Pagina:Leonhard Euler Letter 1765-03-31.pdf/4

artitrariis C e D obtinet?? hos satis Semiplex? intigrale completurros

$x^{\frac {c}{f}}y^{n+2}=Ax^{{\frac {c}{f}}-m-1}+Bx^{m+2}$ excitente $n=b/f-1$ et $m={\frac {c-?f\pm {\sqrt {}}[(c-f)^{?}-?(n+2)af]}{zf}}$

quod adeo serri?dicius ponendo ${\sqrt {}}[c-f]^{2}-4f(\pi ?+2)af)={\sqrt[{(?-f)^{2}-??(b+f)}]{=}}q?$

ita exhiberi potent: $y^{{\frac {b}{f}}+1}=Ax^{\frac {-?1+f-g}{zf}}+Bx^{\frac {-c+f+g}{zf}}$

quei forma si forte? g fiat imaginarium, juxta? $g=hV-1$ alit in ????.

$4^{{\frac {b}{f}}+1}=x^{f-c}{2f}(A\cos {\frac {hlx}{2f}}+B\sin {\frac {hlx}{2f}})$

AEquationis $ay^{p}x^{n-1}dx+6x^{n}y^{p-1}\delta y+cy\delta x+f?x\delta y=0$ integratio per telem nu??.

biplicatorcon? facillime ??ipires uti ja?i oleni est foetc??: ea enim ita repe??.

sentata ${\frac {a\delta x}{x}}+{\frac {b\delta y}{y}}+{\frac {1}{x^{n-1}y^{p-1}}}[{\frac {c\delta x}{x}}+{\frac {a\delta y}{y}}]=0$. priceo membrum integratile?

?edditua per multiplicatorem $x^{a}y^{b}$ et generalius $x^{a?}y^{b?}$ porterius

vero per ???: $x^{n-1}y^{p-1}{\dot {x}}^{vc}y^{vf}$. quos duos multiplicatores uiten se con?quari

oportet: quo fit $??a=vc+\pi -1$ et $?ub?vf+p-1$ unde  ??c et v dart?n.

tum eit integrale: ${\frac {1}{??c}}c^{??ca}y^{cb}+{\frac {1}{v}}x^{ve}y^{vf}=C$ s??

$Ax^{\frac {(p-1)ac}{af-bc}}y^{-{\frac {(n-1)bf}{af-bc}}}={\frac {x^{\frac {(n-1)af}{af-bc}}y^{-{\frac {(p-1)bc}{af-bc}}}}{(n-1)f-(p-1)c}}+{\frac {x^{\frac {(n-1)bc}{af-bc}}y^{-{\frac {(p-1)af}{af-bc}}}}{(n-1)b-(p-1)a}}$.

Pl??a specimin?? ???? modi integrationum per multiplikatores ?ienuci?

Vir celebenime is? Commenta?iis ???? ubi ej?io modi aequationes

differetiles secundi graden expedivi quoe aliis method?i fr???a tractar??

Egregia autern s??nt omnino quoe Tu de ??? argumento mecum benevole

ommunicasti. Vale mihi que sa?et? ?????.

Daba?? Bevolini d. 31 Martii 1765.