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Question Number 156684 by mathdave last updated on 14/Oct/21

$$solvetheD.E$$$$[1+e^{\frac{x}{y}}]dx+e^{\frac{x}{y}}[1-\frac{x}{y}]dy=0$$$$anyonecanhelppls$$

Answered by som(math1967) last updated on 14/Oct/21

$$[1+e^{\frac{x}{y}}]dx=e^{\frac{x}{y}}[\frac{x}{y}-1]dy$$$$\frac{dx}{dy}=\frac{e^{\frac{x}{y}}[\frac{x}{y}-1]}{[1+e^{\frac{x}{y}}]}$$$$let\frac{x}{y}=v$$$$x=vy$$$$\frac{dx}{dy}=v+y\frac{dv}{dy}$$$$v+y\frac{dv}{dy}=\frac{e^v(v-1)}{(1+e^v)}$$$$y\frac{dv}{dy}=\frac{{ve}^v-e^v-v-{ve}^v}{(1+e^v)}$$$$y\frac{dv}{dy}=-\frac{(v+e^v)}{(1+e^v)}$$$$\int \frac{(1+e^v)dv}{v+e^v}=-\int \frac{dy}{y}$$$$\int \frac{d(v+e^v)}{v+e^v}=-\int \frac{dy}{y}$$$$ln\mid v+e^v\mid =-lny+lnC$$$$ln\mid v+e^v\mid =ln\frac{C}{y}$$$$v+e^v=\frac{C}{y}$$$$y(\frac{x}{y}+e^{\frac{x}{y}})=C$$

Answered by mindispower last updated on 14/Oct/21

$$\frac{x}{y}=u\Rightarrow dx=ydu+udy$$$$\iff (1+e^u)(ydu+udy)+e^u(1-u)dy=0$$$$y(1+e^u)du+(u+e^u)dy=0$$$$\Rightarrow \frac{dy}{y}=-\frac{1+e^u}{u+e^u}du$$$$\Rightarrow ln\mid y\mid =-ln(\mid u+e^u\mid )+c$$$$\Rightarrow y=\frac{k}{\mid u+e^u\mid }$$$$\Rightarrow y\mid \frac{x}{y}+e^{\frac{x}{y}}\mid =k$$

Commented by peter frank last updated on 14/Oct/21

$$\mathrm{thank}\mathrm{you}$$

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