Separating variables within square root expressions

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Discussion Overview

The discussion revolves around techniques for manipulating square root expressions in the context of solving differential equations (DEs), specifically focusing on separating variables to facilitate integration. The problem is framed within a more complex thermodynamic scenario.

Discussion Character

  • Exploratory, Technical explanation, Debate/contested, Mathematical reasoning

Main Points Raised

  • One participant inquires about methods to isolate a variable trapped in a square root expression to enable integration with respect to another variable.
  • Another participant presents a coupled system of equations, suggesting a transformation to express the second derivative in terms of the first derivative, leading to a specific ordinary differential equation (ODE).
  • A participant reflects on their initial approach, acknowledging that their earlier suggestion of using the second derivative was misguided and expresses a realization that they can substitute known expressions to simplify the integration process.
  • A later reply indicates that while the definition of the variable y would have been beneficial, the approach taken was appropriate given the circumstances.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the best method for handling the square root expressions, and there are competing views on the validity of different approaches. The discussion remains unresolved regarding the optimal technique for this specific problem.

Contextual Notes

Limitations include the lack of a clear definition for the variable y, which may affect the ability to isolate it effectively. Additionally, there are unresolved mathematical steps related to the transformation of derivatives and the integration process.

timsea81
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Please see attached. This is a general question for a more complex thermo problem, but it fits this form. Are there any tricks to getting expressions out of square roots in solving DEs?

y=y(x)
x=x(t)

I want to get y by itself so I can integrate with respect to x, but it is trapped in a square root expression. My guess of squaring both sides and "squaring" the dx/dt term by making it a second derivative doesn't seem right, but it's all I can think up. Does anyone have any suggestions?
 

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So you have a coupled system:
[tex] \begin{array}{rcl}<br /> \frac{dx}{dt} & = & \sqrt{y-A} \\<br /> \frac{d^{2}x}{dt^{2}} & = & y-A<br /> \end{array}[/tex]
So this gives the ODE
[tex] \frac{d^{2}x}{dt^{2}}=\left(\frac{dx}{dt}\right)^{2}[/tex]
Let [itex]v=dx/dt[/itex], so:
[tex] \frac{dv}{dt}=v^{2}[/tex]
Can you take it from here?
 
Looking back on it I guess my original post was misleading.

The second derivative statement wasn't a given criteria, that was my first guess at how to solve the problem (squaring both sides but for the left hand side the squaring process involves taking it to its second derivative). That doesn't work, I tried it out with some real equations and thought about it some more and concluded it is pure nonsense.

I think I got it though. I know what y=y(x) is so I can plug it into the left hand side and get a complicated expression where x is the only variable and I can integrate. It's ugly, but it works.
 

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The definition of y would have been helpful, but you did the right thing.
 

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