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Math Help - simple question

  1. #1
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    simple question

    hai there, is anyone can show the step-by-step how to solve this equation.please2..

    y''+25y = 10 sin pt whre p is positive value

    y(0) = 0
    y'(0) = 5

    tq
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  2. #2
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    the answer should be p = 5

    and y(t) = 0.3 sin 5t + t cos 5t..

    please anyone show it to me.tq
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  3. #3
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    Quote Originally Posted by llyold View Post
    hai there, is anyone can show the step-by-step how to solve this equation.please2..

    y''+25y = 10 sin pt whre p is positive value

    y(0) = 0
    y'(0) = 5

    tq
    Solve complement and then solve particular via annihilator method.

    m^2+25=0 m=\pm5\mathbf{i}

    y_c=c_1cos(5t)+c_2sin(5t)

    (m^2+p^2)(m^2+25)=0 m=\pm5\mathbf{i}, \pm p\mathbf{i}

    y_p=c_3cos(pt)+c_4sin(pt)

    y'_p=-pc_3sin(pt)+pc_4cos(pt)

    y''_p=-p^2c_3cos(pt)-p^2c_4sin(pt)

    -p^2c_3cos(pt)-p^2c_4sin(pt)+25c_3sin(pt)+25c_4cos(pt)=10sin(pt)

    c_3(-p^2cos(pt)+25sin(pt))+c_4(-p^2sin(pt)+25cos(pt))=10sin(pt)

    I have to get ready to hit the beach so this all I could come up with at the moment.
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