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Variation of parameters inhomogeneous DE help

  1. May 11, 2009 #1
    Ok heres my problem:
    1. Solve the inhomogeneous second order de:
    x^2y" - 3xy' + 4y =x^4


    2. Worked: y(p) = 1/4*x^4

    Given: y(1) = x^2
    y(2) = log(x)*x^2


    3. I just need help getting the roots of the given de so i can determine y(h) of this de. As i have already solved the particular solution y(p) and then the general solution
    y(g) = y(h) + y(p)

    Im thinking this has imaginary roots as i gave it a crack to get y(h) and the best i could think of was (r - 2/x) - 1/x = 0, which doesnt really lead me anywhere i think.

    Do i have to use the quadratic formula to get the roots of this equation? and if so some pointers in the right direction would be much appreciated!

    Thanks
     
  2. jcsd
  3. May 11, 2009 #2

    rock.freak667

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    solve x^2y" - 3xy' + 4y=0 to get two solutions,y1(x) and y2(x), for the variations of parameters method.
     
    Last edited: May 11, 2009
  4. May 12, 2009 #3

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    ??? You are given that two independent solutions to the associated homogeneous equation are x^2 and log(x)x^2.

    You don't need to do that at all, you are given those solutions!
    Of course, the general solution to the entire equation is y(x)= Ax^2+ Blog(x)x^2+ (1/4)x^4 but I suspect you got that yp by "undetermined coefficients" so have not followed the instructions to use "variation of parameters".

    Although, as I said, you do not need to solve the characteristic equation, no it does not have imaginary roots. You can convert a "Euler type" or "equipotential equation" to a "constant coefficients" equation with the change of variables t= log(x) or x= e^t.
    dy/dx= dy/dt dt/dx= (1/x)dy/dt so d^2y/dx^2= d ((1/x)dy/dt)/dx= (1/x) d(dy/dt)/dx- 1/x^2 dy/dx= (1/x)^2 d^2y/dt^2- (1/x^2)dy/dt so x^2d^2y/dt^2- 3x dy/dx+ 4y= x^2(1/x^2)(d^2y/dt^2- dy/dt)- 3x (1/x)dy/dt+ 4y= d^2y/dt^2- 4 dy/dt+ 4y= 0. The characteristic equation for that constant coefficients equation, and so for the equipotential equation, is r^2- 4r+ 4= (r- 2)^2= 0 which has r= 2 as a double root. For a constant coefficients equation, that would mean the solution was y= Ae^(2t)+ Bte^(2t). Since t= log(x), e^(2t)= e^(2log(x))= e^(log(x^2))= x^2 and te^(2t)= log(x)x^2, as "given".
     
    Last edited: May 12, 2009
  5. May 12, 2009 #4
    We were given the solution y(1) = x^2 was a solution, then had to prove y(2) = log(x)x^2 is also a solution.

    Haha your spot on about me getting y(p) via constant coefficients, which i ended up realising just before and ajusted my solution. Seems so stupid now i look back at it.

    I did end up using the change of variables with x= e^t and also got the root r(1)= r(2) = 2 so y= Ae^(2t)+ Bte^(2t) just like you mentioned, and went on to get the general solution.

    All cleared up now, help was much appreciated!!!
     
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