Steps / Historical AI

Leonhard Euler · 1744 · Internet Archive

Methodus inveniendi lineas curvas

All 336 leaves · leaf 85

Transcription

78 DE METHODO MAX. ET MIN. Ex quo jam apparet, si insuper fuerit $R=0$, tum etiam quartam integrationem locum habere, & ita porro. CASUS IV. 66. Si fuerit $M=0$, ita ut fit $dZ=Ndy+Pdp+Qdq+Rdr+Sds+\&c.$ Æquatio pro curva quæsita ante prodiit $0=N-\frac{dP}{dx}+\frac{ddQ}{dx^2}-\frac{d^3R}{dx^3}+\frac{d^4S}{dx^4}-\&c.$ quæ multiplicetur per $dy=pdx$, & tum addatur $dZ-Ndy-Pdp-Qdq-Rdr-Sds-\&c.$ prodibit. $0=dZ-pdP+\frac{pddQ}{dx}-\frac{pd^3R}{dx^2}+\frac{pd^4S}{dx^3}-\&c. -Pdp-Qdq-Rdr-Sds-\&c.$ cujus integrale assignari potest; erit enim $0=A+Z-Pp+\frac{pdQ}{dx}-\frac{pddR}{dx^2}+\frac{pd^3S}{dx^3}-\&c. -Qq+\frac{qdR}{dx}-\frac{qddS}{dx^2}+\&c. -Rr+\frac{rdS}{dx}-\&c. -Ss+\&c.$ vel $0=A+Z-Pp+\frac{pdQ-Qdp}{dx}-\frac{pddR-dpdR+Rddp}{dx^2}+\frac{pd^3S-dpddS+dSddp-Sd^3p}{dx^3}-\&c.$ cujus termini, quomodo ulterius progrediantur, si in $dZ$ insint sequentia differentialia $Tdt, Udu$ &c. sponte patet. CASUS V. 67. Si fit & $M=0$, & $N=0$; ita ut fit $dZ=Pdp+Qdq+Rdr+Sds+\&c.$ Quia est $N=0$, una integratio per casum primum instituitur.
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