How is the equality in this PDE achieved?

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Discussion Overview

The discussion revolves around understanding a specific equality in a partial differential equation (PDE) related to the solution of an ordinary differential equation (ODE) of the form dT/dt = -k*T. Participants explore the derivation of the solution and the relationship between the parameters involved.

Discussion Character

  • Exploratory, Technical explanation, Homework-related

Main Points Raised

  • One participant expresses confusion about how a particular equality in the PDE is achieved and seeks clarification.
  • Another participant identifies the equation as an ordinary differential equation and mentions a parameter "k" in a somewhat informal manner.
  • A third participant provides a detailed derivation of the solution T(t) = Ae^{(ih/E)t}, explaining the steps involved in integrating the equation and separating variables.
  • The same participant suggests that if one did not initially recognize the solution, integrating the equation would lead to the same result.
  • A later reply indicates that the original poster has resolved their confusion independently and apologizes for not mentioning it earlier.

Areas of Agreement / Disagreement

The discussion does not appear to have any significant disagreements, but it includes varying levels of understanding and clarity regarding the derivation process. The initial confusion expressed by the original poster is ultimately resolved without further debate.

Contextual Notes

Some assumptions about the parameters and the context of the PDE may not be fully articulated, and the discussion does not address potential limitations or broader implications of the solution.

Superposed_Cat
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I've got this far on a pde (second last step) but have no idea how they got this equality(I'm a noob), could someone please explain? I was going to put this under homework but it is not homework and it doesn't really fit the template. Thanks in advance.
 

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It is an ordinary diff.eq,
where dT/dt=-k*T, for a "k" with fluffy garments.
 
You can see that T(t)= Ae^{(ih/E)t} does satisfy the equation by differentiating. As arildno says, you can just think of k= (ih/E) and use the fact that the derivtive of Ae^{kt} is kAe^{kt}.

If you are asking "how can we get that solution if we didn't notice that?", you need to integrate to go the other way:
\dfrac{dT}{dt}= -\dfrac{ih}{E}T
separating variables,
\dfrac{dT}{T}= -\dfrac{ih}{E}dt
\int\dfrac{dT}{T}= -\dfrac{ih}{E}dt
ln(T)= -\dfrac{ih}{E}t+ C

Now, take the exponential of both sides:
T= e^{-\frac{ih}{E}t}e^C
and we let A= e^C.
 
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Thanksbut I got it a while ago now. sorry for not mentioning.
 

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