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Solution for a first-order differential equation

  1. Mar 6, 2017 #1
    1. The problem statement, all variables and given/known data
    determine by inspection at least two
    solutions of the given first-order IVP
    dy/dx = 3y2/3
    y(0)=0
    2. Equations:
    integral xa dx= xa+1/(a+1)+constant
    3. The attempt at a solution
    change its form to 1/y2/3 dy/dx =3
    integrate both sides with respect to x
    then it will be
    1/y2/3 dy = 3dx
    now integrate to get y1/3/(1/3)=3x+c
    for y(0)=0 then c=0
    then y1/3/(1/3) =3x then y=x3
    Is that right? does I miss any condition or jumped over some steps?
     
    Last edited: Mar 6, 2017
  2. jcsd
  3. Mar 6, 2017 #2

    haruspex

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    A couple of errors there. Write 1/y2/3 as ya. What is the integral ya.dy?
    That's wrong too, but happens to cancel one of the earlier errors.
     
  4. Mar 6, 2017 #3
    Thanks I corrected them (it was only typing error)
     
  5. Mar 6, 2017 #4

    haruspex

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    That corrected one of the errors in the integration step. One remains. As I suggested, rewrite 1/y2/3 in the form ya before trying to integrate it.

    You are asked for two solutions. Whenever you divide by an expression, what should you check for? At what step did you do a division?
     
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