Consider the differential equation $\cfrac{dy}{dx} = \cfrac{x^{3}-x}{1+e^{x}}$. How to find constant solution $y(x)$ and $\displaystyle\lim_{x \to \infty}y(x),$ where $y(x)$ is the solution such that $y(0) = \cfrac{1}{2}.$ I know if $y$ is a constant solution, then $\cfrac{dy}{dx} = 0$, which implies $$x^{3} - x=0~~\implies x=0,1,-1. $$ Now, how can I show that $\displaystyle\lim_{x \to \infty}y(x)=0~$ such that $y(0) = \cfrac{1}{2}.$
Asked
Active
Viewed 97 times
1
-
But here how can I find exact solution? – Manoj Kumar Aug 19 '21 at 04:02
-
2That's easy, is just $y(x) = \displaystyle \frac12 + \int_0^x \frac{t^3-t}{1+e^t}dt$. – azif00 Aug 19 '21 at 04:06
-
It's hardly obvious that $\int_0^\infty \frac{ t^3 - t }{1 + e^t} d t = -\frac{1}{2}$? – jim Aug 19 '21 at 04:10
-
And the above equation isn't even true, the integral is approximately 4.859. So something's off with the question as written. – Sarvesh Ravichandran Iyer Aug 19 '21 at 04:14
-
1As written the equation has no constant solutions, unless you meant to replace all of the $x$ with $y$ on the right hand side. – Ninad Munshi Aug 19 '21 at 04:18
-
Ok thanks, that means the Professor asked wrong question. – Manoj Kumar Aug 19 '21 at 04:20
-
@TeresaLisbon. If $x\to \infty$ the result is just $\frac{1}{120} \pi ^2 \left(7 \pi ^2-10\right) \sim 4.85973$ as you gave. – Claude Leibovici Aug 19 '21 at 05:13
-
@ClaudeLeibovici Indeed, which would make for a nice question by itself (I mean, I've also found out that the anti-derivative is given by some polylogarithmic type functions, and the idea for the computation "should be" the trick where you multiply top and bottom by $e^{-x}$ and then take $y = e^{-x}+1$ and force polylogs into the reckoning). – Sarvesh Ravichandran Iyer Aug 19 '21 at 05:40
-
There's also this post @ClaudeLeibovici. All in all, a very fruitful day then. – Sarvesh Ravichandran Iyer Aug 19 '21 at 05:44
-
@TeresaLisbon. Thanks for the link. – Claude Leibovici Aug 19 '21 at 05:51
1 Answers
1
Hint.
$$ |\cfrac{x^{3}-x}{1+e^{x}}|\le |(x^{3}-x)e^{-x}| $$
so we have
$$ -(x^{3}-x)e^{-x}\le \frac{dy}{dx}\le (x^{3}-x)e^{-x} $$
and
$$ \int (x^{3}-x)e^{-x} dx = -e^{-x} \left(x^3+3 x^2+5 x+5\right)+C $$
Cesareo
- 33,252