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Separable equations

  • #1

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.
 

Answers and Replies

  • #2
954
117


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)
 
  • #3


I see. I originally thought C was some sort of magic number that stays as is, but your solution is more logical.
 
  • #4
954
117
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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