I'm trying to make sense of this question.

" Find the function x(t) where x(1) = 0 and examine the behavior of x(t) as t -->infinity.

S(2x/2-x^2)dx = S dt " (S is the integral sign)

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- Aug 23rd 2009, 10:46 AMeniuqvwcant make sense of integral question
I'm trying to make sense of this question.

" Find the function x(t) where x(1) = 0 and examine the behavior of x(t) as t -->infinity.

S(2x/2-x^2)dx = S dt " (S is the integral sign) - Aug 23rd 2009, 11:03 AMChris L T521
$\displaystyle \int\frac{2x}{2-x^2}\,dx=\int\,dt$?

Make a u sub: $\displaystyle u=2-x^2\implies-\,du=2x\,dx$.

Thus, we have $\displaystyle -\int\frac{\,du}{u}=\int\,dt\implies -\ln\left|u\right|=t+C\implies \ln\left|2-x^2\right|=-t+C$ $\displaystyle \implies \left|2-x^2\right|=Ce^{-t}\implies 2-x^2=Ke^{-t}\implies x(t)=\sqrt{2-Ke^{-t}}$

Applying the initial condition, we have $\displaystyle 0=\sqrt{2-Ke^{-1}}\implies K=2e$.

Therefore, $\displaystyle x(t)=\sqrt{2-2e^{1-t}}$.

Now, what is $\displaystyle \lim_{t\to\infty}\sqrt{2-2e^{1-t}}$?