Find the range of $y$ in a DE

Question

Consider the equation $$y' = y^2 - y - 2 = (y+1)(y-2).$$ If $y(10) = 0$, find the range of $y(t)$ for $t>10$. That is, find the best $A$ and $B$ such that $A<y(t)<B$ for $t>10$.

From integration by parts, and using $y(10) = 0$, I get the equation $$2e^{3t-30} = \frac{|y-2|}{|y+1|}.$$

Let $f(t) = 2e^{3t-30}$. Since it's for $t>10$, $f(10) = 2$, and we have $2=\frac{|y-2|}{|y+1|}$. Depending on the sign I choose to use, I get either that $y=-4$ or $y =0$. Since $t: 10 \rightarrow \infty$, either $y=-4$ or $y=0$ is the lower bound.

For $t \rightarrow \infty$, we have $\infty = \frac{|y-2|}{|y+1|}$ (I realise this is not really considered a valid equation...). But considering how the numerator and denominator are both $y$ (the largest power, that is), I don't see how it could tend to $\infty$.

Any help would truly be appreciated!

Answer 1

Disclaimer: the below is not a rigorous argument but an outline of one.

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It is much easier to note that for $-1<y<0$ we have

$$y'=(y+1)(y-2)<0$$

That is to say, $y$ is decreasing for this range. One can note that $y=-1$ is attractive, since $y'<0$ for $y>-1$ and $y'>0$ for $y<-1$. That is, it's not possible to leave $y=-1$ once you've reached it. Likewise, there are no such other points for $y\in(-1,0)$, and so $y$ does not have an asymptote above $y=-1$. Thus, the range is given by $(-1,0)$ for $t\in(10,\infty)$.

Answer 2

$$y'=(y+1)(y-2) \implies \int \frac{dy}{(y+1)(y-2)}= \int dx+C$$ $$ \implies \int \frac{1}{3} \left ( \frac{1}{y-2}-\frac{1}{y+1} \right) dy=x+c \implies \ln\frac{(y-2) }{(y+1)}=(3x+3C)$$ $$ \implies \frac{y-2}{y+1}=e^{3x+3C}>0 \implies y<-1 ~~or ~~ y>2.$$ Hence the range is $y \in (-\infty, -1) \cup y\in (2,\infty)$.