1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Differential Equations - am I doing this right

  1. Sep 27, 2009 #1
    1. The problem statement, all variables and given/known data

    What is [tex] y(x) [/tex] [[tex] y [/tex] as a function of [tex] x [/tex]] given that

    2. Relevant equations

    [tex] \frac{dy}{dx} = 3\frac{y}{x} [/tex]

    where you are given the boundary condition that [tex] y = y_0 [/tex] at [tex] x = x_0 [/tex] where [tex] y_0 [/tex] and [tex] x_0 [/tex] are constants.



    3. The attempt at a solution

    Separating variables [tex] \Rightarrow \frac{dy}{y} = \frac{3 dx}{x} [/tex]

    Integrating [tex] \Rightarrow \int{\frac{dy}{y}} = \int{\frac{3 dx}{x}[/tex]

    Integrals give [tex] \Rightarrow ln(y) = 3 ln(x) + C [/tex]

    Given the boundary conditions then [tex] \Rightarrow ln(y_0) = 3 ln(x_0) + C [/tex]

    Therefore [tex] \Rightarrow C = ln(y_0) - 3ln(x_0) = ln(\frac{y_0}{x_0^3}) [/tex]

    Plugging back into [tex] ln(y) = 3 ln(x) + C [/tex] gives
    [tex] ln(y) = 3 ln(x) + ln(\frac{y_0}{x_0^3}) [/tex]

    Therefore [tex] \Rightarrow y = exp\left(ln(x^3) + ln(\frac{y_0}{x_0^3})\right) [/tex]

    [tex] \Rightarrow y = exp\left(ln(x^3 \bullet \frac{y_0}{x_0^3})\right) \Rightarrow y = y_0(\frac{x}{x_0})^3[/tex]
     
  2. jcsd
  3. Sep 27, 2009 #2

    jamesrc

    User Avatar
    Science Advisor
    Gold Member

    Looks good to me.

    One way you can check your work is to find dy/dx from your final equation, and verify that you get your initial equation...
     
  4. Sep 27, 2009 #3
    That's the part where I got worried about whether I did it right or not.

    I think I was getting stucking trying to prove that it was correct, and I just did it.

    Thank you for the assurance and the help! :) Really appreciate it
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook