Why Are Differential Equations So Confusing?

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Discussion Overview

The discussion revolves around the confusion surrounding differential equations, specifically focusing on the equation $$\d{y}{t}=t^2\left(y-y^3\right)^4$$ and the conditions under which $$\d{y}{t}=0$$. Participants explore the implications of this equation, including identifying solutions and analyzing the behavior of functions at specific points.

Discussion Character

  • Exploratory
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • Some participants note that $$\d{y}{t}=0$$ when y is a constant function of t.
  • One participant suggests that $$\d{y}{t}=0$$ can be analyzed using the equation $$t^2\left(y-y^3\right)^4=0$$ to identify turning points on a graph.
  • Participants propose values for t and y that satisfy the equation, including $$t=0$$, $$y=0$$, $$y=-1$$, and $$y=1$$.
  • There is a discussion about eliminating certain choices based on the slope of the solution at specific points, particularly at $$t=0$$ and $$y=0$$.
  • One participant raises the question of whether $$y=-1$$ can be treated as an extraneous root, given that it does not appear in the graph of one of the proposed solutions.
  • Another participant confirms that graph #3 can be eliminated based on the behavior at $$y=0$$, leaving graph #1 as a potential answer, while also questioning the validity of including $$y=-1$$.

Areas of Agreement / Disagreement

Participants express differing views on the implications of the solutions and the treatment of $$y=-1$$, indicating that there is no consensus on how to interpret the results or which graph accurately represents the solution.

Contextual Notes

Participants do not fully resolve the implications of treating $$y=-1$$ as an extraneous root, nor do they clarify the conditions under which the proposed graphs satisfy the original differential equation.

MermaidWonders
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View attachment 7900

I'm so confused about this question :(
 

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What can cause:

$$\d{y}{t}=0$$ ?
 
I'm not sure what kind of answer you're looking for, but all I know is that $\d{y}{t}$ = 0 when y is a constant function of t... 😢
 
MermaidWonders said:
I'm not sure what kind of answer you're looking for, but all I know is that $\d{y}{t}$ = 0 when y is a constant function of t... 😢

Let's look at what we're given:

$$\d{y}{t}=t^2\left(y-y^3\right)^4$$

Now, the reason I asked to consider where $$\d{y}{t}=0$$, is twofold...It is easy to identify on the given graph where there are turning points, and it is easy to use to zero-factor property. So, we then have:

$$t^2\left(y-y^3\right)^4=0$$

What values of $t$ and $y$ satisfy the above?
 
t = 0, y = 0, y = -1, y = 1?
 
MermaidWonders said:
t = 0, y = 0, y = -1, y = 1?

Good, yes. Now, let's look at $t=0$...can we eliminate any of the choices based on the fact that the slope of the solution must be zero when $t=0$?
 
Eliminate #2!
 
MermaidWonders said:
Eliminate #2!

Okay, how about $y=0$? Which, if any of the remaining two choices, can you eliminate with that?
 
MarkFL said:
Okay, how about $y=0$? Which, if any of the remaining two choices, can you eliminate with that?

Now that's a bit tricky, here. Would it be #3, since #3 has a positive slope at y = 0? If so, this would leave us with #1 as the answer, then...

And the other thing is, since y = -1 is one of the solutions to $t^2$$(y - y^3)^4 = 0$, nothing happens at y = -1 for graph #1 since this graph doesn't even extend to y = -1 (with the lowest y value being 0)? Like if graph #1 is the answer to this question, it doesn't really satisfy all of the criteria (solutions)... Unless... we treat y = -1 as an extraneous root to this equation, but does that make sense?
 
Last edited:
  • #10
MermaidWonders said:
Now that's a bit tricky, here. Would it be #3, since #3 has a positive slope at y = 0? If so, this would leave us with #1 as the answer, then...

And the other thing is, since y = -1 is one of the solutions to $t^2$$(y - y^3)^4 = 0$, nothing happens at y = -1 for graph #1 since this graph doesn't even extend to y = -1 (with the lowest y value being 0)? Like if graph #1 is the answer to this question, it doesn't really satisfy all of the criteria (solutions)... Unless... we treat y = -1 as an extraneous root to this equation, but does that make sense?

Anyone??
 
  • #11
MermaidWonders said:
Now that's a bit tricky, here. Would it be #3, since #3 has a positive slope at y = 0? If so, this would leave us with #1 as the answer, then...

And the other thing is, since y = -1 is one of the solutions to $t^2$$(y - y^3)^4 = 0$, nothing happens at y = -1 for graph #1 since this graph doesn't even extend to y = -1 (with the lowest y value being 0)? Like if graph #1 is the answer to this question, it doesn't really satisfy all of the criteria (solutions)... Unless... we treat y = -1 as an extraneous root to this equation, but does that make sense?

Yes, graph #3 can be eliminated with $y=0$, leaving only #1 which fits the other criteria. We can ignore the fact that $y\ne-1$.
 
  • #12
OK, thanks for confirming! :)
 

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