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2nd Order Homogenous ODE (Two solutions?)

  1. May 31, 2012 #1
    2nd Order Homogenous ODE (Two solutions??)

    Alright.

    I understand that if we have a differential equation of the form

    [itex]A\cdot\frac{d^{2}y}{dt}+B\cdot\frac{dy}{dt}+C\cdot y = 0[/itex]

    and it has the solution y1(t), and y2 is also a solution. Then any combination of the two

    yH=C1y1(t)+C2y2(t) is also a solution.

    But, mathmatically speaking, so would a combination with a third "possible" solution y3.

    Now I know there must be some theorem stating that there will only be two solutions for this type of ODE, but can anyone tell me where I can find these?

    ALSO,

    If there are two solutions, why do we always use the combination of the two? Why not just pick one of the solutions and use it? Why "overcomplicate" it and use the combination of the two solutions as the general solution?

    I know there must be a good reason, but I havent found it, and I need someone to point it out to me, or tell me where I can read about it.

    Thank you in advance,
    Rune
     
  2. jcsd
  3. May 31, 2012 #2

    HallsofIvy

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    Re: 2nd Order Homogenous ODE (Two solutions??)

    What you are missing is that the two solutions, [itex]y_1[/itex], and [itex]y_2[/itex], derived from the characteristic equation of the d.e. are independent solutions. That is, if [itex]C_1y_1(t)+ C_2y_2(t)= 0[/itex] for all t, then [itex]C_1= C_2= 0[/itex]. A linear combination of the two might be independent of either of the first two, giving you another two solutions but you would NOT have three independent solutions.

    We do not "just pick one of the two" because that would not give a general solution. For example, the differential equation y''- 3y'+ 2y= 0 has characteristic equation [itex]r^2- 3r+ 2= (r- 2)(r- 1)= 0[/itex] so has solutions [itex]e^{2x}[/itex] and [itex]e^{x}[/itex]. Further, any solution to the differential equation can be written as [itex]Ce^{2x}+ De^x[/itex] for some constants, C and D. If you dropped either of the exponentials you cannot do that.

    The "intial value problem" of finding the unique function, y(x), such that y''- 3y'+ 2y= 0, y(0)= 2, y'(0)= 1 can be solving by knowing that any solution to y''- 3y'+ 2y= 0 can be written in the form [itex]y(x)= Ce^{2x}+ De^x[/itex]. To fit the initial conditions, we must have [itex]y(0)= Ce^0+ De^0= C+ D= 2[/itex] and [itex]y'(0)= 2Ce^0+ De^0= 1[/itex]. The two equations, C+D= 2 and 2C+ D= 1, have solution C= -1 and D= 3. That is, the solution to the intial value problem is [itex]y(x)= -e^{2x}+ 3e^x[/itex]. Without both functions, we could not get that.

    It is not "overcomplicating" to use both, it is necessary.
     
  4. Jun 1, 2012 #3
    Re: 2nd Order Homogenous ODE (Two solutions??)

    Thank you very much Halls of Ivy!

    After sitting down and talking a bit I now see the idea behind the general solution. Thank you.

    Next question, however, is: Why do we need two initial conditions?

    From a physical point of view I see it intuitively, for example in an RLC circuit we would need the initial energy of the system to compute the behavior. But im looking for a more mathmatical "proof"/reason.

    Im sure im just overlooking something again :)

    EDIT: Ahaaa... We need two initial conditions because we have two unknown constant.. DUH! :) Great, great. I have it now :)

    Thank you in advance,
    Rune
     
    Last edited: Jun 1, 2012
  5. Jun 1, 2012 #4
    Re: 2nd Order Homogenous ODE (Two solutions??)

    Khan Academy explains it very well: 2nd Order Linear Homogeneous Differential Equations 1
     
  6. Jun 1, 2012 #5

    HallsofIvy

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    Re: 2nd Order Homogenous ODE (Two solutions??)

    I feel I should point out that we don't need "two initial conditions". But, because solving a second order d.e. is equivalent to integrating twice, we do get two "constants of integration" and need two additional conditions to determine those. It is possible that those conditions are to give y and y' at some point ("initial conditions") or to give y at different points ("boundary conditions") or some combination- giving y at one point, y' at another.
     
  7. Jun 2, 2012 #6
    Re: 2nd Order Homogenous ODE (Two solutions??)

    The theorem you are asking about would be an existence/uniqueness theorem that says that if you specify the initial position and velocity ( x(0), x'(0)), then there is a unique solution with those values. This means that your general solution will have two unknown constants to give you the freedom to match to the two specified conditions.

    The reason that you need to specify both the initial position and initial velocity goes something like this. Think of a second order ODE as a particular expression of Newton's second law:

    Force = mass*acceleration

    acceleration = x''. Force is a quantity that depends on the location and velocity of the object. In other words Force might look like F = Bx'+Cx.

    So you have mx'' = Bx'+Cx.

    Suppose the initial position is x_0 and the initial velocity is v_0. Here is how you can use the differential equation to track how x and v change over time.

    Over a short time interval [itex]\Delta t[/itex], the position will change to [itex]x_1=x_0+v_0\Delta t[/itex] because v_0 is the initial velocity. By plugging in x=x_0, x'=v_0 into Newton's 2nd Law (i.e. the differential equation), you can find the initial acceleration a_0. Then estimate [itex]v_1=v_0+a_0\Delta t[/itex]. So now you know that at time [itex]t=\Delta t[/itex], x = x_1 and v=v_1. By plugging these values into the Diff. Eq. you can then find the acceleration a_1. Well, then you repeat the same process to find the next pair of values (x_2,v_2) at time [itex] t=2\Delta t[/itex]. By repeating this process indefinitely you can trace out the values of x and v for all time.

    This is a numerical procedure and it goes by the name Euler's method, but conceptually it tells the reason why you need to know both the initial position and the initial velocity. And it also shows that you don't need to know anything more than that. This is why there are two unknown constants in your general solution. Because that gives you exactly the freedom you need to fit your solution to the two initial conditions.
     
    Last edited: Jun 2, 2012
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