Finding an interval in which the IVP has a unique solution.

[theBEAST]

Homework Statement

Here is the question with the solution from the textbook:

I don't get how looking at (2/t) and 4t tells us that the solution must be on the interval 0 to infinity. Don't we have to set y' to zero and solve for how the directional fields behave and thus find out how y behaves as t -> infinity?

I don't think I am visualizing this correctly and thus I don't understand how the answer works. Could anyone please explain?

Thanks

 

[STEMucator]

Notice your initial condition, y(1) = 2, which is positive, yes? This corresponds to your interval of t being strictly positive ( Since if t = 0, then our equation is undefined ).

So 0 < t < ∞

 

[theBEAST]

Notice your initial condition, y(1) = 2, which is positive, yes? This corresponds to your interval of t being strictly positive ( Since if t = 0, then our equation is undefined ).

So 0 < t < ∞

So because of the discontinuity at t = 0, the solution will never be negative if the initial condition is positive?

 

[STEMucator]

So because of the discontinuity at t = 0, the solution will never be negative if the initial condition is positive?

Yes indeed.

 

[theBEAST]

What about this question? Why is the negative answer from taking the root of y^2 rejected?

Is it because y has a discontinuity at 0 and thus because our initial condition y(2) = 3 starts in the positive y the solution must vary between 0 < y < infinity?

I just want to clarify this one because it is slightly different when compared to my original question in that the discontinuity is y instead of t or x.

Thank you!

 

[STEMucator]

You would want the solution corresponding with the positive initial condition, so yes you ignore the negative case.