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Diff. Eq.'s with right sides

  1. Jun 20, 2007 #1
    1. The problem statement, all variables and given/known data
    y''-5y'+4y = f(x)
    where I need to find a
    -general solution if f(x) = 0
    -particular solution if f(x) = 8(x^2) - 25
    -particular solution if f(x) = 9e^(x)
    -general solution if f(x) = (3e^x) - (8x^2) + 25

    2. Relevant equations
    I just completed a few second order differential equations, using the method of variation of parameters to find general solutions, but this sort of question puzzles me. The other questions I have done provided one general solution and I worked with that.

    3. The attempt at a solution
    I presume the method of variation is used for this but I have some problems on how to exactly apply it. For f(x)=0 can I use y=0 as a solution to the differential equation and how do the other f(x) alter my approach?
    Any help or advice would be great. Thanks in advance.
    Last edited: Jun 20, 2007
  2. jcsd
  3. Jun 20, 2007 #2


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    The method of variation of parameters is only really used for inhomogeneous equations. For the first equation (which reduces to y''-y=0) you can simply take the solution to be e^mx, where m is found by substituting into the equation.

    Then use these solutions in the procedure for the other equations. You really need to show some work before we can delve into this any further.
  4. Jun 20, 2007 #3
    Oh stupid me I wrote the f(x) equation wrong. I edited to the right one, sorry about that Cristo. Does what you suggested still apply?
  5. Jun 20, 2007 #4


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    Yes, it's the standard method for any homogenous second order ODE.
  6. Jun 20, 2007 #5
    Alright for the first one this is how I went:
    y''-5y'+4y = 0, y1= e^mx

    y'=V'(e^mx) + V[m(e^mx)]
    y''=V''(e^mx) + V'[m(e^mx)] + V[m^2(e^mx)]

    Sub. back in I get:
    V''(e^mx) + V'[m(e^mx)] + V[m^2(e^mx)] - 5* (V'(e^mx) + V[m(e^mx)]) + 4[V(e^mx)]=0

    This is what I did for previously worked questions and it worked for me, though all of them canceled the V out so just V' and V'' was left.
    Is this right so far? It doesn't look like it to me. Thanks in advance and cheers to cristo for the advice so far.
  7. Jun 20, 2007 #6


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    OK, so sub y_1 and its derivatives into the differential equation to obtain the characteristic equation for m: m^2-5m+4=0.

    Now, can you solve this for m, giving two values; m_1 and m_2? The solution to your equation is then y=Ae^(m_1x)+Be^(m_2x), where A and B are constants of integration. This is then the solution to the differential equation.
  8. Jun 20, 2007 #7
    Alright so if m^2-5m+4=0 then (m-4)(m-1)=0, and hence m1=4 and m2=1.
    I'm just not not 100% sure you got to m^2-5m+4=0 so quickly and is the rest of my working out correct?

    So I presume the solution to the first equation is y=Ae^(4x)+Be^(x). Do I use a similar approach to the other questions?
  9. Jun 20, 2007 #8


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    You should not be trying to do non-homogeneous differential equations until you can easily handle homogenous ones!

    y"- 5y'+ 4y= 0 is a homogeneous linear differential equation with constant coefficients. The standard way of solving such an equation is just what you started with- but without the "V":
    Since e^mx is never 0, divide the entire equation by it to get m^2- 5m+ 4= 0, the "characteristic equation" for your differential equation. After you have done a few of those you should be able to write down the characteristic equation for such a differential equation without having to do the substitution!

    Yes, it is true that m^2- 5m+ 4= (m-4)(m-1) so m= 4 and m= 1 are solutions to that equation which tells you that both e^(x) and e^(4x) are solutions to y"- 5y'+ 4y= 0. You should know enough o fthe theory to know that the set of all solutions of a linear homogeneous second order equation form a 2 dimensional vector space and so you know how to form the general solution to the equation.

    The y(x)= Ve^mx where V is a function of x (the "variation of parameters" method) requires that you already know the solutions to the homogeneous equation: you need, for these problems, y(x)= u(x)e^x+ v(x)e^(4x).

    A simpler method, that, however, only works when the right hand side is an exponential, sine or cosine, polynomial, or combination of those, is the "undetermined coefficients" method. In that case you can make a good "guess" as to what the specific solution must be and just try to find coefficients that will work.

    For example, for f(x)= 8x^2- 25 try y(x)= Ax^2+ Bx+ C. Put that into your equation and determine what A, B, C must be to make the equation true. For f(x)= 9e^x, try y= Axe^x. (Do you see why you need that "x" multiplying e^x? If not review your text book.) For the last, you want to combine those previous answers.
  10. Jun 20, 2007 #9
    Oh fantastic, I'm beginning to see what's going on now. No I don't know why the "x" is being multiplied to e^2 but I will go hunt for the answer now.
    Thanks a lot guys, especially HallsOfIvy for the thorough and very helpful response.
  11. Jun 20, 2007 #10


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    What would happen if you tried only y(x)= Ae^x as a solution to the entire equation? What happens if you put y(x)= Axe^x?
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