Explain why any solution y=y(x) to the differential equation must have -2<y(x)<1 for all x?

Question

If the initial value problem is y' = y^2+y-2, and y(xo) = yo, what is this question asking?? I understand the equilibrium solutions are -2 and 1, that I have already solved, but I don't know why this is. Could someone rephrase?

Answer 1

Given:

y' = y² + y - 2

y( x₀ ) = y₀

Statement:

Any solution y = y(x) to the given Ordinary Differential Equation must have output values greater than 1 and output values less than -2 for all values of the input variable x.

Solution:

y' = ( y + 2 ) ( y - 1 )

Now, you could go through the calculus and find the "exact" solution for the initial condition (I did for the fun of it), it's kind of neat but it's lengthy.

y(x) = [ y₀ + 2 + 2 ( y₀ - 1 ) * exp( 3 ( x- x₀ ) ] / [ y₀ + 2 - ( y₀ - 1 ) * exp( 3 ( x- x₀ ) ]

But ultimately, if you analyze the above differential equation for the absolute minimum/absolute maximum, you'll notice something.

For the absolute min/max to exist, the derivative of the function will equal 0.

So,

0 = ( y + 2 ) ( y - 1 )

Like you've noted, the roots to that are:

y = -2, and y = 1

But what that interprets to is that the slope will zero out when the output of the function, y(x), approaches either -2 or 1.

However, if you go thru an inequality analysis of the roots like so:

( y + 2 ) ( y - 1 ) > 0

y + 2 > 0 y > -2

y - 1 > 0 y > 1

∴ For y' > 0, then it MUST be true that y > -2 & y > 1

This means for the function to increase, the outputs will be greater than -2 (which consequently means greater than +1, as well)

( y + 2 ) ( y - 1 ) < 0

y + 2 < 0 y < -2

y - 1 < 0 y < 1

∴ For y' < 0, then it MUST be true that y < -2 & y < 1

This means for the function to increase, the outputs will be less than 1 (which consequently means less than -2, as well)

Now, what happens is that these overlap between output values from -2 to 1.

However, that would mean the function is both INCREASING and DECREASING between y = -2 and y = +1.

For -2 < y < 1, y' > 0 & y' < 0....

This can only happen if there are asymptotes @ y = -2 and y = +1 and the function graphs for everything of y > 1 and for everything y < -2.

Which means whatever the solution is, the output values will be greater than 1 ( y > 1 ) or will be less than -2 ( y < -2 ) but cannot plot for anything in between ( -2 < y < 1 )

So, as I looked back thru the solution, I noticed the beginning inequality has the inequality signs flipped in the beginning prompt. I believe there is a typo because the plot I created for this problem does not do what the prompt implies.

The prompt displayed originally from you implies that the solution never reaches any output values greater than 1 nor less than -2, and that is the EXACT opposite of what happens.

I hope this helps! Please message me in the comments if you have any questions, comments, or concerns with how I handled this problem or if you are confused anywhere because there is a lot to take in.

EDIT: Depending on your choice of y₀, you can actually alter that...if you make y₀ restricted to: -2 < y₀ < 1 than so does y(x) get restricted to -2 < y(x) < 1

However, if you make y₀ restricted to y₀ > 1 or y₀ < -2, then you get everything that I asserted in the main part of my solution and y will also be restricted to y > 1 or y < -2, respectively.

Sorry, this was a loaded problem.

Answer 2

The problem is that the solution to this ODE does NOT have to meet that requirement for y ... It does, however, have to have y in that range to be a stable solution, or a solution that converges to y'=0 as x increase s