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Differential Equations of form y'(x)=f(ax+by+c)

  1. Apr 17, 2010 #1
    1. The problem statement, all variables and given/known data
    My professor states that a differential equation of form y'(x)=f(ax+by+c) can be reduced to a separable equation by substituting in v=ax+by+c, but I don't see how.

    Edit: more specifically: y'(x)= sqrt(3x -4y +2)


    2. Relevant equations
    y'(x)=f(ax+by+c)
    v=ax+by+c

    3. The attempt at a solution

    If v=ax+by+c, then dv/dx = a + b*dy/dx
    Then dv/dx = a + b*f(v)

    But this isn't a separable differential equation... the constant a is in the way.

    Edit: more specifically:
    dv/dx = 3 - 4*sqrt(v)
     
    Last edited: Apr 17, 2010
  2. jcsd
  3. Apr 17, 2010 #2
    Does

    [tex]\displaystyle\int \frac{dv}{a + bf(v)} = \displaystyle\int dx[/tex]

    not work?
     
  4. Apr 17, 2010 #3
    yeah, i was about to post that i'm a moron for not realizing such a solution a few moments ago...

    But now I'm stuck on how to solve

    [tex]\displaystyle\int \frac{dv}{3 - 4\sqrt{v}} = \displaystyle\int dx[/tex]
     
    Last edited: Apr 17, 2010
  5. Apr 17, 2010 #4
    I just solved that particular integral by hand. I don't know if I should give away the method I used yet. It's an easy straightforward one, once you see what to do. No trigonometric substitution or anything like that, obviously.
     
  6. Apr 17, 2010 #5
    I'll give it another shot. If i still can't figure it out in a while, I'll ask again?

    Edit:

    I think I solved it, but it's a little messy

    [tex]
    \displaystyle\int \frac{dv*(3 + 4\sqrt{v})}{9 - 16v} = \displaystyle\int dx
    [/tex]

    [tex]
    \displaystyle\frac{-3*\ln{|9-16v|}}{16} + \frac{\sqrt{v}}{2} + c = x
    [/tex]

    [tex]
    \displaystyle\ln{|64y - 48x -23|} - \frac{8\sqrt{3x - 4y + 2}}{3} - \frac{16x}{3} + c = 0
    [/tex]

    Is this good?
     
    Last edited: Apr 17, 2010
  7. Apr 18, 2010 #6
    That method seems a little messy.

    [tex]x = \int \frac{dv}{3-4\sqrt{v}} = \int \frac{3+4\sqrt{v}}{9-16v}dv[/tex]

    [tex]x = \int \frac{3}{9-16v}dv + 4 \int \frac{\sqrt{v}}{9-16v}dv[/tex]

    [tex]x = - \frac{3}{16}\ln|9-16v| + 4 \int \frac{\sqrt{v}}{9-16v}dv[/tex]

    I fear the second integral is not equal to sqrt(v)/2.

    My suggestion would have been to use the substitution u = sqrt(v), then the integral becomes

    [tex]x = \int \frac{2u}{3-4u}du[/tex]

    which is the same as

    [tex]x = \int \left( - \frac{1}{2} + \frac{3/2}{3-4u}\right) du[/tex]

    I think. Might want to double check the division.
     
  8. Apr 18, 2010 #7
    Ahh substitution. It seems obvious now lol thanks.

    But can you tell me how you brought the 1/2 out of the last equation to get rid of the extra u? I can see that the last two equations are equivalent, but I don't understand the process. Did you just guess at taking -1/2 out? Or is there some basic rule I'm forgetting? On another problem right now i need to get the integral of (2u + 1)du/(5u + 5). Would the same steps apply there?

    Edit: Scratch helping me solve (2u + 1)du/(5u +5), I figured it out using the same method you showed me. Still, could you tell me if that process has a name? I don't feel like I've ever been taught it before. Or is it just supposed to be obvious to the student?
     
    Last edited: Apr 18, 2010
  9. Apr 18, 2010 #8
  10. Apr 18, 2010 #9
    Cheers
     
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