Differential Equations Problems

[MermaidWonders]

View attachment 7900

I'm so confused about this question :(

 

[MarkFL]

What can cause:

\(\displaystyle \d{y}{t}=0\) ?

 

[MermaidWonders]

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...

 

[MarkFL]

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:

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

Now, the reason I asked to consider where \(\displaystyle \d{y}{t}=0\), 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:

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

What values of $t$ and $y$ satisfy the above?

 

[MermaidWonders]

t = 0, y = 0, y = -1, y = 1?

 

[MarkFL]

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 $t=0$?

 

[MermaidWonders]

Eliminate #2!

 

[MarkFL]

Eliminate #2!

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

 

[MermaidWonders]

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?

 

[MermaidWonders]

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??

 

[MarkFL]

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 $y\ne-1$.

 

[MermaidWonders]

OK, thanks for confirming! :)