} 128. By the equation Xy“ ý + X’ym+1 ;=X"y" , we integrate. by dividing by Xy, and making y
Example. 1. Integrate the fluxional equation y + yr -axt
129. When a fluxional equation involves the second or higher powers of : and y, as in the equation ya-aj=0, we may find the value of Y by solving an algebraic equation. In the present case
Ş=Ła, so that y+aš=0, and also y-a:=0; hencey+ar+c=0, and y-ar+c'=0, are two primitive equations, from either of which the Auxional equation may be derived; as also from their product (y+ar+c)(y-ar+c)=o.
Example. Solve the fluxional equation yo-ar i=0.
je. 130. When the equation contains only one of the variab.e quantities, s for example, we may deduce from it Y=X, a function of : ; and hence y=fXš. But if it be more easy to resolve the equation in respect
of then putting y=p, we may find s-P, some function of p, and hence r=; and since y=px, therefore y=pP, and y=fpP=pP-SPp. The relation between x and y is now found by
y eliminating p, by means of the two equations x=P, and y=pP-SPA
For example, let we+ay=6V (**+*). Making = 4, we have =b (1+po)-ap=P, and y=bpv (1+p)
-ap- ospv (1 +p). The fluent of pv (1+p?) may be found by former rules.
There are but very few cases remaining, in which the general 80lution of Aluxional equations is possible.
We shall terminate this article with several examples of the inverse method of tangents.
Prob. I. To find th curve whose subtangent 4.4 Separa
ting the variables, we shall obtain nizmy, and taking the duento
y alrmly +(n-m) lc;t from which we easily deduce yur" , the equation required.
Prob. II. Required the curve of which the subtangent Y = 10+
By first separating the variables, we find that Y=
aa + 1.1 69
by integrating, that yy=(au +5*), an equation to the hyperbola.
Prob. III. To find the curve in which
By the problem we have . Say = syi,
=, or yu =x* a"-n.
y Prob. IV. To find the curve BM, in which the space ABMP=the arc BM mul. tiplied by a constant quantity a, or such that jyš=asv ( +%*). Separating the variables, we obtain ay
; and therefore (yy-aa)
yyProb. V. To find the curve AM, in which the radius of curvature MC="MN.
*=19+v(.xy=-aa) is the equation required.
• Because from the article quoted it appears that MN=Y
shall have ös_ve
= Therefore *.1 Y=1 P
which is the fluxional equation of the first order (cy) of the curve required.
If n=m, we have x=yy 100~yy), and s=d'Ev(cc-yy), an equation to the circle. If m=2n, we have ==yvy, . ,
✓(c-y) to the cycloid.
Prob. VI. Required a curve BM, such that drawing through the vertex A a right line AO, making with the axis an angle of 45", we may always have the following proportion.
As the ordinate PM is to the subtangent PT, so is a given line a to OM.
By the enunciation of the problem it appears that y: .::a:y-t, and
X ax=(ym) y. Let y-rzz, and we T
a shall have ya = and integrating Y=lam, and
We might have found this fluent directly, by comparing the equator 3+ yy, with those in articles 126 and 127. The least ordinate BP will be found by making ? =c, and then we find BD
SAD<al ; and the space DBMP=xy-1 yy+nal . +faata