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?

ChrisLorient
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DFranklin
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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). 
ChrisLorient
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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).
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).Last edited by DFranklin; 3 days ago at 16:47. 
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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).
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
to ?
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
to ?
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.
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.
After transforming eq (1) you want to express in terms of , and substitute this in for in eq (2) after you transform THAT one. Tidy it up and you get yourself in a position where you can easily solve for , before inverting it as required.Last edited by RDKGames; 3 days ago at 17:26. 
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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.
[It's sort of the equivalent of having a differential equation
and saying "well, the derivative of x is 1, so we can write this as "; ).Is the method correct though?
for some function s.
You can then apply the Laplace transform to find x(t).
The other thing to be aware is that ; 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.
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.Last edited by DFranklin; 3 days ago at 17:39. 
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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
and saying "well, the derivative of x is 1, so we can write this as "; ).
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,
for some function s.
You can then apply the Laplace transform to find x(t).
The other thing to be aware is that ; 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.
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?. 
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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. 
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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. 
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(Original post by DFranklin)
The results will be the same. 
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(Original post by ChrisLorient)
so is there nothing in the way this is worded that requires Laplace explicitly?
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 Alevel) 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 Alevel) don't really indicate that Laplace transforms are something you should be using yet.
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 + 15/7Last edited by ChrisLorient; 3 days ago at 12:01. 
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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
But I have NO idea what you've done past that.
From the third line, you should have:
RHS is just and the LHS quadratic is just
So then you simply have
So then
Of course, you'd want to express the RHS as partial fractions before inverting the side.Last edited by RDKGames; 3 days ago at 12:29. 
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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 Alevel) don't really indicate that Laplace transforms are something you should be using yet.
Is that correct?Last edited by ChrisLorient; 3 days ago at 12:46. 
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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.
So taking it over gives 1, no?Last edited by ChrisLorient; 1 day ago at 12:40.
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