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which form should I use y = -c|x^3| or y = -c|x^3|

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PG1995

Active Member
Hi

Please have a look on the attachment. I have integrated the "solution" which is a differential equation to see if I get back y just out of curiosity. As you can see, I ended up with:
1: y = -c|x^3|, or, 2: y = c|x^3|

The original form was y=c1*x^3. So which form should I use, "1" or "2"? I think it's "2" because the original form didn't have a negative sign "-". But what about the absolute value? Please help me with it. Thanks a lot.

Regards
PG
 

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MrAl

Well-Known Member
Most Helpful Member
Hi

Please have a look on the attachment. I have integrated the "solution" which is a differential equation to see if I get back y just out of curiosity. As you can see, I ended up with:
1: y = -c|x^3|, or, 2: y = c|x^3|

The original form was y=c1*x^3. So which form should I use, "1" or "2"? I think it's "2" because the original form didn't have a negative sign "-". But what about the absolute value? Please help me with it. Thanks a lot.

Regards
PG

Hi,

Actually, when you start with a differential equation if you knew what the answer was then you wouldnt need to do anything. So when you do this experiment you have to assume that you didnt know what the starting equation was. So you start with the differential equation.

That said, if you find more than one proposed solution, you start with the most general form and test that in the original differential equation. The most general form that works in the original differential equation (that you integrated) is the best solution. If all the solutions work, then all the solutions are valid.
Many times in real life there are some other constraints to consider that limits the actual solutions that are valid, even though they also work in the original equation.
 

PG1995

Active Member
Hi MrAl

Both solutions, 1: y = -c|x^3|, or, 2: y = c|x^3|, satisfy the differential equation, dy/dx=3y/x. I started with y=c1*x^3 (let's call "3") which is also a solution to the differential equation in addition to "1" and "2". But when I integrated the differential equation I only ended up with "1" and "2", there was no third solution. Isn't there any difference between y=c|x^3| and y=c1*x^3? I think there is. Please help me with it. Thank you.

Best wishes
PG
 

MrAl

Well-Known Member
Most Helpful Member
Hi,

Let me see if i am following you here...

You started with
dy/dx=3*y/x
and change the form to
dy/y=3*dx/x
Then integrate both sides
ln(y)+C1=3*ln(x)+C2
so
y=x^3*e^(C2-C1)

So where did you get that c or c1 from and why did you use absolute value signs for x^3 ???

Oh ok, so you got c from e^(C2-C1) right?
They why did you add the abs value?
 
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PG1995

Active Member
You started with
dy/dx=3*y/x
and change the form to
dy/y=3*dx/x
Then integrate both sides
ln(y)+C1=3*ln(x)+C2
so
y=x^3*e^(C2-C1)

Your solution is correct but you see it requires x>=0 because In(-ve) isn't defined. That's why I used abs values in the highlighted line. Please let me know if I'm thinking a long the right lines. Your solution is exactly the same as "3" and this is what I had at the start of the problem. Were my two solutions, "1" and "2", incorrect where I used abs values? Thank you for the help. It has been very kind of you.

Best regards
PG
 

MrAl

Well-Known Member
Most Helpful Member
Hi,

Oh really? :) Lets see what happens here.

Lets work on a simpler equation first that is only slightly different.

dy/dx=y/x

From what you were saying before, for this one you would come up with solutions:
1. y=c*|x|
2. y=-c*|x|

Im not saying whether or not any of these are right or wrong, so what you should do is test both solutions in the original equation as i was saying before, and see if they work. Also, there may be a simplification if you look at this carefully.
But there is one other possibility. That is:
3. y=c*x

Now we have three possible solutions for dy/dx=y/x (any one or more might be right), so what you need to do is test all three solutions in the original equation and see what you can find out. See if you can find out something that is interesting.

Not sure if this will help:
ln|y|+C1=ln|x|+C2
|y|=C*|x|
 
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PG1995

Active Member
Thank you, MrAl.

For the original problem, these were the solutions I came up with: 1: y = -c|x^3|, 2: y = c|x^3|, 3: y = cx^3.

For "1" and "2", we can write: y=(+/-)c*|x^3|. As "c" is a constant, therefore "(+/-)c" could be written simply as "c". It becomes: y=c*|x^3|. If we simply assume that x>=0, then it could further be written as: c*x^3.

I'm sorry if I'm speaking nonsense. Right now I'm really sleepy. :(

Many thanks
PG
 

MrAl

Well-Known Member
Most Helpful Member
Hi again PG,


Yes, that's good. We can condense plus and minus c to just c because after all it's just a constant and can take on positive or negative values. That leaves us with:
y=c*x (or y=c*x^3 in your case)
and
y=c*abs(x) (or y=c*abs(x^3) in your case)

We're down to two possible solutions now, so the next step would be to insert each one into the original diff equation and see if one or both or none works.
I'd like to show you this and then you can do it with your case and see what you can find out.

So we'll look at the dy/dx=y/x first, with possible solutions y=c*x and y=c*abs(x).
Im going to use the symbols "!=" for "not equal to".

For y=c*x we have dy/dx=c and y/x=c, so that must be a solution because dy/dx=y/x.
For y=c*|x| we have dy/dx=c*signum(x) and y/x=c*signum(x), but there's a catch here because dy/dx!=y/x for all x. In fact, dy/dx!=y/x for x=0, so this may invalidate the solution unless we include that exception with the solution.
dy/dx!=y/x for x=0 because lim x-->0+ (dy/dx) != lim x-->0- (dy/dx).

So we can write:
y=c*x

as a solution, but we cant write:
y=c*|x|

as a solution unless we also include the exception like this:
y=c*|x|, x!=0

Most textbooks present the result of dy/dx=y/x as simply y=c*x, but textbooks sometimes dont include an entire discussion that would be necessary. A solution in some forums is a solution that fits for *all* x, but in fact a solution only has to be valid over the range in which it is specified in real life. If you are working a course you'd have to find out what the instructor expects to get a good mark on this :)

Now it's your turn to do your original problem again and check for any inconsistencies.
 
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PG1995

Active Member
Thanks a lot, MrAl.

Okay. Now it's my turn! :)

First some words about absolute value function and sign function are in order.

absolute value function:
absolutefucn-1.jpg

360pxAbsolute_valuesvg-1.png


sign function:
signfucton-1.jpg

Signum_function-1.svg



For the original problem we had three solutions: 1: y = -c|x^3|, 2: y = c|x^3|, 3: y = c*x^3. The three solutions can be reduced to only two as I said in my last post: 1: y = c*x^3, 2: y = c|x^3|.

The differential equation: [LATEX]\frac{dy}{dx}=\frac{3y}{x}[/LATEX]

is satisfied for y = c*x^3. Therefore, we turn to y=c*|x^3|.

As it turns out y=c*|x^3| is also a valid solution. You were saying that derivative of y = c|x| isn't defined at x=0; it's obvious from the graph that its derivative isn't defined at x=0. But derivative of y=c*|x^3| is defined at x=0, isn't it?

urfuct-1.jpg
 

MrAl

Well-Known Member
Most Helpful Member
Hi again,


Yes! And so you end up with two solutions anyway :)

So, "are we done yet?" (that's also the title of a cool movie too)

Absolute value signs are great to get rid of if possible. Can we eliminate them?
 
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PG1995

Active Member
Hi MrAl

As they say there is no end to pursuit of knowledge! :) So, I'm not sure if we are done yet!

By the way, I have checked out the details of the movie "Are We Done Yet?". I don't know how I missed it. Although there are two remakes available of the original starring Cary Grant, whenever I get time I'm going to watch the original because I have always liked Cary Grant. You might like Bringing Up a Baby.

I don't really know how to get rid of absolute value signs. I can only think of two choice and not sure if they are correct; you either use a piece-wise function, or restrict the value 'x' over the interval x>=0 then it makes |x|=x.

Many thanks for the help.

With best regards
PG
 
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MrAl

Well-Known Member
Most Helpful Member
Hi,

Im in a little bit of a hurry right now, so why dont you try to define it in two piece wise functions and see what you can find out. It's interesting i assure you :)
 

PG1995

Active Member
Hi again, MrAl :)

Here it is:
absolutepiecewise-1.jpg


So, what this means is that we have only two solutions: y=-cx^3 and y=cx^3. But can't we just replace -c in y=-cx^3 with +c? This would mean we have only one solution?

With best regards
PG
 

MrAl

Well-Known Member
Most Helpful Member
Hi,

Yes, isnt that interesting? c*x^3 covers (of course) c*x^3 solutions with c either plus or minus, and it also covers solutions c*|x^3| for c plus or minus.
Pretty amazing i think.

This is kinda pickie stuff i know, but take heart, in real life it's a little simpler to figure out what works and what doesnt once we get some kind of physical situation.
 
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