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Solving a second order diff. equation

  1. Sep 16, 2008 #1
    1. The problem statement, all variables and given/known data


    I need to solve the following:

    [itex]\frac{d^{2} x}{dt} + 2 b \frac{dx}{dt} + (b^2 - \frac{db}{dt}) x = 0[/tex]

    2. The attempt at a solution

    Isn't this equivalent to??::

    [itex]x^{''} + 2bx^{'}+ (b^2 - b^{'})x = 0 \iff x^2 + 2x +1 = 0[/itex] If I let b = 1.

    which gives me the solution [tex]\lamda = -1[/tex]

    Thus giving me the particular solution

    [tex]x(t) = c \cdot e^{-t}[/tex]

    Have I understood correctly? There is no precondition regarding either t or b.

    Best Regards

  2. jcsd
  3. Sep 16, 2008 #2


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    Welcome to PF!

    Hi Alphaboy! Welcome to PF! :smile:

    (type tex instead of itex, and it comes out bigger :wink:)
    Yes … technically , that's correct …

    for any constant b, x = C e-bt is a solution.

    But where does that get you? :confused:

    Is this a part of some bigger problem? :smile:
  4. Sep 16, 2008 #3
    Re: Welcome to PF!

    Hello tiny-tim,

    I was refered to this forum by a friend and then I got stuck with the bigger problem which will follow below, I decided to check this forum out :)

    by the way there is typoo is my original equation:

    [itex]\frac{d^{2}x}{dt^2} - 2b \frac{dx}{dt} + (b^2 - \frac{db}{dt})x = 0[/itex]
    (the right one)

    Here is the greater problem:

    Let [itex]\phi \subset \mathbb{R}[/itex] be an open interval, and [itex]b \in C^{1}(\phi; \mathbb{C})[/itex] where [itex]C^{1}[/itex] is the area on the solutions for

    [itex]\frac{d^{2}x}{dt^2} - 2b \frac{dx}{dt} + (b^2 - \frac{db}{dt})x = 0[/itex]

    (1) Show that if [itex](\phi, x)[/itex] solves the equation [itex]\frac{dx}{dt} - bx = 0[/itex] then [itex]x \in C^{2}(\phi;\mathbb{C})[/itex] thus [itex](\phi,x)[/itex] is the solution of the original equation.

    I invision the following way of showing this:

    In for the solution x to belong to the greater Disk [itex]C^2[/itex] and show that the equation [itex]\frac{dx}{dt} - bx = 0[/itex] have a commen solution Disk then:

    [itex]\frac{dx}{dt} - bx = 0 \iff \frac{dx}{dt} = bx [/itex] and by substituting this into

    [itex](bx)^{''} - 2b (bx)^{'} + (b^2 - \frac{db}{dt})x = 0[/itex]

    then I re-write the above to become

    [itex](bx)^2-2b^2 \cdot x -1 = 0[/itex] which be assuming b = 1

    [itex]x^2-2 \cdot x -1 = 0[/itex] which gives the solution

    [itex]\lambda = \pm (\sqrt{2} \pm 1)[/itex]

    Then the two equation mentioned originally must have their solution within the same Disk and thusly [itex](\phi,x)[/itex] defines the solution for both
    [itex](bx)^{''} - 2b (bx)^{'} + (b^2 - \frac{db}{dt})x = 0[/itex]


    [itex]\frac{dx}{dt} - bx = 0[/itex]


    Finding the particular solution for
    [itex]\frac{d^{2}x}{dt^2} - 2b \frac{dx}{dt} + (b^2 - \frac{db}{dt})x = 0[/itex]


    [itex]\lambda = \pm (\sqrt{2} \pm 1)[/itex]


    [tex]x(t) = (C_{1}+C_{2}) \cdot e^{\pm (\sqrt{2} \pm 1) \cdot (t)}[/tex]

    Mister Tim have caught the essentials now?

    There is one more question in the entire problem, but I leave that out for now.

    Then [itex]C_{1} = \frac{1}{e^{\sqrt{2} +1}}[/itex] and [itex]C_{2} = \frac{1}{e^{-\sqrt{2} - 1}}[/itex]

    if and only if t = 1

    [tex]x(t) = (\frac{1}{e^{\sqrt{2} +1}} + \frac{1}{e^{-\sqrt{2} - 1}}) \cdot e^{\pm (\sqrt{2} \pm 1) \cdot (t)} = 0[/tex]

    How does this look, this the complete solution for the first equation isn't it?

    best regards
  5. Sep 17, 2008 #4


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    Hello Alphaboy! :smile:

    (please don't do fractions in itex … they're too small! :frown:)
    I don't follow what you've done …

    you can't put b = 1. :confused:

    All you need is, if x' = bx,

    then x'' = b'x + bx' = b'x + b2x,

    so x'' - 2bx' + (b2 - b')x = … ? :smile:
  6. Sep 17, 2008 #5
    Sorry Mr. Tiny Tim I made such a stupid error in the post above that I should be send to Guantanamo :sad:

    Anyway using Your calculations

    x'' - 2bx' + (b2 - b')x = bx' + b^2x - 2bx' + b^2x - bx' = 0

    Putting this into order.

    2bx^2 - 2bx' = 0

    and since

    x' = bx


    2bx^2 -2b(bx) = 0

    Therefore 2bx^2 - 2bx^2 = 0 <-> 0 = 0

    Thusly using terminology put forward in my task.

    Since describes the solution (phi, x) for x' - bx = 0 and then x' = bx as proved above is also a solution forx''(t) - 2bx(t) + (b^2 - b'(t))x = 0 then [tex]x \in C^2(\phi; \mathbb{C})[/tex]

    connecting this with part(2)

    Find the general solution for for

    x''(t) - 2bx(t) + (b^2 - b'(t)) = 0

    Which must imply that any for any belonging to C^1


    x''(t) - 2bx(t) + (b^2 - b'(t))x = 0 <-> x^2 - 2bx - (b^2-b) = 0

    thus the solution being

    [tex]\lambda = \frac{2bx \pm \sqrt{(-2bx)^2 - 4 \cdot - (b^2 -b)}}{2} [/tex]

    which gives the the solutions [tex]\lambda = \pm 2bx[/tex]

    Then the general solution must be

    [tex]x(t) = C_1 \cdot e^{2bx} + C_2 \cdot e^{-2bx}[/tex]

    This must be it? Isn't it?

    Final question

    Given the equation unhomogenous equation

    [tex]x''(t) - 2bx(t) + (b^2 - b'(t))x = e^{\int_{t_0}^{t} (b(s) ds)[/tex]

    solve it then x(t_0) = x'(t_0) = 0

    I know that [tex]x^2 - 2bx + (b^2-b) = e^{t - t_0}[/tex]

    Which gives us

    [tex]x(t) = C_1 \cdot e^{2bx} + C_2 \cdot e^{-2bx} + \ldots[/tex]

    Don't I substitute (t-t0) for X in the original differential equation in order to obtain for the inhomogenous version?


    Best Regards
    Last edited: Sep 17, 2008
  7. Sep 17, 2008 #6


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    Hi Alphaboy! :smile:
    (Guantanamo Bay, actually … Guantanamo itself is a rather nice Cuban fishing village … the song "Guantanamera" means "girl from Guantanamo" :smile:)

    But it's ok … I reported you to the Department of Homeland Security, and they say it didn't actually endanger the free world …

    but they are keeping an eye on you … :wink:

    Why are you going back to the original equation? :cry:

    The question itself tells you exactly what to do … solve x' = bx.

    Assuming that b(t) and x(t) are independent (you didn't say), that means:

    x'(t)/x(t) = b(t), so … ? :smile:

    (though I must confess I don't yet see how that gives you all the solutions :confused:)
  8. Sep 17, 2008 #7
    I don't know if reason why we are talking past each other (I fear) is that my professor maybe is using his own reasoning (like talking about solutions for diff. eqn belong to their own respective sets??)))

    Anyway trying to solve x' = bx I get that [tex]x(t) = e^{b(t)} + C[/tex] and since [tex]x \in C^{1}(\phi; \mathbb{C})[/tex] and that [tex] b \in C^{2}(\phi;\mathbb{C})[/tex]

    then since [tex]x' - bx = 0[/tex] gives a solution where b and x both are in C^1 and C^2, then those must also describe the solution for both x'(t) - bx = 0 and

    x'' -2bx' + (b^2 - b')x = 0


    Their curves cross each other???

    If I am not dealing with a specific 'b' how else to I get the general solution for the original eqn???

    Sincerely a confusses Alphaboy
  9. Sep 17, 2008 #8


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    Hi Alphaboy! :smile:
    No, it's [tex]x(t) = C e^{\int b(t) dt}[/tex] :smile:
  10. Sep 17, 2008 #9
    Hello again Tim,

    This is then the solution for x'(t) - bx = 0 don't I need to show here too that this is also the solution for

    [tex]x''(t) - 2bx'(t) + (b^2- b'(t))x = 0*[/tex]?? Sorry for the stupid question now. Is this then done by using the fact that there is analogy between every [tex]b \in C^{1}[/tex] and every possible x in [tex]C^2[/tex]???

    Is the solution for for x' - bx = 0 connected to finding the general solution for (*)??

    Best Regards
  11. Sep 17, 2008 #10


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    No … you don't need to prove anything … you've already done it in part (1).
    Yes … it is the general solution (or at least a sort of sub-general solution, see below …)

    Perhaps I'm missing something, but the hint in the question seems only to prove that every solution of x' = bx is a solution of the main equation, yet not the other way round.

    So it produces a "family" of solutions … but not, I think, the most general family. :confused:
  12. Sep 17, 2008 #11
    What I don't get about question (2) Which says Find the general solution of (*) Isn't that in fact part of question(1) in the answer for question(2)?

    Is there a specific theorem here what I use to prove that

    (2: answer) Such that The generalized solution of (*) is supposedly [tex]x(t) = C\cdot e^{\int b(t) dt}[/tex]? because describes the 'family of solutions for (*) since b and x are members of the same family [tex](\phi; \mathbb{C})[/tex]??

    Thank You for all Your help

    Best Regards
    Last edited: Sep 17, 2008
  13. Sep 17, 2008 #12


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    Well, that's what I don't get either …

    as I said at the end of my last post, C eB(t), where B' = b, seems to be only one family of at least two families of a general solution, and I can't yet see any way of getting the other family. :smile:

    (Don't forget, you haven't dealt with part (3) yet:

    x'' - 2bx' + (b2 - b')x = eB(t))
  14. Sep 17, 2008 #13
    But what about limit on b(t)?? Since the original (3) says find the solution for

    [tex]x''(t) - 2bx(t) + (b^2 - b'(t)) = e^{\int_{t_0}^{t} b(s) ds}[/tex]

    then [tex]x(t_0) = x'(t_0) = 0[/tex]

    Then [tex]x_{p}(t) = e^{B(t)} - e^{B(t_0)} [/tex]

    Is the solution here for (3) then

    [tex]x(t) = C \cdot e^{\int b(t) dt} +x_{p}(t) = e^{B(t)} - e^{B(t_0)}[/tex]???

    Best Regards
  15. Sep 17, 2008 #14


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    No, that doesn't work at all …

    you need to find a "particular solution" for [tex]x''(t) - 2bx(t) + (b^2 - b'(t))x = e^{\int_{t_0}^{t} b(s) ds}[/tex] …

    which is basically guesswork …

    just spend a few hours trying out guesses! :smile:
  16. Sep 17, 2008 #15
    Hi again,

    If I set x t to be [tex]x = e^{\int b(s) \ ds}[/tex] ?? what do I regarding the bound on rhs of inhomogenous equation??

    since x(t_0) - x'(t_0) = 0 then x(t_0) - x'(t_0) = 0??

    remove it??

    Sincerely Alphaboy
    Last edited: Sep 17, 2008
  17. Sep 17, 2008 #16


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    Hello Alphaboy! :smile:

    I'm sorry … I'm not following you at all …

    the question gives the bounds (for the ∫) of t0 and t …

    if you change the lower bound from t0 to t1, say, then that's the same as multiplying by

    [tex]C\ =\ e^{-\int_{t_0}^{t_1} b(s)\,ds}[/tex]

    your next line just repeats itself … :confused:

    and remove what?
  18. Sep 17, 2008 #17
    Hello again tim, and thanks for your answer.

    I meant the bound, sorry I am messed up.

    Anyway if I choose [tex]x = e^{\int b(t) \ ds}[/tex] and insert this to obtain the expression

    [tex](e^{\int b(t) \ ds})^2 -2b(e^{\int b(t) \ ds}) + (b^2 - b) \cdot (e^{\int b(t) \ ds}) = \ e^{-\int_{t_0}^{t_1} b(s)\,ds}[/tex]

    But don't I need to derive b for a complete solution?

    then condition for the eqn is x(t_0) = x'(t_0) =0....

    Best Regards
  19. Sep 17, 2008 #18


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    But the LHS is zero, isn't it? :confused:
  20. Sep 17, 2008 #19
    Hi Tiny Tim,

    Its here that I am confused. Because if I set the RHS to zero, then I the homogenous solution from (2) and solution for inhomogenous system?

    If [tex]e^\int(b(s) ds)[/tex] is a solution for the homogenous system. How do I find solution for the inhomogenous without having suitable b?

    since the inhomogenous system being

    [tex]x''(t) - 2b x'(t) + (b^2 - b'(t))x = e^{\int_{t_0}^{t} b(s) \ ds[/tex] which needs to be solved. And if I reuse our solution for (2) and our choosen x = ... which I insert in place of x in the inhomogenous equation. But then I arrive a equation where rhs is still dependent the interval [t0,t] and lhs is only dependent on t ???

    Sincerely Alphaboy
  21. Sep 18, 2008 #20


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    Hi Alphaboy! :smile:

    I don't know why you're worrying about the RHS being "dependent the interval [t0,t]".

    If you write [tex]B(t)\ =\ \int_{t_0}^{t} b(s) \ ds[/tex], then it's just

    [tex]x''(t) - 2b x'(t) + (b^2 - b'(t))x = e^{B(t)}\ \text{with}\ B' \ =\ b[/tex]

    (or [tex]x''(t) - 2B' x'(t) + (B'^2 - B''(t))x = e^{B(t)}[/tex])

    with no interval to worry about :smile:

    You now need a "particular solution" …

    I must admit I can't yet see one …

    just make intelligent guesses, starting with things like x = teB(t), and then trying the general substitution x = y(t) eB(t) :wink:
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