Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Bit hazy on reduction of order

  1. Mar 31, 2004 #1
    Although on the whole my Boyce & DiPrima book is fairly good in of itself and compared with others, but of-coarse some areas are lacking.

    vt'' - 3v' = 0

    I suppose I integrate twice, but then what?
     
  2. jcsd
  3. Mar 31, 2004 #2

    matt grime

    User Avatar
    Science Advisor
    Homework Helper

    vt''-3v'? You might want to check that, v and t are both functions of some paramater?
     
  4. Mar 31, 2004 #3
    More specifically, v(t)*t'' - 3v(t) = 0

    I'd go insane if I kept that (t) in there while differentiating and combining. :P
     
  5. Mar 31, 2004 #4

    matt grime

    User Avatar
    Science Advisor
    Homework Helper

    And that factorizes as:

    v(t)*(t''-3)=0

    not sure that's right either as now the second v is not primed.
     
  6. Mar 31, 2004 #5
    v''(t)*t - 3v'(t) = 0

    Sorry, this is it now.
     
  7. Mar 31, 2004 #6

    matt grime

    User Avatar
    Science Advisor
    Homework Helper

    Well, one solution is v(t)=t^3, don't you have to use this to get the other solution?


    Edit, of course it isn't. There's the constant solution, duh, and that gives the answer in the next post.
     
    Last edited: Apr 2, 2004
  8. Apr 1, 2004 #7
    The general solution

    Hi;
    The general solution is:
    [tex]v(t)=C_1t^4+C_2[/tex]
    Good luck,
    Max.
     
  9. Feb 12, 2008 #8
    add the 3v' to the other side getting:
    vt''=3v'
    divide v to the other side and then integrate with respect to t and v, based on the side of the equation, being sure to keep the integration constants in.
     
  10. Feb 13, 2008 #9

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    No, you cannot treat second derivatives like that. If you really had a single equation in two dependent variables, you will need another equation to get an explicite solutions just as you cannot solve one numeric equation in two variables.

    The problem is, after editing, tv"j- 3v'= 0.
    Matt Jacques, since you explicitely refer to "reduction of order", I think you mean this: set u= v' so that u'= v" and you have the first order equation tu'- 3u= 0. Now, you can do exactly what bowzerman suggested, since this is now a first order equation.
    t du/dt= 3u so du/u= 3t dt. ln|u|= (3/2)t2+ C and then
    [tex]u= C_1 e^{(3/2)t^2[itex].
    Since u= dv/dt, we have
    [tex]dv= C_1 e^{3/2}t^2 dt[/tex]
    and only have to integerate again. I'll leave that to you!
     
    Last edited: Feb 15, 2008
  11. Feb 13, 2008 #10

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    No, you cannot treat second derivatives like that. If you really had a single equation in two dependent variables, you will need another equation to get an explicite solutions just as you cannot solve one numeric equation in two variables.

    The problem is, after editing, tv"- 3v'= 0.
    Matt Jacques, since you explicitely refer to "reduction of order", I think you mean this: set u= v' so that u'= v" and you have the first order equation tu'- 3u= 0. Now, you can do exactly what bowzerman suggested, since this is now a first order equation.
    t du/dt= 3u so du/u= 3t dt. ln|u|= (3/2)t2+ C and then
    [tex]u= C_1 e^{(3/2)t^2[tex].
    Since u= dv/dt, we have
    [tex]dv= C_1 e^{3/2}t^2 dt[/tex]
    and only have to integerate again. I'll leave that to you!
     
  12. Feb 14, 2008 #11
    yeah, sorry, I was just wrote that one too quick. Forgot to clarify.
     
  13. Feb 14, 2008 #12
    oh, and shouldn't that be:
    t du/dt = 3u
    1/u du = 3/t ?
    ln(u)=3ln(t) +C
    u= c*t^3

    , then sub, integrate, and then plug in?
     
    Last edited: Feb 14, 2008
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Bit hazy on reduction of order
  1. Reduction of order (Replies: 2)

  2. Reduction of Order (Replies: 2)

Loading...