Separable Equations: Solving with Constant C

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Homework Help Overview

The discussion revolves around solving separable equations, specifically focusing on the manipulation of constants during the solution process. Participants are examining the relationship between constants in the context of a differential equation solution.

Discussion Character

  • Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants are attempting to understand the transformation of constants in the equation, particularly the reasoning behind defining a new constant C as half of C1. Questions arise regarding the nature of constants in the context of the solution.

Discussion Status

Some participants express confusion about the definition and manipulation of constants, while others provide clarification on the flexibility of defining new constants for convenience. The discussion appears to be productive, with participants engaging in reasoning and clarification.

Contextual Notes

There is an underlying assumption that the constants can be redefined without loss of generality, which is a common practice in mathematical problem-solving. Participants are navigating this concept within the framework of their homework assignment.

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Homework Statement



step 1. 2((2/3)y^(3/2) = 2x^(1/2) + C1

step 2. (2/3)y^(3/2) - x^(1/2) = C, where C = 1/2C1



The Attempt at a Solution



I don't understand the where C = 1/2C1 - what is that? I understand everything else, except that.
 
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step 1: 2((2/3)y^(3/2) = 2x^(1/2) + C1

step 1.33: 2((2/3)y^(3/2) - 2x^(1/2) = C1
(bring 2x^(1/2) term over)

step 1.67: ((2/3)y^(3/2) - x^(1/2) = C1 /2
(divide both sides by 2)

step 2. (2/3)y^(3/2) - x^(1/2) = C, where C = (1/2)C1
(replace C1 /2 with C)
 


I see. I originally thought C was some sort of magic number that stays as is, but your solution is more logical.
 
The C's are just constants, so we are free to define new constants in terms of the old ones to make it convenient.
 

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