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Math Help - 2nd order DE

  1. #1
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    2nd order DE

    Y''+Y'+Y = e^3x + cos(5x)

    how to solve this DE??? Great Help
    Last edited by mr fantastic; November 18th 2010 at 04:12 AM. Reason: Re-titled.
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  2. #2
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    Quote Originally Posted by scifreak View Post
    Y''+Y'+Y = e^3x + cos(5x)

    how to solve this DE??? Great Help
    Step 1: Can you solve Y'' + Y' + Y = 0? Please show all your work and say where you're stuck if you can't solve it.
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  3. #3
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    i solved it upto Y"+Y'+Y= 0. it can be solved using the general second order DE solution steps. But the problem is the Right hand side of the equation. How to solve that part? I mean how to find the Particular Integral ..... The answer for the first part of is

    Yc=e^(-0.5x) (Acos(root(3)/2) + B sin (root(3)/2) )


    my problem again is as to how to find the Particular integral
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  4. #4
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    Quote Originally Posted by scifreak View Post
    i solved it upto Y"+Y'+Y= 0. it can be solved using the general second order DE solution steps. But the problem is the Right hand side of the equation. How to solve that part? I mean how to find the Particular Integral ..... The answer for the first part of is

    Yc=e^(-0.5x) (Acos(root(3)/2) + B sin (root(3)/2) )


    my problem again is as to how to find the Particular integral
    Use the method of undetermined coefficients to find a particular solution of the form y = a e^{3x} + b \cos (5x) + c \sin (5x).
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  5. #5
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    Thank you sir, Problem solved. I will solve it and if i have further problems i will notify you
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