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Question Number 160516 by peter frank last updated on 30/Nov/21
Answered by aleks041103 last updated on 06/Dec/21
$${u}={x}/{y}\Rightarrow{y}={x}/{u} \\ $$$$\Rightarrow{y}'=\frac{{u}−{u}'{x}}{{u}^{\mathrm{2}} }=\frac{\mathrm{1}}{{u}}−\frac{{u}'}{{u}^{\mathrm{2}} }{x} \\ $$$$\Rightarrow\frac{\mathrm{1}}{{u}}−\frac{{x}}{{u}^{\mathrm{2}} }\:\frac{{du}}{{dx}}={e}^{{u}} \\ $$$$\Rightarrow\frac{{x}}{{u}^{\mathrm{2}} }\:\frac{{du}}{{dx}}=\frac{\mathrm{1}}{{u}}−{e}^{{u}} \\ $$$$\Rightarrow\frac{{du}}{{u}−{u}^{\mathrm{2}} {e}^{{u}} }=\frac{{dx}}{{x}} \\ $$
Commented by aleks041103 last updated on 06/Dec/21
$${and}\:{it}\:{goes}\:{on}\:{somehow}... \\ $$$${we}\:{need}\:{to}\:{solve}: \\ $$$$\int\frac{{du}}{{u}−{u}^{\mathrm{2}} {e}^{{u}} }=? \\ $$