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VCE Stuff => VCE Mathematics => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Specialist Mathematics => Topic started by: /0 on July 11, 2009, 10:32:58 pm

Title: solving de's
Post by: /0 on July 11, 2009, 10:32:58 pm: Is it necessary to make $y$ the subject or can it be $x$ too? Sometimes putting $x$ as the subject takes a lot less working. (I've lost my spesh thread again)
Title: Re: solving de's
Post by: TrueTears on July 11, 2009, 10:34:08 pm: Depends on what the question asks for, eg "Find an expression for x in terms of blah blah". If not stated, either is alright.

cool story bro
Title: Re: solving de's
Post by: kamil9876 on July 11, 2009, 10:44:41 pm: 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:

$(x-1)f^{[4]}(x) + f^{[3]}(x) + x^2 f''(x)=0$

It's going to be very hard to verify that $x=...$ 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: $y=x^2$ )

*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.
Title: Re: solving de's
Post by: /0 on July 11, 2009, 10:58:34 pm: hmmm yeah interesting point, but I don't think y in terms of x will always be a function, e.g. $\frac{dy}{dx} = y$

You can have $x=\log_e|y|+C$ or $y=\pm Ae^x$ . The second one is quicker for subbing in but the first doesn't need the $\pm$ sign... lol
I dunno
Title: Re: solving de's
Post by: TrueTears on July 11, 2009, 11:01:40 pm: 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...
Title: Re: solving de's
Post by: kamil9876 on July 11, 2009, 11:07:47 pm: Quote from: /0 on July 11, 2009, 10:58:34 pm
hmmm yeah interesting point, but I don't think y in terms of x will always be a function, e.g. $\frac{dy}{dx} = y$

You can have $x=\log_e|y|+C$ or $y=\pm Ae^x$ . The second one is quicker for subbing in but the first doesn't need the $\pm$ 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'
Title: Re: solving de's
Post by: /0 on July 11, 2009, 11:29:45 pm: yeah but isn't $y = \pm Ae^x$ a more general solution?

e.g. if they said $y(3) = 3$ then you would get two solutions for $y=\pm Ae^x$ but only 1 for $y=Ae^x$
Title: Re: solving de's
Post by: TrueTears on July 12, 2009, 12:04:19 am: The $\pm$ is absorbed by the A (constant)
Title: Re: solving de's
Post by: kamil9876 on July 12, 2009, 12:34:51 am: Quote from: /0 on July 11, 2009, 11:29:45 pm
yeah but isn't $y = \pm Ae^x$ a more general solution?

e.g. if they said $y(3) = 3$ then you would get two solutions for $y=\pm Ae^x$ but only 1 for $y=Ae^x$

No you wouldn't, think about it:

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

$log_e |x| +c$
$=log_e Ax$

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:

$\int_{b}^{a} log_e x dx$
$=log_e (\frac{a}{b})$

and so
$<br />\int_{-2}^{-3} log_e x dx$
$=log_e (\frac{3}{2})$

and no need to worry about modulus as it was cancelled out together with the A, or c.
Title: Re: solving de's
Post by: /0 on July 12, 2009, 12:51:01 am: thanks guys
Title: Re: solving de's
Post by: Mao on July 12, 2009, 01:34:46 am: For the case $\frac{dy}{dx} = y$ , it is actually a first order homogeneous linear DE in disguise: $\implies \frac{dy}{dx} - y = 0$
Title: Re: solving de's
Post by: /0 on July 12, 2009, 02:58:39 pm: Quote from: Mao on July 12, 2009, 01:34:46 am
For the case $\frac{dy}{dx} = y$ , it is actually a first order homogeneous linear DE in disguise: $\implies \frac{dy}{dx} - y = 0$

Ah yeah true

Just another question,

Should you say $A = \pm e^{-c}$ or is it ok to say $A = e^{-c}$
Title: Re: solving de's
Post by: /0 on July 20, 2009, 09:15:06 pm: I asked my teacher and he said you should say

$A = \pm e^{-c}$
Title: Re: solving de's
Post by: dcc on July 20, 2009, 09:21:49 pm: But what do you think? Is it important that you notice the two possible 'choices'?
Title: Re: solving de's
Post by: /0 on July 20, 2009, 09:28:07 pm: I think each way of defining A is undesirable.
$A = e^{-c}$ limits A to the positive real numbers,
$A = \pm e^{-c}$ 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.
Title: Re: solving de's
Post by: dcc on July 20, 2009, 09:35:42 pm: Quote from: /0 on July 20, 2009, 09:28:07 pm
I'll just go with what the teacher says though.

I just died a little on the inside.
Title: Re: solving de's
Post by: hyperblade01 on July 20, 2009, 09:45:32 pm: Quote from: /0 on July 20, 2009, 09:28:07 pm
I'll just go with what the teacher says though.

Asked a similar question today; my teacher said to just stick with the positive A value as the context usually removes the negative one...
Title: Re: solving de's
Post by: /0 on July 20, 2009, 10:00:50 pm: Quote from: dcc on July 20, 2009, 09:35:42 pm
Quote from: /0 on July 20, 2009, 09:28:07 pm
I'll just go with what the teacher says though.

I just died a little on the inside.

haha
or maybe something better (but infinitely more risky):

$A = e^{-c}\ \veebar \ -e^{-c}$

xor
Title: Re: solving de's
Post by: dcc on July 20, 2009, 10:11:48 pm: You could write your solution (because, thats really what they want, a SOLUTION to the differential equation) with the accompanying statement that $A \in \mathbb{R}$ . Because any A will work, right? (except trivial shit like 0).
Title: Re: solving de's
Post by: /0 on July 20, 2009, 10:52:33 pm: hah yeah true, screw writing $A=\pm_\pm^\pm e^{-abcdefghijklmnopqrstuvwxyz}$

I'm just gonna givem the ol' $A \in \mathbb{R}$
Title: Re: solving de's
Post by: TrueTears on July 20, 2009, 11:39:20 pm: i lol at this thread