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Math Help - is this derivative correct?

  1. #1
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    is this derivative correct?

    y = e^-0.4x(C cos0.2x + Dsin0.2x)
    y' = -0.4e^-0.4x(-0.2C sin0.2x + 0.2 D cos0.2x)

    This from trying to work out a particular solution to a inhomogeneous equation

    Paul B
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  2. #2
    GAMMA Mathematics
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    Quote Originally Posted by poundedintodust View Post
    y = e^-0.4x(C cos0.2x + Dsin0.2x)
    y' = -0.4e^-0.4x(-0.2C sin0.2x + 0.2 D cos0.2x)

    This from trying to work out a particular solution to a inhomogeneous equation

    Paul B
    You have to follow the product rule.

    y = e^{-0.4x}[C cos(0.2x) + Dsin(0.2x)]

    y'=-0.4e^{-0.4x}[C cos(0.2x) + Dsin(0.2x)]+e^{-0.4x}[0.2Dcos(0.2x)-0.2Csin(0.2x)]
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  3. #3
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    if y = 3 and x=0

    should i be left with

    0.4C + 0.2D = 3?
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  4. #4
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    Quote Originally Posted by poundedintodust View Post
    if y = 3 and x=0

    should i be left with

    0.4C + 0.2D = 3?
    If y(0)=3, then C=3
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  5. #5
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    I am a bit confused now so i think Ill post whole problem and maybe you could tell me where I am going wrong.

    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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