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Question Number 145300 by imjagoll last updated on 04/Jul/21
$$\:\:\:\:\:\:\:\mathrm{yy}'\:=\:\mathrm{x}\:\mathrm{e}^{\frac{\mathrm{x}}{\mathrm{y}}} \: \\ $$
Answered by puissant last updated on 04/Jul/21
$$\mathrm{y}'=\frac{\mathrm{x}}{\mathrm{y}}\mathrm{e}^{\frac{\mathrm{x}}{\mathrm{y}}} \:\:\mathrm{put}\:\:\:\mathrm{t}=\frac{\mathrm{x}}{\mathrm{y}} \\ $$$$\Rightarrow\frac{\mathrm{dy}}{\mathrm{dt}}\:=\:\mathrm{te}^{\mathrm{t}} \Rightarrow\:\mathrm{dy}\:=\:\mathrm{te}^{\mathrm{t}} \mathrm{dt} \\ $$$$\Rightarrow\int\:\mathrm{dy}\:=\:\int\:\mathrm{te}^{\mathrm{t}} \mathrm{dt} \\ $$$$\int\mathrm{te}^{\mathrm{t}} \mathrm{dt}\:=\:\mathrm{te}^{\mathrm{t}} −\mathrm{e}^{\mathrm{t}} +\mathrm{k}\:=\:\mathrm{e}^{\mathrm{t}} \left(\mathrm{t}−\mathrm{1}\right)+\mathrm{k} \\ $$$$\Rightarrow\:\mathrm{y}=\:\mathrm{e}^{\mathrm{t}} \left(\mathrm{t}−\mathrm{1}\right)+\gamma.... \\ $$
Commented by imjagoll last updated on 04/Jul/21
$$\mathrm{if}\:\mathrm{t}=\frac{\mathrm{x}}{\mathrm{y}}\:\Rightarrow\mathrm{y}=\frac{\mathrm{x}}{\mathrm{t}}\: \\ $$$$\Rightarrow\frac{\mathrm{dy}}{\mathrm{dx}}\:=\:\frac{\mathrm{1}}{\mathrm{t}}−\frac{\mathrm{x}}{\mathrm{t}^{\mathrm{2}} }\:\frac{\mathrm{dt}}{\mathrm{dx}}\: \\ $$