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Homework Help: 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


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    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
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