Question

Find the general solution of the differential equation dy/dx+1x=e^y/x

Answer 1

Step-by-step explanation:

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Answer 2

$\quad\quad\bold{x\:e^{y}=c\:(1-e^{y})}$

Step-by-step explanation:

The given differential equation is

$\quad\:\:\mathrm{\dfrac{dy}{dx}+\dfrac{1}{x}=\dfrac{e^{y}}{x}}$

$\to \mathrm{\dfrac{dy}{dx}+\dfrac{1}{x}-\dfrac{e^{y}}{x}=0}$

$\to \mathrm{\dfrac{dy}{dx}+\dfrac{1-e^{y}}{x}=0}$

$\to \mathrm{\dfrac{dy}{1-e^{y}}+\dfrac{dx}{x}=0}$

$\to \mathrm{\dfrac{e^{-y}\:dy}{e^{-y}-1}+\dfrac{dx}{x}=0}$

$\to \mathrm{-\dfrac{d(e^{-y}-1)}{e^{-y}-1}+d(logx)=0}$

Integrating we get

$\quad\:\:\mathrm{-\int \dfrac{d(e^{-y}-1)}{e^{-y}-1}+\int d(logx)=0}$

$\to \mathrm{-log(e^{-y}-1)+logx=logc}$

where c is constant of integration

$\to \mathrm{log\dfrac{x}{e^{-y}-1}=logc}$

$\to \mathrm{\dfrac{x}{e^{-y}-1}=c}$

$\to \mathrm{\dfrac{x\:e^{y}}{1-e^{y}}=c}$

$\to \boxed{\bold{x\:e^{y}=c\:(1-e^{y})}}$

This is the required general solution.

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