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Introductuin to differential equations

  1. Oct 18, 2007 #1
    1. The problem statement, all variables and given/known data
    use the concept that y = c, -inf < x < inf is a constant function if and only if y' = 0 to determine whether the specified differential equation has any constant solutions:

    y'' + 4y' + 6y = 10


    3. The attempt at a solution

    What throws me off in this particular problem is y''. If y' = 0 then y'' = 0 as well.
    Only now when I integrate y'' to y' its y' = c1 so c1 = 0 and y' still = 0.

    then I basically substitute my results in the equation to get:

    (0) + 4(0) + 6(c) = 10

    6c = 10

    [c = 10/6, c1 = 0] are the constant solutions to the equation.

    Does this make sense?

    Thanks.
     
  2. jcsd
  3. Oct 18, 2007 #2

    Dick

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    You are thinking too hard. You don't have to integrate anything. Set y''=y'=0 and solve the equation for y. y=0 is NOT a solution.
     
  4. Oct 18, 2007 #3
    ahh.. so

    y = (10 - 4y' - y'')/6

    y = (10 - 4(0) - 0)/6

    [y = 10/6] ?

    Thanks for the help! Please let me know if this is right.
     
  5. Oct 18, 2007 #4
    The great thing about ODE's is that you can usually quite readily check the answer by substituting your solution back into the differential equation. Give it a try and you'll find that your solution is correct.
     
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