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Second Order Linear Homogeneous differential equations

  1. May 9, 2008 #1
    hey I am having a little trouble with this topic. Here are the questions I was set.

    a) Find the general solution of d^2y/dt^2 - 2(dy/dt) + y = 0. Verify your answer.

    b) Solve the initial value problem y'' + 4y' + 5y = 0; y(0)= -3, y'(0) = 0

    c) Find a DE that has the given functions as solutions: y1 = e^5x , y2 = e^-2x

    Any help would be greatly appreciated.
     
  2. jcsd
  3. May 9, 2008 #2
    Let's look at the first two problems. How can you take a second order linear differential equation and turn it into a system of first order equations? Hint: Let dy/dt = v.
     
  4. May 9, 2008 #3
    would you turn it into v^2 + v + 1 = 0 then solve for the roots?
     
  5. May 9, 2008 #4
    Not quite. We're looking for 2 first order equations derived from the second order equation. If we let dy/dt = v, that's one equation of the system. Note that:

    [tex] \frac{d^2y}{dt^2} = \frac{d}{dt} \frac{dy}{dt} = \frac{dv}{dt} [/tex]

    Now solve the given second order equation for dv/dt, and you'll have your first order system.
     
  6. May 9, 2008 #5
    i'm not sure that I follow. I can understand how you get dv/dt, but I dont know what to do next
     
  7. May 9, 2008 #6
    Half of your first order system is already known, that is, dy/dt = v.

    Now lets look at the equation that you were given. We have a d^2y/dt^2 term in the given second order equation. As shown above, this equates to dv/dt. You want to solve for dv/dt. We also have a dy/dt term in the second order equation, but we've already decided that this equates to "v." Once you plug in these terms, you'll have an equation for dv/dt. Combining this with dy/dt = v, you'll have a partially decoupled system of first order linear equations. Can you show me what the second order equation will look like once you solve for dv/dt?
     
  8. May 9, 2008 #7

    Vid

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    Make a guess that y = e^(rx) for some r. Then solve for r by solving the resulting quadratic polynomial to get two values for r (two different solutions). Since it's a linear diff eq any constant times the solution is a solution, and the sum of two solutions is a solution, the general solution is just the span of the two different solutions you've found. Use the initial conditions to find the exact solution you're looking for.
     
    Last edited: May 9, 2008
  9. May 9, 2008 #8
    ok I am really confused sorry. I might just spend a few days on this myself, its very confusing for me.
     
  10. May 9, 2008 #9
    It doesn't sound like he's in a guessing kind of mood.

    Shadow, try to look through the textbook examples that show how to construct a first order linear system from a second order equation. If we let dy/dt = v, we'll end up with a system of 2 first order equations; one for dy/dt, and one for dv/dt. After you set up the system, you can use eigenvaules to solve for y(t).

    Setting up that second equation is just a matter of plugging in the right terms and solving algebraically for dv/dt. Just post if you need help or further clarification.
     
    Last edited: May 9, 2008
  11. May 9, 2008 #10

    Defennder

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    Actually I don't think we need to introduce the concept of converting a 2nd order linear ODE into a first order one. He may not even have learnt that yet. When I did an intro DE course, I learnt how to solve 2nd order linear ODE without knowing how to convert it into a system of 2 linear 1st order ODE. I did that without having to explicitly use the concept of eigenvalues, even though the theory underlying the method indeed made use of such.

    Now, you have the equation y'' - 2y' + y = 0. Start by asking yourself, given that this is a homogenous equation, what is the characteristic equation of this DE? Once you have equation, solve for [tex]\lambda[/tex]. Note the values of lambda. Are they both real and distinct, repeated roots or complex? What should you do if they fall into each of the three preceding categories?
     
  12. May 9, 2008 #11
    ummm i read somewhere that for a root thats the same, as it is in this case, you simply do
    C1e^m1x+C2e^32x=0

    excuse my lack of code knowledge
     
  13. May 9, 2008 #12

    Defennder

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    Ok, you're right that the characteristic equation has 1 repeated root. I don't understand your answer for the general solution in that case. Where did '32' come from? I assume that 'm1' is the value of lambda which you have found by solving the roots of the characteristic equation. Your answer is only partly correct. The other half is incorrect.
     
  14. May 10, 2008 #13
    oh sorry it should be C1e^m1x+C2e^m2x=0
     
  15. May 10, 2008 #14

    Defennder

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    That's not quite correct. That is the general solution in the case by which you get real and distinct roots. And why did you equate the general solution to 0?
     
  16. May 10, 2008 #15
    i'm not sure, thats just what i read on the internet. this topic isnt covered in our textbook.
    so what do i do?
     
  17. May 10, 2008 #16

    Hootenanny

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    Since the ODE is second order, you need two linearly independent solutions to form your general solution. If your characteristic equation has a repeated root, then the usual second solution is of the form,

    [tex]y = Cxe^{\lambda x}[/tex]

    Such that your overall general solution is of the form,

    [tex]y = C_0e^{\lambda x} + C_1xe^{\lambda x}[/tex]

    Since we have just 'plucked' this solution out of the air, one would need to verify it before it would be accepted as a solution. I haven't got time to present the theory behind the second linearly independent solution now, but if I get time later I will. Alternatively, I'm sure that there are plenty resources on the Internet that would detail the theory behind it.
     
  18. May 10, 2008 #17
    ok, for part a i got an answer of

    y=e^t - 2e^t

    is this correct? if so how would I verify it.

    and what about parts b/c
     
  19. May 10, 2008 #18
    and for part c i got an answer of

    y'' - 3y' - 10y = 0
     
  20. May 11, 2008 #19

    Hootenanny

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    No, unfortunately it is not correct. Remember that you are looking for the general solution (since you are given no boundary conditions) and therefore you should have some undetermined constants.

    Since this thread has gone nowhere I think this would be a good point to ascertain your level ov knowledge. What do you actually know about ODE's? How much have you studied them?
     
  21. May 11, 2008 #20
    none, this is a research assignment. So I am trying to get some help off you guys
     
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