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Math Help - Inhomogenous D.e,,help needed!

  1. #1
    Member i_zz_y_ill's Avatar
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    Inhomogenous D.e,,help needed!

    par (iv) (B)* im stuck on, pls bare with me
    q1 A solution is sought to te differential eqn

    y(doubledot)+4(dot)+4y=2e^-2t
    (i)find the complementary function
    (ii)Explain why an expression of either of the fomrs ae^-2t or ate^-2t cannot be a particular integral of the D.e. State a correct form for the partcular integral and hence show the the general solution is:

    y=(A+Bt+t^2)e^-2t
    (iii)It is given that y=0 when t=1 and that y=0 when t=3. Calculate the constants A and B and write the particular solution in factorised form.
    (iv)By reference to the previous answer, write dwn the particular solution in the cases
    (A) y=0 when t=2 and y=0 when t=7
    (B)*y=0 only when t=1

    the answer in (iii) was (t-1)(t-3)e^-2t
    I can't really see how this relates to (B).
    I just put in y=0 and t=1 getting A+B+1=0 in which case A and B can be anything adding up to -1. The mark scheme i have however gives a mark for the answer being y=e^-2t(t-1)^2 in which case I manipulated this aswer and found that A would equal 1 and B would equal -2.(which contradicts what i thought) Obviously I am missing something either from 3 or something fundamental or this Question is very badly worded or ultimately wrong. Any inputs would be highly appreciated thanks!
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  2. #2
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    Quote Originally Posted by i_zz_y_ill View Post
    par (iv) (B)* im stuck on, pls bare with me
    q1 A solution is sought to te differential eqn

    y(doubledot)+4(dot)+4y=2e^-2t
    (i)find the complementary function
    (ii)Explain why an expression of either of the fomrs ae^-2t or ate^-2t cannot be a particular integral of the D.e. State a correct form for the partcular integral and hence show the the general solution is:

    y=(A+Bt+t^2)e^-2t
    (iii)It is given that y=0 when t=1 and that y=0 when t=3. Calculate the constants A and B and write the particular solution in factorised form.
    (iv)By reference to the previous answer, write dwn the particular solution in the cases
    (A) y=0 when t=2 and y=0 when t=7
    (B)*y=0 only when t=1

    the answer in (iii) was (t-1)(t-3)e^-2t
    I can't really see how this relates to (B).
    I just put in y=0 and t=1 getting A+B+1=0 in which case A and B can be anything adding up to -1. The mark scheme i have however gives a mark for the answer being y=e^-2t(t-1)^2 in which case I manipulated this aswer and found that A would equal 1 and B would equal -2.(which contradicts what i thought) Obviously I am missing something either from 3 or something fundamental or this Question is very badly worded or ultimately wrong. Any inputs would be highly appreciated thanks!
    First of all, I think part (iv) requires finding brand new values for A and B. So your starting point is

    y = (A + Bt + t^2) e^{-2t}.

    You require (t - 1)^2 = A + Bt + t^2 since this is the only way that y = 0 only when t = 1. So A = 1 and B = -2.

    Incidentally, for the first bit you require (t - 1)(t - 7) = A + Bt + t^2 since this is the only way that y = 0 when t = 1 or t = 7 ......
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  3. #3
    Member i_zz_y_ill's Avatar
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    I can't just say i required (t-1)^2 i thought it could be like anything. When for (A) for instance two simultaneous eqns were calculated and A dn B were strictly limited, with (B) however heres only one imultaneos eqn A + B + 1 = 0

    in which case there could be numerous answers to (B) do you see what i mean?
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  4. #4
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    Quote Originally Posted by i_zz_y_ill View Post
    I can't just say i required (t-1)^2 i thought it could be like anything. When for (A) for instance two simultaneous eqns were calculated and A dn B were strictly limited, with (B) however heres only one imultaneos eqn A + B + 1 = 0

    in which case there could be numerous answers to (B) do you see what i mean?
    A particular solution has no arbitrary constants in it.

    The only way y = (A + Bt + t^2) e^{-2t} can be zero only for t = 1 is in the way that I've already said.
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