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April 19, 2024, 11:40:07 pm

Author Topic: solving de's  (Read 3305 times)  Share 

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solving de's
« on: July 11, 2009, 10:32:58 pm »
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Is it necessary to make the subject or can it be too? Sometimes putting as the subject takes a lot less working. (I've lost my spesh thread again)

TrueTears

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Re: solving de's
« Reply #1 on: July 11, 2009, 10:34:08 pm »
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Depends on what the question asks for, eg "Find an expression for x in terms of blah blah". If not stated, either is alright.

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Re: solving de's
« Reply #2 on: July 11, 2009, 10:44:41 pm »
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It's much more conventional practice to make y the subject, I'm not sure myself if there is some rule but here are my arguments for putting y as subject:

*If you get a complicated high order differential equation like:



It's going to be very hard to verify that is a solution without transposing first.

*y in terms of x will indeed be a function, but it is quite likely that x is not a function of y (e.g: )

*I think by SOLVING it means that which FUNCTION has the property that it's 4th derivative with respect to x multiplied by x-1.... equals 0.


Although I agree that if the question does not mention SOLVE but isntead says "find the time at which the particle is at x=3" then I do endorse cutting down the computations as it is still answering all the the quesiton requests.

I'm not sure on any of this though hah.
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Re: solving de's
« Reply #3 on: July 11, 2009, 10:58:34 pm »
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hmmm yeah interesting point, but I don't think y in terms of x will always be a function, e.g.

You can have or . The second one is quicker for subbing in but the first doesn't need the sign... lol
I dunno

TrueTears

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Re: solving de's
« Reply #4 on: July 11, 2009, 11:01:40 pm »
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Id just stick to the x expression?

If it wants the y, just solve for y and you're done. What's the problem?

EDIT: I think it's easier to leave it in terms of y for 2nd order homogeneous DE's since the general equation is in terms of x...
« Last Edit: July 11, 2009, 11:05:19 pm by TrueTears »
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kamil9876

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Re: solving de's
« Reply #5 on: July 11, 2009, 11:07:47 pm »
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hmmm yeah interesting point, but I don't think y in terms of x will always be a function, e.g.

You can have or . The second one is quicker for subbing in but the first doesn't need the sign... lol
I dunno


ahh yeahh haha, btw, the +/- is just as arbitrary as the A, hence you can just write it as one constant and it would be just as much of a constant as C.

And yes, I don't really know what is the convention, but I'm just speaking out of experience that making y the subject is what is implicit in the mentioning of 'solving de'
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Re: solving de's
« Reply #6 on: July 11, 2009, 11:29:45 pm »
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yeah but isn't a more general solution?

e.g. if they said then you would get two solutions for but only 1 for

TrueTears

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Re: solving de's
« Reply #7 on: July 12, 2009, 12:04:19 am »
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The is absorbed by the A (constant)
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kamil9876

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Re: solving de's
« Reply #8 on: July 12, 2009, 12:34:51 am »
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yeah but isn't a more general solution?

e.g. if they said then you would get two solutions for but only 1 for

No you wouldn't, think about it:

if y(3)=3 then and you can only get the positive constant. The +/- is there just so that you kids don't freak out when you have . In fact as TT said it's good practice for the - to be 'absobred'. You can absorb the modulus in a similair way:




Where if x is negative then the A is negative, and if x is positive then the A is positive and this does the modulus's job.

e.g in definite integrals:




and so



and no need to worry about modulus as it was cancelled out together with the A, or c.
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Re: solving de's
« Reply #9 on: July 12, 2009, 12:51:01 am »
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thanks guys

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Re: solving de's
« Reply #10 on: July 12, 2009, 01:34:46 am »
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For the case , it is actually a first order homogeneous linear DE in disguise:
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Re: solving de's
« Reply #11 on: July 12, 2009, 02:58:39 pm »
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For the case , it is actually a first order homogeneous linear DE in disguise:

Ah yeah true

Just another question,

Should you say or is it ok to say


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Re: solving de's
« Reply #12 on: July 20, 2009, 09:15:06 pm »
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I asked my teacher and he said you should say


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Re: solving de's
« Reply #13 on: July 20, 2009, 09:21:49 pm »
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But what do you think?  Is it important that you notice the two possible 'choices'?


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Re: solving de's
« Reply #14 on: July 20, 2009, 09:28:07 pm »
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I think each way of defining A is undesirable.
limits A to the positive real numbers,
means that if the correct value of A is of one sign, then the incorrect, oppositely signed value of A also comes along for the ride.

I'll just go with what the teacher says though.