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Question Number 92464 by jagoll last updated on 07/May/20

$$\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{{\mathrm{e}}^{\mathrm{y}}}{{\mathrm{x}}^2}-\frac{1}{\mathrm{x}}$$

Commented by mr W last updated on 07/May/20

$$howdoyouget$$$$\frac{dy}{e^y}=\frac{dx}{x^2}-\frac{dx}{x}from$$$$\frac{dy}{dx}=\frac{e^y}{x^2}-\frac{1}{x}?$$$$itisnot$$$$\frac{dy}{dx}=\frac{e^y}{x^2}-\frac{e^y}{x}!$$

Commented by mathmax by abdo last updated on 07/May/20

$$\Rightarrow \frac{dy}{e^y}=\frac{dx}{x^2}-\frac{dx}{x}let{e}^y=t\Rightarrow y=ln\left(t\right)\Rightarrow dy=\frac{dt}{t}$$$$\Rightarrow \frac{dt}{t^2}=\frac{dx}{x^2}-\frac{dx}{x}\Rightarrow -\frac{1}{t}=-\frac{1}{x}-ln\left(x\right)+c\Rightarrow$$$$\frac{1}{t}=\frac{1}{x}+lnx+c\Rightarrow t=\frac{1}{\frac{1}{x}+lnx+c}\Rightarrow y=-ln(\frac{1}{x}+lnx+c)$$

Commented by i jagooll last updated on 16/May/20

$$\mathrm{mr}\mathrm{Abdo}.\mathrm{no}\mathrm{sir}.\mathrm{it}\mathrm{no}\mathrm{correct}$$$$$$

Commented by i jagooll last updated on 16/May/20

$$\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{{\mathrm{e}}^{\mathrm{y}}-\mathrm{x}}{{\mathrm{x}}^2}\Rightarrow \frac{\mathrm{dy}}{{\mathrm{e}}^{\mathrm{y}}-\mathrm{x}}=\frac{\mathrm{dx}}{{\mathrm{x}}^2}$$$$\mathrm{how}\mathrm{get}\frac{\mathrm{dy}}{{\mathrm{e}}^{\mathrm{y}}}=\frac{\mathrm{dx}}{{\mathrm{x}}^2}-\frac{\mathrm{dx}}{\mathrm{x}}$$

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