Methodus inveniendi lineas curvas

AD CURVAS INVENIENDAS ABSOLUTA. 69 erit $ \frac{ydp}{q^2} = \frac{ydx}{q} - \frac{ydy}{pq} $; ideoque $ \frac{adx}{p} = \frac{pdy}{q^2} - \frac{2ydq}{q^3} $, vel $ \frac{adp}{p^2} = \frac{dy}{q} - \frac{2ydq}{q^2} $. Si ponatur constans $ a=0 $, haec aequatio fiet integrabilis, eritque $ y=bq^2 $ & $ q=\sqrt{\frac{y}{b}} = \frac{dp}{dx} = \frac{pdp}{dy} $, unde fit $ pdp = dy \sqrt{\frac{y}{b}} $; atque integrando $ \frac{p^2}{2} = \frac{2}{3}y \sqrt{\frac{y}{b}} + c $, feu, mutatis constantibus, $ p^2 = \frac{y\sqrt{y}+a}{b\sqrt{b}} $, & $ p = \sqrt{\frac{y^{3/2}+a^{3/2}}{b^{3/2}}} = \frac{dy}{dx} $; hincque $ dx = dy \sqrt{\frac{b^{3/2}}{y^{3/2}+a^{3/2 [TRUNCATED]