ecuatie diferentiala cu limita solutiei 0
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Cezar Lupu
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ecuatie diferentiala cu limita solutiei 0
Se considera \( a(.), b(.):I=(0,\infty)\to\mathbb{R} \) doua functi continue care definesc ecuatia diferentiala \( \frac{dx}{dt}=-a(t)x+b(t) \). Sa se demonstreze ca daca \( a(t)\geq c>0 \) oricare ar fi \( t\in I \) si \( \lim_{t\to\infty} b(t)=0 \), atunci \( \lim_{t\to\infty}\varphi(t)=0 \), pentru orice \( \varphi(.) \) solutie a ecuatiei diferentiale definite mai sus.
Last edited by Cezar Lupu on Mon Oct 08, 2007 12:30 am, edited 1 time in total.
An infinite number of mathematicians walk into a bar. The first one orders a beer. The second orders half a beer. The third, a quarter of a beer. The bartender says “You’re all idiots”, and pours two beers.
\( \frac{dx}{dt}=-a(t)x+b(t) \) \( \Rightarrow \varphi(t)=e^{-A(t)}(\int{e^{A(t)}bt)dt+K), A(t)=\int{a(t)}dt \) si avem ca \( \lim_{t\to\infty}\varphi(t)=\lim_{t\to\infty}\frac{\int_0^t{e^{A(s)}b(s)ds+k}}{e^{A(t)}} \) si cu L'Hospital \( \lim_{t\to\infty}\varphi(t)=lim_{t\to\infty}\frac{e^{A(t)}b(t)}{a(t)e^{A(t)}}=\lim_{t\to\infty}\frac{b(t)}{a(t)}=0, a(t)\geq{c}>0 \Rightarrow \lim_{t\to\infty}a(t)\geq{c}>0 \)