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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

$$u=x/y\Rightarrow y=x/u$$$$\Rightarrow y^{\prime }=\frac{u-u^{\prime }x}{u^2}=\frac{1}{u}-\frac{u^{\prime }}{u^2}x$$$$\Rightarrow \frac{1}{u}-\frac{x}{u^2}\frac{du}{dx}=e^u$$$$\Rightarrow \frac{x}{u^2}\frac{du}{dx}=\frac{1}{u}-e^u$$$$\Rightarrow \frac{du}{u-u^2{e}^u}=\frac{dx}{x}$$

Commented by aleks041103 last updated on 06/Dec/21

$$anditgoesonsomehow...$$$$weneedtosolve:$$$$\int \frac{du}{u-u^2{e}^u}=?$$

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