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Math Help - particular soultion of a initial value problem

  1. #1
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    particular soultion of a initial value problem

    i know this is more than likely a calculus problem but not sure if my algebra is right so hence the re posting over here. If its the wrong way to post it sorry!

    Starting with:

    5(d^2y/d^2x) + 4(dy/dx) + y = 0

    I have find a particular solution to the initial value problem where

    y(0) = -2, y'(0)=3

    I know a general solution is

    y = e^-0.4x(C cos0.2x + Dsin0.2x)

    after finding the roots.

    So substituting in the values into the derivative

    y' = -0.4e^-0.4x(-0.2C sin0.2x + 0.2 D cos0.2x)

    ends with 0.4C+0.2D =3

    So C = 3 and D = 9

    Doing the same with the general solution with x=-2

    y = e^-0.4x(C cos0.2x + Dsin0.2x)

    I end up with 1( C x 1 + D x 0) in other words C = -2

    My confusion comes from the fact that C should be the same in both instances??

    Cheers for the help
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  2. #2
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    Quote Originally Posted by poundedintodust View Post
    Starting with:

    5(d^2y/d^2x) + 4(dy/dx) + y = 0

    I have find a particular solution to the initial value problem where

    y(0) = -2, y'(0)=3

    I know a general solution is

    y = e^-0.4x(C cos0.2x + Dsin0.2x)

    after finding the roots. (I haven't checked that, so I'll assume it's correct.)

    So substituting in the values into the derivative

    y' = -0.4e^-0.4x(-0.2C sin0.2x + 0.2 D cos0.2x)

    ends with 0.4C+0.2D =3 No! What you are told is that y'(0)=3. This means that you put x=0 in the expression for y', which tells you that -0.4(0.2 D)=3 (since e^0 and cos0 are both 1, and sin0 is 0).

    So C = 3 and D = 9

    Doing the same with the general solution with x=-2 That should be y=-2 when x=0

    y = e^-0.4x(C cos0.2x + Dsin0.2x)

    I end up with 1( C x 1 + D x 0) in other words C = -2 This time you've got it right!

    My confusion comes from the fact that C should be the same in both instances??
    ..
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