MAT244-2014F > MAT244 Math--Lectures

Problem 1.1.28

**Victor Ivrii**:

Xinyi Li,

your explanation is superficial. One can note that if $y(t_0)=0$ then $y(t)\to +\infty $ as $t\to +\infty$. Everything was explained before: there is solution $y=-\frac{1}{2}e^{-t}$ tending to $0$ as $t\to +\infty$ and all other solutions tend to $\pm \infty$.

**Xinyi Li**:

Thanks for explaining.

Yeah, i think i am using too easy theories to look at the questions.

Thanks for pointing out, now I can see the whole picture.

Big helps~

**Chenxi Lai**:

Hi, i've seen somebody post this question. while i dont think the graph is right. I think the y=-e^(-t) is a equilibrium solution. so the graph may look in following form.

**Xinyi Li**:

Based on my understanding, the directional field is purely described by this equation only: y^'=e^(-t)+y

From the equation we can see that, when y=0 and tâ†’+infinity yâ€™ will goes to 0. That is the only place that the directional field will have horizontal lines meaning zero slope. The answer should be something like the attachment.

For the general solution, clearly the prof has helped us confirmed is:

So if we set c = 0, then we will easily find an equilibrium solution for the ODE. I think you have a little misunderstanding with the directional field. You can look at the original post, professor Victor has given a pretty clear answers. Hope this can help.

**Victor Ivrii**:

Chenxi Lai, if you saw that somebody already started this topic, you should not start a new one, but continue in the existing one! The question has been settled already: there is a solution which tends to $0$ at $+\infty$, all solutions above it tend to $+\infty$ and all solutions below it â€” to $-\infty$.

PS Try not to post crappy snapshots from you cell

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