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Solve the following linear first-order equation.

$(1+e^x)y '+e^xy=0$

I resolved:

$a_0(x)\acute{y}+a_1(x)y=g(x) => \acute{y}+p(x)y=Q(x)$

$\acute{y}+\frac{{e}^{x}}{1+{e}^{x}}y=0 , Q(x)=0, P(x)=\frac{{e}^{x}}{1+{e}^{x}}$

Integral factor in building:

$\mu (x)=exp\int P(x)d\acute{x}=exp\int\frac{{e}^{x}}{1+{e}^{x}} dx=exp\int\frac{e^x}{u}*\frac{du}{e^x}$

$1+e^x=u \rightarrow e^xdx=du \rightarrow dx=\frac{dx}{e^x} \rightarrow exp\int \frac{du}{u}=exp(ln\left|u\right| )=u-1+e^x$

$(1+e^x)\acute{y}+e^xy=0 \Rightarrow d((1+e^x)y)=0 \rightarrow \int d((1+e^x)y)=0 \rightarrow 1+e^xy=0 \rightarrow y=0$

Is my solved correctly?

Thanks for any help :)

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I wonder where is the $y'$ in original equation...... – ABC Jun 1 '13 at 8:06
Thank you for your attention :) @user007 – Software Jun 1 '13 at 8:14
Glad to see you around, @Software! +1 (And please know, I do not "hate" you; not in the least! We all make mistakes!) ;-) – amWhy Jul 22 '13 at 17:36
@amWhy •.¸(¯`'•.¸*♫♪$\Large\color{blue}{Thank~you~@amwhy~You~have~a~kind}$$\color{red}{‌​♥}$ ♫♪*¸.•'´¯)¸.• – Software Jul 22 '13 at 18:06
No No.You're very kind.You are so lovely. are you good? Are you better now? – Software Jul 22 '13 at 18:21
up vote 1 down vote accepted

notice:$$(1+e^x)\acute{y}+e^xy=0 \Rightarrow d((1+e^x)y)=0 \rightarrow \int d((1+e^x)y)=c \rightarrow 1+e^xy=c\rightarrow y=\frac{c}{1+e^x}$$

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