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$\frac{d}{dt}\left(\frac{x'(t)}{x(t)}\right)=x(t)-x^2(t)$ where $x'(t)=\frac{d}{dt}x(t)$ What reasoning (if it exists) I can apply to solve this differential equation? Thanks.

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A good motivation for Julián's method is to notice that the $\frac{x'}{x}$ term is just $(\ln x)'$, so the desired function is the exponential of another. –  Eugene Shvarts Jul 4 '12 at 13:50

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Let $x=e^u$; then $u$ satisfies de differential equation $$ u''=e^u-e^{2u}. $$ Multiply by $u´$ and integrate once to get $$ \frac12(u')^2=e^u-\frac12\,e^{2u}+C_1. $$ This is a differential equation in separeted variables, whose solution is $$ \int\frac{du}{\sqrt{2\,e^u-e^{2u}+2\,C_1}}=\pm t+C_2. $$

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If we let $x=e^u$, then $\dfrac{d}{du}(e^u/e^u)=0=e^u-e^{2u}$. Aren't we thinking of $x$ as a function of $u$? So what does $u''$ mean (does it mean $-\dfrac{1}{x^2}$)? I'm just curious and wanting to learn. =) Shouldn't one write $(xx''-(x')^2)/x^2=e^u-e^{2u}$? Here on the LHS, I just took the derivative of $x'/x$ with respect to $u$. –  math-visitor Jul 6 '12 at 7:21
    
@math-visitor $x$ and $u$ are unknowns functions of the independent variable $t$. The definition of $u$ is $x(t)=e^{u(t)}$. –  Julián Aguirre Jul 6 '12 at 9:39
    
Oh, I see. I was speculating that, but I wasn't sure. Thank you Julian! –  math-visitor Jul 6 '12 at 18:59

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