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!

Two questions on differential equations

  1. Nov 29, 2003 #1
    Any thoughts on how to to any of these!! (I'm sorry if i'm insulting you but y' = dy/dx, etc!)

    1)Find a continous solution with continous first derivative of the system:

    y'' + 2y' + 2y = sin x + f(x)

    subject to y(-pi/2)=y(pi)=0, where

    f(x)= 0 (x<or= 0)
    =x^2 (x>0)

    2)You may ignore the first bits (i am including them just incase they're relavent for the last bit)

    Verify y = x+1 is a soln of

    (x^2-1)y'' + (x+1)y' - y = 0 *

    Writing y=(x+1)u show that u'=0

    hence show the gen soln of * is

    y=K[0.25(x+1)ln((x-1)/(x+1))-0.5] + K'(x+1)

    where K and K' are arbitary constants.
     
  2. jcsd
  3. Nov 29, 2003 #2

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    You have the equation y'' + 2y' + 2y = sin x + f(x)
    subject to y(-pi/2)=y(pi)=0.

    I take it that you know how to solve non-homogeneous linear differential equations with constant coefficients and that the problem is just that f(x)= 0 if x<or= 0 and f(x)=x^2 if x>0.

    Think of this first as the equation y'' + 2y' + 2y = sin x. Find the general solution to that equation. That will involve 2 unknown constants. Call that solution Y1(x).

    Now, find the general solution to the non-homogeneous equation y''+ 2y'+ 2y= sin x+ x2. That will also involve 2 unknown constants. Call that solution Y2(x).

    Use the fact that Y1(-pi/2)= 0, Y2(pi)= 0, Y1(0)= Y2=0 (the solution must be continuous at x=0) and Y1'(0)= Y2'(0) (the derivative of the solution must be continuous at x= 0) to get 4 equations to solve for the 4 unknown constants.

    For the second question "Verify y = x+1 is a soln of (x^2-1)y'' + (x+1)y' - y = 0", you should be able to do that part. Find the first and second derivatives of x+1, put them into the equation and see what happens!
    "Writing y=(x+1)u show that u'=0."
    You can't prove that- it's not true. Unless I've made a stupid mistake (it's rather late!) y'= (x+1)u'+ u and y''= (x+1)u''+ 2u so the equation reduces to (x-1)u"+ 3u'= 0.
     
  4. Nov 30, 2003 #3
    Thanks for the advice, I've tried it out and it works fine!

    I made a very careless typing error for the second part:

    "Writing y=(x+1)u show that u'=0"

    I actually meant: u'=du/dx !!

    Sorry!

    Is partial diffrentiation the way to go inorder to show:
    du'/dx + [(3x-1)/(x^2-1)]u' = 0 ?
     
  5. Nov 30, 2003 #4

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    Surely, you don't mean "show that u'= du/dx"??? Isn't that how it's defined?

    Partial differentiation has nothing to do with it. y is a function of the single variable x, u is a function of the single variable x. The product rule is all you need.
     
  6. Nov 30, 2003 #5
    sorry, i am being very careless. I'll write it out again!

    Verify y = x+1 is a soln of

    (x^2-1)y'' + (x+1)y' - y = 0 *

    Writing y=(x+1)u show that u' = du/dx satisfies:

    du'/dx + [(3x-1)/(x^2-1)]u' = 0

    Hence show the gen soln of * is

    y=K[0.25(x+1)ln((x-1)/(x+1))-0.5] + K'(x+1)

    where K and K' are arbitary constants.

    Yes ofcourse, I'll try this (I am really concerned... i miss basic steps :frown:)

    Thank you Halls
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Two questions on differential equations
Loading...