22 DE INT EGRATIONE 



Problema 5. 



37. Si propofita fit aequatio diflerentialis : 

 ?j>* d x -\- Qjdx -\-Kdj — o 

 vbi P , Q. et R denotent functiones quascunque ipfius#, 

 inuenire multiplicatorem , qui eam reddat integrabilem. 



Solutio. 



Erit ergo M =: ?j n -+- Q/ et NzzR, hincquc 



Quare pofito multiplicatore quaefito L ct dhzzzpdx 

 -\-qdy^ erit ex ante inuentis : 



Fingatur L— S/" , exiftente S fun&ione ipfius £ tan- 

 tum , erit p=^J — , et ^zrwSj m "~', quibus valori- 



bus fubftitutis , prodibit : 



^-«P>—~ Z «Q~«Py-'+(i-f|- 

 Quae aequatio vt fubfiftere poflit, fumi debet mzzz— /?, 

 ac fiet 



RdS , -. dR roti d_S (i-n)Q .d« dR 



Sdx — t 1 ~~ n )\l.^dx1 ieu S R -" "R 



Vnde cum integrando proueniat S^R^—^nr^erit, ob 

 mzzz -», multiplicator quaefuus : 



1— - R -« R 



ct aequatio integralis erit 



v \—n [ Q_dx 



f—J'-*» — + /!^5^— )/^*- Conft. 



Coroll. 1. 



