Methodus inveniendi lineas curvas

82 DE METHODO MAX. ET MIN. $\frac{6(1+pp)r}{qq} + \frac{(1+pp)^2 r r}{q^4} = \int dx(\frac{9pp(1+pp)^2}{q} - \frac{6(1+pp)^3 r}{q^3} + \frac{(1+pp)^4 r r}{q^5})$, ex quo $Z$ erit functio ipsarum $p, q$ & $r$; unde differentiando prodibit: $dZ = \frac{18pdp(1+pp)(1+3pp)}{q} - \frac{9ppdq(1+pp)^2}{qq} - \frac{6dr(1+pp)^3}{q^3} - \frac{36prdp(1+pp)^2}{q^3} + \frac{18rdq(1+pp)^3}{q^4} + \frac{2rdr(1+pp)^4}{q^5} + \frac{8rrpdp(1+pp)^3}{q^5} - \frac{5rrdq(1+pp)^4}{q^6}$. Comparatione ergo cum forma generali instituta, erit $M=0$; $N=0$; $Z = \frac{9pp(1+pp)^2}{q} - \frac{6(1+pp)^3 r}{q^3} + \frac{(1+pp)^4 r r}{q^5}$. $P = \frac{18p(1+pp)(1+3pp)}{q} - \frac{36pr(1+pp)^2}{q^3} + \frac{8rrp(1+pp)^3}{q^5}$. $Q = - \frac{9pp(1+pp)^2}{qq} + \frac{18r(1+pp)^3}{q^4} - \frac{5rr(1+pp)^4}{q^6}$. $R = - \frac{6(1+pp)^3}{q^3} + \frac{2r(1+pp)^4}{q^5}$. Cum nunc fit $M=0$ & $N=0$, solutio cadit in Casum quintum, eritque aequatio pro curva quaesita haec, $0 = Ap - B - Z + Qq + Rr - \frac{q dR}{dx}$ quae, factis substitutionibus, transit in hanc $0 = Ap - B - \frac{18pr(1+pp)^2}{q} - \frac{16pr(1+pp)^3}{q^3} + \frac{6rr(1+pp)^4}{q^5} - \frac{2dr(1+pp)^4}{q^4 dx} + \frac{36p(1+pp)^2}{q}$; quae aequatio nimis est complicata, quam ut ejus ulteriores integrationes suscipi queant. Ceterum apparet hanc aequationem esse differentialem quarti ordinis, ita ut per quatuor residuas integrationes quatuor constantes adhuc ingrediantur: ex quo sex data oportebit esse puncta, per quae curva transeat, ut Problema determinetur. CAPUT