Methodus inveniendi lineas curvas

IN RESOLVENDIS QUAESTIONIBUS. 143 $+\frac{2yydy}{n}+ndy$; unde erit $\frac{ydy}{n\sqrt{(yy+n^2)}} - \frac{Vdn}{y} - \frac{4Vdy}{n} = d.(\frac{y\sqrt{(yy+n^2)}}{n} - \frac{2Vy}{n} - \frac{nV}{y}) = \frac{ydy}{n\sqrt{(yy+n^2)}} - \frac{yydn}{n^2\sqrt{(yy+n^2)}} - \frac{2Vdy}{n} - \frac{2ydV}{n} + \frac{2Vydn}{n^2} + \frac{nVdy}{yy} - \frac{Vdn}{y} - \frac{ndV}{y}$; hincque $\frac{yydn}{n^2\sqrt{(y^2+n^2)}} - \frac{2Vdy}{n} + \frac{2ydV}{n} - \frac{2Vydn}{n^2} + \frac{nVdy}{yy} - \frac{Vdn}{y} - \frac{ndV}{y}=0$. Verum, est generaliter $dV=-Ldx -V[L]dx$; unde erit $dV=\frac{yydy}{n^2\sqrt{(yy+nn [TRUNCATED]