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First order differential equation

  1. Aug 17, 2014 #1
    1. The problem statement, all variables and given/known data
    I'm starting college this autumn(physics) and I started learning some calculus on my own, basic stuff like first order differential equation and so on.Recently i stumbled on something that i don t understand.I was reading the course and re-solving the given examples for myself.So you have the equation

    [tex] \frac{dy}{dx} = \frac{ \frac{y}{x} - 1 }{ \frac{y}{x} + 1 } [/tex]
    2. Relevant equations

    calculus

    3. The attempt at a solution
    Given that the equation is already solved but i didn't understand one of the steps if the solutions i'll show you where i get stuck.

    So you substitute y=tx and here comes the boom, by doing this substitution they say the result is
    [tex] \frac{t+1}{t2+1}dt =-\frac{dx}{x} [/tex] and from here on it s quite simple, you just integrate both sides and get the result.But i didn't understand the steps from substituting y=xt to getting to that result.I ll show you how i tried to solve it but got to a different result.

    So if y=tx --> dy/dx=tdx and it s clear that if you substitute dy/dx in that equation you don t get their answer.

    My main problem is that, in high school we didn't use this notation, we used f'(x) instead of dy/dx and so on, so i'm having a bit of trouble getting used with this notation, but these are the one used in physics so now i have to squash my brains a little and try to get used to them.

    If someone can explain that step to me and maybe give me a few tips on how to make the transition between these two notation easier it would be great!

    Thank you
     
    Last edited: Aug 17, 2014
  2. jcsd
  3. Aug 17, 2014 #2

    HallsofIvy

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    "Here comes the boom"?

    Seeing the "y/x" on the right, I think it would be obvious to set [itex]t= y/x[/itex] which is the same as [itex]y= xt[/itex]. Differentiating both sides with respect to x,
    [tex]\frac{dy}{dx}= x\frac{dt}{dx}+ t[/tex].
    And the whole point of that substitution is that
    [tex]\frac{\frac{y}{x}- 1}{\frac{y}{x}+ 1}= \frac{t- 1}{t+ 1}[/tex]
    So the equation becomes
    [tex]x\frac{dt}{dx}+ t= \frac{t- 1}{t+ 1}[/tex]
    [tex]x\frac{dt}{dx}= \frac{t-1}{t+ 1}- t= \frac{t- 1}{t+ 1}- \frac{t(t+ 1)}{t+1}[/tex]


    Combine the fractions on the right and you have a relatively simple "separable" equation.

    (Personally, I thing [itex]dy/dx[/itex] is a better notation than y' but you can translate back:
    y'= (y/x- 1)/(y/x+ 1). Let t= y/x so that y= xt and y'= xt'+ t. The equation becomes [itex]xt'+ t= (t- 1)/(t+ 1)[/itex]. [itex]xt'= (t- 1)/(t+ 1)- t= (t- 1)/(t+ 1)- (t(t+ 1))/(t+ 1)[/itex]. Of course, when you learned integration, you must have learned that if y'= f(x) then [itex]y= \int f(x)dx[/itex] which follows more easily from [itex]dy/dx= f(x)[/itex], [itex]dy= f(x)dx[/itex].)
     
    Last edited: Aug 17, 2014
  4. Aug 17, 2014 #3
    Ah... Got it! thank you
     
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