then making =z, we readily perceive that will be some func tion Z of z. y Therefore we shall have x+Zy=o, or zy+yz+Zy=o, and separating, we shall find. y For example, (ax+by) x =(mx+ny) y becomes on making=z; an equation easily intey y az2+(b―m) 2—n ́ , grated by what has been already explained. Examples for Practice. 1. Solve the fluxional equation xx+yy=2yx. 126. Let there now be proposed the equation (ax+by+c) x + (mx+ny+p) y = 0; we must first make ax+by+c=u, and mx + ny + p≈, and we shall find r = y nu-bz+bp-cn ----- an-mb and az—nu+mc―ap; substituting, we have (nu-mz) ù+(az-bu) ż=o, an-bm of which the fluent may be found by the last article. Let again axy + byx + xTM y′′ (fxy + gyx) =o=a + -), by dividing each term by xy. If we make ya x'=z, gx y=t, we shall have a + bx, and fy + Again = 4 Let now ap―fq=m, bp—gq=n, and we shall find p="g-f ¿+++1i=0; integrating qz¬"+pt=C, or (bn—am) (3° 2' 4 127. Let now g+Pyx=aQx, P and Q being functions of r Suppose, according to the method of Bernoulli, that y=Xz, X being some other function of x, then X+zX+PX zx=aQx. If we now make xX+PXzx=o, we shall have or X=e, and consequently aQre. Therefore separating the variables and integrating z=aƒ (e112 Qx), or ye1rå—aƒ (e11i Qs) +C. For example, the equation y+yx=axTM è gives y=ae~ (C+ƒexTM x) =a Ce*+a { x ̃—mxTM1+m (m—1) xTM-2_&c. } 128. By the equation Xy" y+X'yTM+1 ¿=X” ya å, we integrate. by dividing by Xy, and making y Example. m-n+1 1. Integrate the fluxional equation y+yx=ax3 129. When a fluxional equation involves the second or higher powers of and y, as in the equation y-a-o, we may find the value of by solving an algebraic equation. In the present case I =±a, so that y+ax=o, and also y-ax=o; hence y+ax+c=o, and y-ax+c=o, are two primitive equations, from either of which the fluxional equation may be derived; as also from their product (y+ax+c) (y—ax+c')=o. Example. Solve the fluxional equation y-ar s2=o. 130. When the equation contains only one of the variab.e quantities, z for example, we may deduce from it =X, a function of x; and hence y=fX. But if it be more easy to resolve the equation in respect of r, then putting p, we may find P, some function of p, and hence P; and since y=px, therefore y=pP, and y=SpP=pP-ƒPp. The relation between x and y is now found by eliminating p, by means of the two equations P, and y=pP-/Pp x=P, For example, let xx+ay=b√(x2+y2). Makings, we have s=b (1+p3)~ap=P, and y=bp/(1+p2)—§ ap2— bsp√(1+p3). The fluent of p/(1+p) may be found by former rules. There are but very few cases remaining, in which the general solution of fluxional 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 y 2. Separa ting the variables, we shall obtain namy, and taking the fluents I y nlr=mly+(n-m) le;t from which we easily deduce y"="c", the equation required. Prob. II. Required the curve of which the subtangent y y By first separating the variables, we find that =. by integrating, that yy = aa + xx xx and aa+xx ; (aa+x), an equation to the hyperbola. Σ=1Y+√(yy—aa) is the equation required. B • Because from the article quoted it appears that MN=; appears that MN=y{(21+y')"} while MC=(¿1+y1)' shall have ÿ—— Py, and m Therefore 19=1 P i= m √(p2+1) y= P y p(p+1) +y" y which is the fluxional equation of the first order 20 √(cm —ym) of the curve required. If n=m, we have x=yỷ (œ—ÿy) ̄, and x=c'±√(cc—yy), an equation to the circle. If m=2n, we have x =±√, an equation tor C (c-y) and integrating 1=1, and a P We might have found this fluent directly, by comparing the equayy, with those in articles 126 and 127. The least or dinate BP will be found by making ∞, and then we find BD =AD=al and the space DBMP=xy-yy+aal+faara |