Solve Simple ODE: Prove y=Cx^k

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Homework Help Overview

The discussion revolves around a simple ordinary differential equation (ODE) related to the boiling of a mixture of two liquids, X and Y. The problem involves proving a relationship of the form y = Cx^k, where C and k are constants, based on the proportionality of the quantities of the liquids in the mixture.

Discussion Character

  • Exploratory, Conceptual clarification, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the initial setup of the problem, questioning how to express the quantities X and Y in terms of x and y. There is also exploration of the relationship between the rates of change of these quantities.

Discussion Status

The conversation is ongoing, with some participants attempting to clarify the relationships between the variables involved. There is an indication that one participant feels ready to proceed with solving the ODE, although no consensus or final approach has been established yet.

Contextual Notes

Participants are navigating the definitions and relationships within the problem, with some uncertainty about the implications of the proportionality stated in the problem. The discussion reflects a focus on understanding the setup rather than moving directly to a solution.

John O' Meara
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A mixture of two liquids X and Y, is boiling. At any time t there is present in the mixture a quantity x of one and y of the other. The ratio of the amounts of X and Y passing off at anyone time is proportional to the ratio of the quantities still in the liquid state. Prove that [tex]y=Cx^k \\[/tex], where C and k are constant.
I really don't know how to start this off, I have [tex]\frac{X}{Y} \propto \frac{x}{y} \\[/tex]. What is the next step? Do I express X in terms of x and Y in terms of y? Thanks for helping.
 
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Well, what is X/Y? X and Y are just labels. x and y are the quantities. Read the problem again and tell me what it says about the relation between x,y and x',y'.
 
Does it say x'/y'=k*x/y, where x=x(t) e.t.c.
 
Sure. Now can you solve that ODE?
 
I think I can.
 

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