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how to solve differential equation like this

$y'=ye^{-y'}$

I have no idea to solve problem like this

is there any special function or special way to solve ?

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    $\begingroup$ My guess is this would be a place to start: en.wikipedia.org/wiki/Lambert_W_function $\endgroup$ – Mike May 26 '13 at 16:19
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    $\begingroup$ That's the second evil looking DE you've posted; where are these coming from? $\endgroup$ – Sharkos May 26 '13 at 16:20
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Hint: When $f(y,y')=0$ and you can write $y$ with the respect to $y'$, i.e.;$$y=g(y')$$ then differentiate form both sides with repect to $x$ and then out $p=y'$. You will find a separable OE: $$pdx=g'(p)dp$$ I think it is better to indicate the solution as a parametric general solution: $$x=\int \frac{g'(p)}pdp,~~~y=g(p)$$

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  • $\begingroup$ can you explain more your hint ? $\endgroup$ – user79560 May 26 '13 at 16:27
  • $\begingroup$ @rama.rnsh: It is just a hint. Do it to see if it works or not. $\endgroup$ – mrs May 26 '13 at 16:29
  • $\begingroup$ $y=y'e^{y'}$ implies that $$y' = y'' e^{y'}(1+y')$$ so that if $p=y'$ we have $$1=e^{p}(1+1/p)\times p'$$ Then $t=\int dp e^p (1+1/p)= \mathrm{Ei}(p)+e^p +t_0$. I doubt this can be inverted for $p$ and integrated. $\endgroup$ – Sharkos May 26 '13 at 16:40
  • $\begingroup$ $1+$ I hope to see you before I go to bed (when you awaken, perhaps?) $\endgroup$ – amWhy May 27 '13 at 0:09
  • $\begingroup$ $\Large\color{green}{✓}^\color{red}{❁}$@BabakS. $\endgroup$ – Software May 27 '13 at 8:56

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