Separable Equations: Solving with Constant C

[robertjford80]

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.

 

[Fightfish]

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)

 

[robertjford80]

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

 

[Fightfish]

The C's are just constants, so we are free to define new constants in terms of the old ones to make it convenient.