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    Where am I going wrong?

    Where do I use x(0)?

    Am I meant to eliminate one and then sub in for the other? How?
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    How have you got L(x) = 2/s?

    More generally; the obvious thing to do seems to be to eliminate L(x(s)). However, it seems to me you might as well so this by eliminating X in the original DEs (leaving an easily solvable DE only involving y).
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    (Original post by DFranklin)
    How have you got L(x) = 2/s?

    More generally; the obvious thing to do seems to be to eliminate L(x(s)). However, it seems to me you might as well so this by eliminating X in the original DEs (leaving an easily solvable DE only involving y).
    Same equation, but it is x' in (1), as opposed to y'.

    Ended up with this.

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    Sorry, I can barely read your attachment (in some critical points) and I can't work out what you're trying to do.

    (The general principle would be to eliminate one of L(x) or L(y) and then invert. Which I suspect you're trying to do, but I can't actually make out exactly how you're trying to do it).
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    (Original post by DFranklin)
    Sorry, I can barely read your attachment (in some critical points) and I can't work out what you're trying to do.

    (The general principle would be to eliminate one of L(x) or L(y) and then invert. Which I suspect you're trying to do, but I can't actually make out exactly how you're trying to do it).
    Sorry. Hoping this is better.

    Took (1) and got X in terms of Y. Then substituted this into my (2).

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    How on earth have you gone from s\tilde{x}(s) - 1  + \tilde{x}(s) - \tilde{y}(s) = 0

    to 2 \tilde{x}(s) - \tilde{y}(s) = \frac{1}{s} = 1?

    The only thing i can imagine is that you think you can treat s as though it is 1. Which seems crazy (and unless I have had a mathematical equivalent of a stroke, entirely wrong), but otherwise I have no idea what your reasoning is.
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    (Original post by DFranklin)
    How on earth have you gone from s\tilde{x}(s) - 1  + \tilde{x}(s) - \tilde{y}(s) = 0

    to 2 \tilde{x}(s) - \tilde{y}(s) = \frac{1}{s} = 1?

    The only thing i can imagine is that you think you can treat s as though it is 1. Which seems crazy (and unless I have had a mathematical equivalent of a stroke, entirely wrong), but otherwise I have no idea what your reasoning is.
    Tables say 1/s is 1, but I am assuming I only use that at the end when inverting.

    Is the method correct though?

    No idea how to eliminate from ODE.
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    (Original post by ChrisLorient)
    Tables say 1/s is 1, but I am assuming I only use that at the end when inverting.

    Is the method correct though?

    No idea how to eliminate from ODE.
    \mathcal{L}^{-1}(\frac{1}{s})=1 so yes you only use it when inverting!

    After transforming eq (1) you want to express Y(s) in terms of X(s), and substitute this in for Y(s) in eq (2) after you transform THAT one. Tidy it up and you get yourself in a position where you can easily solve for X(s), before inverting it as required.
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    (Original post by ChrisLorient)
    Tables say 1/s is 1, but I am assuming I only use that at the end when inverting.
    Yes, you can only do that at the end (when you are actually inverting) - it is totally wrong to do it as you have done in your working here.

    [It's sort of the equivalent of having a differential equation

    \dfrac{d}{dx}(xy) = 1 and saying "well, the derivative of x is 1, so we can write this as \dfrac{d}{dx} (y)= 1"; ).

    Is the method correct though?
    More or less, as long as you don't assume s = 1. You should be able to eliminate one of L(x), L(y), so that you can write, say,

    L(x)(s) = f(s) for some function s.

    You can then apply the Laplace transform to find x(t).

    The other thing to be aware is that \mathcal{L}(y') = s\tilde{y}(s) - y(0); this is how you take account of the initial conditions (I think you asked this in an earlier post).

    No idea how to eliminate from ODE.
    You can't easily do it here (you could in your first example).

    If I was actually to try to solve these, I would simply differentiate the first equation (to get something with x'', x' and y', and then use this plus the original 2 equations to get a 2nd order DE only involving x. It's not terribly "pretty" (and doesn't use Laplace transforms!) but it's usually the easiest method IMHO.
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    (Original post by DFranklin)
    Yes, you can only do that at the end (when you are actually inverting) - it is totally wrong to do it as you have done in your working here.

    [It's sort of the equivalent of having a differential equation

    \dfrac{d}{dx}(xy) = 1 and saying "well, the derivative of x is 1, so we can write this as \dfrac{d}{dx} (y)= 1"; ).

    More or less, as long as you don't assume s = 1. You should be able to eliminate one of L(x), L(y), so that you can write, say,

    L(x)(s) = f(s) for some function s.

    You can then apply the Laplace transform to find x(t).

    The other thing to be aware is that \mathcal{L}(y'<img src="images/smilies/wink.png" border="0" alt="" title=";)" class="inlineimg" /> = s\tilde{y}(s) - y(0); this is how you take account of the initial conditions (I think you asked this in an earlier post).

    You can't easily do it here (you could in your first example).

    If I was actually to try to solve these, I would simply differentiate the first equation (to get something with x'', x' and y', and then use this plus the original 2 equations to get a 2nd order DE only involving x. It's not terribly "pretty" (and doesn't use Laplace transforms!) but it's usually the easiest method IMHO.
    I was given initial conditions for both x and y. Would this make a difference in attempting to do it without Laplace?

    Thanks for the rest. I will attempt it again and post back.
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    (Original post by ChrisLorient)
    I was given initial conditions for both x and y. Would this make a difference in attempting to do it without Laplace?.
    Not really. You'll get a 2nd order linear ode in x, and your condition for x(0) won't be enough to determine the arbitrary constants. But just leave them arbitrary; substitution back to find y in terms of those constants, and then the conditions on x(0),y(0) will be enough to find the constants.
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    (Original post by DFranklin)
    Not really. You'll get a 2nd order linear ode in x, and your condition for x(0) won't be enough to determine the arbitrary constants. But just leave them arbitrary; substitution back to find y in terms of those constants, and then the conditions on x(0),y(0) will be enough to find the constants.
    How would the result from each approach differ? Laplace uses s ant t, but the above would be in x? Just trying to understand why they are different.
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    (Original post by ChrisLorient)
    How would the result from each approach differ? Laplace uses s ant t, but the above would be in x? Just trying to understand why they are different.
    The results will be the same.
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    (Original post by DFranklin)
    The results will be the same.
    OK, so is there nothing in the way this is worded that requires Laplace explicitly?

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    (Original post by ChrisLorient)
    so is there nothing in the way this is worded that requires Laplace explicitly?
    No. Diff the 1st equation to get x''+x'-y' = 0.
    Use the 2nd equation to replace y' in the above by e^-t - 2x - 4y.
    Then use the original 1st equation to replace y by x''+x'.
    You end up with an equation looking something like:

    Ax'' + Bx' + Cx = D e^-t (where A, B, C, D are constants).

    Find the general solution of this satisfying x(0) = t ( you will have 1 arbitrary constant).
    Then use the first equation to find y. You can then use y(0) = 2 to determine the last constant.

    Why did you think this is to do with Laplace transforms? With all respect, your posts (and the fact you labelled the thread as A-level) don't really indicate that Laplace transforms are something you should be using yet.
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    (Original post by DFranklin)
    No. Diff the 1st equation to get x''+x'-y' = 0.
    Use the 2nd equation to replace y' in the above by e^-t - 2x - 4y.
    Then use the original 1st equation to replace y by x''+x'.
    You end up hwith an equation looking something like:

    Ax'' + Bx' + Cx = D e^-t (where A, B, C, D are constants).

    Find the general solution of this satisfying x(0) = t ( you will have 1 arbitrary constant).
    Then use the first equation to find y. You can then use y(0) = 2 to determine the last constant.

    Why did you think this is to do with Laplace transforms? With all respect, your posts (and the fact you labelled the thread as A-level) don't really indicate that Laplace transforms are something you should be using yet.
    OK, thanks.

    Using Laplace though, I have got this so far.

    Attachment 703506

    Is that correct? I now need to manipulate it using tables to get x(t)? For example, third fraction is 1/3 e^-t ?

    I get x(t) = 1/3 e^-t + 1/3 t + 15/7
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    (Original post by ChrisLorient)
    OK, thanks.

    Using Laplace though, I have got this so far.

    Is that correct? I now need to manipulate it using tables to get x(t)? For example, third fraction is 1/3 e^-t ?

    I get x(t) = 1/3 e^-t + 1/3 t + 15/7
    That's not correct, note how your result doesn't yield x(0)=1. If you REALLY want to be doing Laplace here, then your approach is correct throughout most of it, up to the third line from the bottom.

    But I have NO idea what you've done past that.

    From the third line, you should have:

    \displaystyle (s^2+5s+6)X(s)=6+s+\frac{1}{s+1}

    RHS is just \frac{s^2+7s+7}{s+1} and the LHS quadratic is just (s+2)(s+3)

    So then you simply have \displaystyle X(s)=\frac{s^2+7s+7}{(s+1)(s+2)(  s+3)}

    So then \displaystyle x(t)= \mathcal{L}^{-1} \left( \frac{s^2+7s+7}{(s+1)(s+2)(s+3)} \right)

    Of course, you'd want to express the RHS as partial fractions before inverting the side.
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    (Original post by DFranklin)
    No. Diff the 1st equation to get x''+x'-y' = 0.
    Use the 2nd equation to replace y' in the above by e^-t - 2x - 4y.
    Then use the original 1st equation to replace y by x''+x'.
    You end up with an equation looking something like:

    Ax'' + Bx' + Cx = D e^-t (where A, B, C, D are constants).

    Find the general solution of this satisfying x(0) = t ( you will have 1 arbitrary constant).
    Then use the first equation to find y. You can then use y(0) = 2 to determine the last constant.

    Why did you think this is to do with Laplace transforms? With all respect, your posts (and the fact you labelled the thread as A-level) don't really indicate that Laplace transforms are something you should be using yet.
    A = 1, B = 5, C = 6, D = 1

    Is that correct?
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    (Original post by ChrisLorient)
    A = 1, B = 5, C = 6, D = 1

    Is that correct?
    Agreed for A, B, C, but I get D = -1; one of us has a sign error.
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    (Original post by DFranklin)
    Agreed for A, B, C, but I get D = -1; one of us has a sign error.
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    So taking it over gives 1, no?
 
 
 
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