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dy-dx-x-y-ln-x-y-




Question Number 128395 by bramlexs22 last updated on 06/Jan/21
   (dy/dx) = ((x/y)) ln ((x/y))
$$\:\:\:\frac{{dy}}{{dx}}\:=\:\left(\frac{{x}}{{y}}\right)\:\mathrm{ln}\:\left(\frac{{x}}{{y}}\right) \\ $$
Answered by mr W last updated on 07/Jan/21
let x=yu  (dx/dy)=u+y(du/dy)  1=(u+y(du/dy))uln u  ((1−u^2 ln u)/(uln u))=y(du/dy)  ((uln u du)/(1−u^2 ln u))=(dy/y)  ln y=∫((uln u du)/(1−u^2 ln u))  ...
$${let}\:{x}={yu} \\ $$$$\frac{{dx}}{{dy}}={u}+{y}\frac{{du}}{{dy}} \\ $$$$\mathrm{1}=\left({u}+{y}\frac{{du}}{{dy}}\right){u}\mathrm{ln}\:{u} \\ $$$$\frac{\mathrm{1}−{u}^{\mathrm{2}} \mathrm{ln}\:{u}}{{u}\mathrm{ln}\:{u}}={y}\frac{{du}}{{dy}} \\ $$$$\frac{{u}\mathrm{ln}\:{u}\:{du}}{\mathrm{1}−{u}^{\mathrm{2}} \mathrm{ln}\:{u}}=\frac{{dy}}{{y}} \\ $$$$\mathrm{ln}\:{y}=\int\frac{{u}\mathrm{ln}\:{u}\:{du}}{\mathrm{1}−{u}^{\mathrm{2}} \mathrm{ln}\:{u}} \\ $$$$… \\ $$

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