The range for y if t>=10 is [-4, 0].

[JessicaHelena]

Homework Statement

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.

Relevant Equations

N/A

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 realize 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!

 

[Mark44]

Homework Statement:: 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.
Homework Equations:: N/A

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

I get a solution that's quite different. A simpler approach might be to separate the DE, and then use partial fraction decomposition. After separation you end up with $\int \frac{1}{(y+1)(y - 2)} dy = \int dt$. The integral on the left can be evaluated without much work.

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 realize 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!

 

[LCKurtz]

Is it really necessary to actually solve the DE? What about looking at the slope of the direction field? You have $y' = (y+1)(y-2)$ so the slope is negative when $-1 < y < 2$ and positive otherwise. What could the graph possibly do if $y(10)=0$?

 

[ehild]

Homework Statement:: 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.
Homework Equations:: N/A

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.

Your solution is correct, but It is given that y=0 when t=10. In what interval should be y ? What is the y(t) function then? Is it monotonous?

For $t \rightarrow \infty$, we have $\infty = \frac{|y-2|}{|y+1|}$ (I realize 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 $\infty$.

y can tend to -1 and then your f(t) tends to infinity. So what is the range for y if t>=10?