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Math Help - differential equations

  1. #1
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    differential equations

    Can anyone help with this?

    x''+9x'+20x=29+20t+6e^(-2t)

    i) Obtain the solution x(t) as the sum of the solutions obtained when each component is considered individually in turn.
    ii) Solve the equation directly
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by boab View Post
    Can anyone help with this?

    x''+9x'+20x=29+20t+6e^(-2t)

    i) Obtain the solution x(t) as the sum of the solutions obtained when each component is considered individually in turn.
    they are asking you to solve

    x'' + 9x\ + 20x = 29

    x'' + 9x\ + 20x = 20t

    and

    x'' + 9x\ + 20x = 6e^{-2t}

    separately, then add the solutions together

    the method of undetermined coefficients is the way to go here

    ii) Solve the equation directly
    see here
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  3. #3
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    Thanks for your help but it hasn't really cleared anything up for me. I'm still not sure how to go about it. Can you offer any further help?

    Oh, and thanks again for your help with another problem I posted
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  4. #4
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by boab View Post
    Thanks for your help but it hasn't really cleared anything up for me. I'm still not sure how to go about it. Can you offer any further help?

    Oh, and thanks again for your help with another problem I posted
    take x'' + 9x' + 20x = 6e^{-2t} for example.

    we must find the homogeneous solutions first. that is, solve x'' + 9x' + 20x = 0

    we assume a solution of x = e^{mt}

    the characteristic equation is m^2 + 9m + 20 = 0 .........you should know how to get this.

    \Rightarrow (m + 5)(m + 4) = 0

    \Rightarrow m = -5 or m = -4

    Thus, the homogeneous solutions are: x = Ae^{-5t} + Be^{-4t}.

    Now we find a particular solution. It must be of the form of the original right hand side, but not the same as the solution to the homogeneous. there is no danger of the latter case here.

    so, assume x_p = Ce^{-2t} is a particular solution.

    \Rightarrow x_p' = -2Ce^{-2t}

    and x_p'' = 4Ce^{-2t}

    plug these into the original differential equation, we get:

    4Ce^{-2t} - 18Ce^{-2t} + 20Ce^{-2t} = 6e^{-2t}

    now simplify and equate coefficients to solve for C.

    the solution to the equation will be of the form:

    x(t) = Ae^{-5t} + Be^{-4t} + x_p


    the others are solved in exactly the same way
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  5. #5
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    Thanks again. I'll freely admit this is a part of the course which just isn't clicking (as you've probably gathered!). I'll see how I get on, but thanks again for your help.
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