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Homework Help: Which Method Do I Use?

  1. Sep 26, 2006 #1
    Could somone help me out and tell me what method to use to solve:

    (t^3)(y^2)dt + (t&4)(y^-6)dy = 0

    Also, the equation dy/dx - y/x = (x^2)sin2x is homogeneous, right?

    Thanks,
    Dao Tuat
     
  2. jcsd
  3. Sep 26, 2006 #2
    for the top one try seperating the variables
     
  4. Sep 26, 2006 #3
    Ok, so I seperated them and integrated and ended up with:
    y=[7^(6/7)]/(7(ln(t)-c))

    Does this look right?

    Thanks,
    Dao Tuat
     
    Last edited: Sep 27, 2006
  5. Sep 27, 2006 #4
    Can anyone help me out with the second equation? I'm completly lost.
     
  6. Sep 27, 2006 #5
    that equation for post #3 is not what i got. What did your seperation of variables look like?

    EDIT: I'm assuming that equation you have is either [tex] y=\frac {7^\frac {6}{7}}{7(ln(t)-c)}[/tex] or [tex] y=7^\frac{6/7}{7(ln(t)-c))}[/tex]
     
    Last edited: Sep 27, 2006
  7. Sep 27, 2006 #6
    Yes, it looked like the first one, which is the same as
    y=7*sqrt[-1/(7[ln(t)+c])], correct?

    What I did was end up with -[(t^3)/(t^4)]dt=[(y^-6)/(y^2)]dy after seperating

    I then integrated and got -ln(t)+c=-1/(7y^7), and then solved for y.

    For the other equation, dy/dx - y/x = (x^2)sin2x, I got:
    y=[(xsin2x)/4]-[([x^2]cos2x)/2]
     
  8. Sep 28, 2006 #7
    Do these answeres and approach look right?
     
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