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 Equation Substitution

  1. Oct 5, 2009 #1
    1. The problem statement, all variables and given/known data

    Solve the given differential equation by using an appropriate substitution.

    2. Relevant equations

    [tex]x\frac{dy}{dx} = y + \sqrt{x^{2} - y^{2}}[/tex], x > 0

    3. The attempt at a solution

    [tex]x\frac{dy}{dx} = y + \sqrt{x^{2} - y^{2}}[/tex]

    [tex]xdy = (y + \sqrt{x^{2} - y^{2}})dx[/tex]

    [tex]y = ux[/tex]

    [tex]u = \frac{y}{x}[/tex]

    [tex]dy = udx + xdu[/tex]

    [tex]x[udx + xdu] = (ux + \sqrt{x^{2} - u^{2}x^{2}})dx[/tex]

    [tex]xudx + x^{2}du = uxdx + x\sqrt{1 - u^{2}}dx[/tex]

    [tex]\frac{du}{\sqrt{1 - u^{2}}} = \frac{dx}{x}[/tex]

    [tex]\frac{du}{\sqrt{1 - u^{2}}} - \frac{dx}{x} = 0[/tex]

    [tex]\int\frac{du}{\sqrt{1 - u^{2}}} - \int\frac{dx}{x} = 0[/tex]

    [tex]sin^{-1}(u) - ln|x| = ln|c|[/tex]

    [tex]sin^{-1}(\frac{y}{x}) - ln|x| = ln|c|[/tex]

    [tex]e^{sin^{-1}(\frac{y}{x})} - x = c[/tex]
     
    Last edited: Oct 5, 2009
  2. jcsd
  3. Oct 5, 2009 #2
    Hi KillerZ,

    Look perfect to me except the very end. You have:

    [tex]
    arcsin(u) - ln|x| = ln|c|
    [/tex]

    the next step isn't quite right, it should go:

    [tex]
    arcsin(u) = ln|x| + ln|c|
    [/tex]

    [tex]
    arcsin\left(\frac{y}{x}\right) = ln|cx|
    [/tex]

    [tex]
    e^{arcsin\left(\frac{y}{x}\right)} = cx
    [/tex]

    but other than that its perfect, I think you'll probably realise the mistake when you see, just a momentary lapse of judgement. I will say you didn't really need to do this

    [tex]
    \frac{du}{\sqrt{1 - u^{2}}} - \frac{dx}{x} = 0
    [/tex]

    in fact its usally best to keep the integration of different variables on different sides, just as you did in you step before this:

    [tex]
    \int \frac{1}{\sqrt{1 - u^{2}}} \ du \ = \ \int \frac{1}{x} \ dx
    [/tex]

    I suppose the mistake at the end is a perfect example of why. Have fun KillerZ :D
     
    Last edited: Oct 5, 2009
  4. Oct 5, 2009 #3

    rock.freak667

    User Avatar
    Homework Helper

    From this line

    [tex]
    sin^{-1}(\frac{y}{x}) - ln|x| = ln|c|
    [/tex]

    if you raise both sides to e then you need to use

    ea+b=ea*eb
     
  5. Oct 5, 2009 #4
    Ok thanks I just messed up my log rules.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Differential Equation Substitution
Loading...