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    for this question, why is P.I t cube instead of t square, is there a way of knowing the power of t straight away?
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    (Original post by Iconic_panda)
    for this question, why is P.I t cube instead of t square, is there a way of knowing the power of t straight away?
    What's the question?
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    (Original post by Iconic_panda)
    for this question, why is P.I t cube instead of t square, is there a way of knowing the power of t straight away?
    I'll guess that you already have a t squared term in your CF. You therefore have to multiply by another t.

    Similarly, if you already had a e^t term in the CF (just that, not multiplied by some more algebra), then you'd have your PI as te^t.
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    this question here
    (Original post by RDKGames)
    What's the question?
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    my CF only have t in there, no t square, can you please check the question below?
    (Original post by tiny hobbit)
    I'll guess that you already have a t squared term in your CF. You therefore have to multiply by another t.

    Similarly, if you already had a e^t term in the CF (just that, not multiplied by some more algebra), then you'd have your PI as te^t.
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    (Original post by Iconic_panda)
    this question here
    The auxiliary equation has a repeated root, 2, which is also in 3te^{2t}

    The general rule of thumb, is that if your auxiliary eq. has a repeated root \gamma while your RHS of the ODE has e^{\gamma t}, then the suggested P.I. is t^2e^{\gamma t}. Here, we have the RHS as te^{\gamma t} therefore we go up one in the power of t on our P.I. as well, hence we use t^3e^{\gamma t}

    Have a good look here starting 'In a nutshell...' (particularly the P.I. choice section) to get a solid understanding of how to deal with all types of ODE's at A-Level.
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    (Original post by Iconic_panda)
    this question here
    For the particular integral.

    TRY  x= ate^{2t}

    find x'(t) = 2ate^{2t} + ae^{2t}
    x''(t) = 4ate^{2t} + 4ae^{2t}

    sub in ODE to produce a vanishing LHS : 0=0

    oh no!

    so we try
    x=3t^2e^{2t}
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    ah right!
    what if it keeps on not working hahha, do we keep going up the power?
    (Original post by NotNotBatman)
    For the particular integral.

    TRY  x= ate^{2t}

    find x'(t) = 2ate^{2t} + ae^{2t}
    x''(t) = 4ate^{2t} + 4ae^{2t}

    sub in ODE to produce a vanishing LHS : 0=0

    oh no!

    so we try
    x=3t^2e^{2t}
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    I got another question, what if the RHS is a combination of two, let say 5 sinx + e^x. would we try PI as a sin x+ b cos x +c e^x, i.e the combination of two as well?

    (Original post by RDKGames)
    The auxiliary equation has a repeated root, 2, which is also in 3te^{2t}

    The general rule of thumb, is that if your auxiliary eq. has a repeated root \gamma while your RHS of the ODE has e^{\gamma t}, then the suggested P.I. is t^2e^{\gamma t}. Here, we have the RHS as te^{\gamma t} therefore we go up one in the power of t on our P.I. as well, hence we use t^3e^{\gamma t}

    Have a good look here starting 'In a nutshell...' (particularly the P.I. choice section) to get a solid understanding of how to deal with all types of ODE's at A-Level.
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    (Original post by Iconic_panda)
    I got another question, what if the RHS is a combination of two, let say 5 sinx + e^x. would we try PI as a sin x+ b cos x +c e^x, i.e the combination of two as well?
    Yep
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    thank you
    (Original post by RDKGames)
    Yep
 
 
 
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