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Homework Help: Differential equation of second order

  1. Jun 10, 2009 #1
    I would very much appreciate if anyone can help me with this problem. I got stuck at the end.
    I have to find
    1.the complementary function, particular integral and the general solution.
    2. find complete solution when t=0, q=0, dq/dt=0.

    -equation 2(d^2q)/(dt^2)+200dq/dt+5000q = 250
    -in D operator form it looks like (2D^2+200d+5000)y=250
    - subsituting m for D i receve repeated root m= -50
    - so the complemenary function (At+B)e^(-50t)
    - right side is constant so i subsituted constant into equation and solved that k= 0.05
    - so the general solution is q=(At+B)e^(-50t)+0.05 right??

    Now when t=0. q=0 , dq/dt=0
    1) 0 = (A(0)+B)e^0+0.05
    B= -0.05

    And now i have problem
    2) dq/dt= -50(At+B)e^(-50t) what the solution should be. I should find A but it dissapears because of t=0

    I have no idea what to do with it.
    1. The problem statement, all variables and given/known data

    2. Relevant equations

    3. The attempt at a solution
  2. jcsd
  3. Jun 10, 2009 #2


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    This is not true, don't forget the product rule.
  4. Jun 10, 2009 #3
    would that be like?:

    First i multiplied A and B by that e.

    (At+e^-50t)+ (Be^-50t)

    so the chain rule gives me

    [(At)'(e^-50t)+ (e^-50t)'(At)] + (Be^-50t)'

    [A(e^-50t)-(50e^-50t(At)) ] -50Be^-50t t=0

    0= A - 50-50(-0.05)
    A= 47,5???

    Is that right?
  5. Jun 10, 2009 #4


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    Your equation is inhomogeneous. You've found the homogeneous solution. Now you must find the particular solution. Since the inhomogeneous term is just a constant I believe your particular solution will be as well.
    y = y_h + y_p.

    Always verify that your function is indeed a solution...even if you know you "did it right". The verification step should have reminded you that you forgot the particular part.

    NOW you can substitute your initial conditions and resolve the values of your arbitrary constants.
  6. Jun 10, 2009 #5


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    He already calculated the particular solution, 0.05=250/5000.

    The second equality is not correct. for t=0, the second term gives?
  7. Jun 10, 2009 #6
    Hmm, would that be

    0= A-0-50(-0.05)
    A= -2.5??

    So the final solution is q= (-2.5-0.05t)e^-50t???
  8. Jun 10, 2009 #7


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    You're being extremely sloppy.

    A=-2.5 is correct.

    This is not correct. First off the particular solution disappeared and secondly you put the value for A in B and the value for B in A. Stop being sloppy!

    Also when you find a solution you can check if it's correct or not all by yourself by just putting it into the differential equation and see if the equality holds. You should really teach yourself to check your answers like that.
  9. Jun 10, 2009 #8
    I am really sorry for being sloopy. I am nervous. Thank you for your help and comments.

    I checked the solution and I think i should look like:

    q= (-2.5t-0.05)e^-50t+ 0.05

    and when t=0

    q= -0.05 +0.05

  10. Jun 10, 2009 #9


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    Your solution is now complete and correct!
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