- #1
MermaidWonders
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Why Are Differential Equations So Confusing?
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In summary: Yes, graph #3 can be eliminated with $y=0$, leaving only #1 which fits the other criteria. We can ignore the fact that $y\ne-1$.
- #2
MarkFL
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What can cause:
\(\displaystyle \d{y}{t}=0\) ?
\(\displaystyle \d{y}{t}=0\) ?
- #3
MermaidWonders
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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... 
- #4
MarkFL
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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:
\(\displaystyle \d{y}{t}=t^2\left(y-y^3\right)^4\)
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?
- #5
MermaidWonders
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t = 0, y = 0, y = -1, y = 1?
- #6
MarkFL
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MermaidWonders said:
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 $t=0$?
- #7
MermaidWonders
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Eliminate #2!
- #8
MarkFL
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MermaidWonders said:
Okay, how about $y=0$? Which, if any of the remaining two choices, can you eliminate with that?
- #9
MermaidWonders
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MarkFL 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?
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- #10
MermaidWonders
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MermaidWonders said:
Anyone??
- #11
MarkFL
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MermaidWonders said:
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
MermaidWonders
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OK, thanks for confirming! :)
FAQ: Why Are Differential Equations So Confusing?
What are differential equations?
Differential equations are mathematical equations that describe the relationship between a function and its derivatives. They are used to model many physical, biological, and economic phenomena.
What is the purpose of solving differential equations?
The purpose of solving differential equations is to find a function that satisfies the equation and can be used to make predictions or analyze a system.
What are some common methods for solving differential equations?
Some common methods for solving differential equations include separation of variables, substitution, and using integrating factors. More complicated equations may require numerical methods or computer simulations.
What are initial value problems?
Initial value problems are a type of differential equation problem where the values of the function and its derivatives are known at a specific point. The goal is to find the function that satisfies the equation at all points.
What are boundary value problems?
Boundary value problems are a type of differential equation problem where the values of the function are known at multiple points. The goal is to find the function that satisfies the equation at all points between the given boundaries.