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Ive seen one of my TA's as well as my professor do this just the other day and I havent come across it before or at least I dont recall.
Heres an example of it from my Fluids Course.
obviously both sides were integrated twice to come up with the answer but I am curious as to what rule allows this to be done. I had messed this up on a quiz where we had to prove something the other day because i tried separation of variables which must not work for 2nd Order DE's or something.
I wouldnt consider my math background strong even though ive been through Differential Equations and a math course for engineers which is essentially Diff EQ's again. I dont recall seeing an equation like this in Diff Eq. Could anyone explain this to me?
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This is just indefinite integrals.
Take the first equation and integrate both side with respect to y and you get:
d/dy=-Cy+C1
Then integrating both sides again with respect to y:
d2/dy2=-1/2Cy^2+C1y+C2
This is all the possible functions whose second derivative with respect to y is -C.
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On November 24 2009 01:54 Zortch wrote:
This is just indefinite integrals.
Take the first equation and integrate both side with respect to y and you get:
d/dy=-Cy+C1
Then integrating both sides again with respect to y:
d2/dy2=-1/2Cy^2+C1y+C2
This is all the possible functions whose second derivative with respect to y is -C.
so you know its indefinate integrals because theres no other variable in the equation?
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On November 24 2009 01:54 Zortch wrote:
This is just indefinite integrals.
Take the first equation and integrate both side with respect to y and you get:
d/dy=-Cy+C1
Then integrating both sides again with respect to y:
d2/dy2=-1/2Cy^2+C1y+C2
This is all the possible functions whose second derivative with respect to y is -C.
He pretty much summed it up himself.
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Its an indefinite integral as opposed to a definite integral because of the arbitraty constants.
But what was the original problem?
Was it this:
Solve d2y/dy2=-C for y?
because then just integrate both side w/r y twice and poof!
This is just basic calculus, I'm not sure what the question is. (not to sound snooty or anything)
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On November 24 2009 02:03 Zortch wrote:
Its an indefinite integral as opposed to a definite integral because of the arbitraty constants.
But what was the original problem?
Was it this:
Solve d2y/dy2=-C for y?
because then just integrate both side w/r y twice and poof!
This is just basic calculus, I'm not sure what the question is. (not to sound snooty or anything)
Well the original problem not the one i provided had to do with finding the shear center of a beam or something. I cant find the specific problem atm. The problem I posted was given as a velocity field and we integrated to substitute into other equations etc. I was just curious because I dont remember being able to integrate both sides like that. I Specifically remember doing separation of variables with just dy/dx even with only one variable in the equation. I realize now that it comes out to be the same thing as integrating both sides as is. I guess I want to know if I can conclude that I only have to do separation of variables if 2 variables are present in the equation. If not I can always just integrate both sides?
I guess im wondering why I was told to do problems like this that were first order by separation of variables even though its obvious now that i dont have to.
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You are assuming that
1. C is a constant w.r.t. y
2. u is a function of y only (hence the "du/dy" instead of "partial u/partial y")
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On November 24 2009 02:11 nosliw wrote:
You are assuming that
1. C is a constant w.r.t. y
2. u is a function of y only (hence the "du/dy" instead of "partial u/partial y")
Ah, yes this is true. Good point.
In general, if you have:
d^nu(y)/dy^n=f(y), then just integrate n times on both sides to get u(y).
Thats just exactly what that equation means.
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Just a shot in the dark, but maybe try to keep in mind that d2u/dy2 is in fact d(du/dy)/dy. It should seem clearer that you can then go integrate d(du/dy) = -Cdy which gives du/dy = -Cy + C_1 and from there it should be clear where to go.
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you can multiply both sides by dy2 if you want
d2u=-C*dy2
d(du)=-C*dy*dy
then integrate both sides with indefinite boundaries (remember that du on the left side acts as a variable now, and integrated function =1):
du+D1=-C*y*dy+D2
since D1 and D2 are any numbers, these can be written as one symbol C1=D2-D1:
du=-C*y*dy+C1
again integrate with indefinite boundaries, this time left side's variable is u not du
u+D3=-C*y*y/2+C1*y+D4
then we take C2=D4-D3
u=-C*y*y/2+C1*y+C2
that's seroiusly one of the easiest diff eqs i've ever seen, what kind of equations did you have in those previous courses?
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LOL that isn't even a differential equation. It's just basic calculus.
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On November 24 2009 02:23 ZBiR wrote:
you can multiply both sides by dy2 if you want
d2u=-C*dy2
d(du)=-C*dy*dy
then integrate both sides with indefinite boundaries (remember that du on the left side acts as a variable now, and integrated function =1):
du+D1=-C*y*dy+D2
since D1 and D2 are any numbers, these can be written as one symbol C1=D2-D1:
du=-C*y*dy+C1
again integrate with indefinite boundaries, this time left side's variable is u not du
u+D3=-C*y*y/2+C1*y+D4
then we take C2=D4-D3
u=-C*y*y/2+C1*y+C2
that's seroiusly one of the easiest diff eqs i've ever seen, what kind of equations did you have in those previous courses?
Obviously i misunderstood something.
I know how to use bernoulli, laplace transforms, etc. It must be one of those that are so simple i didnt recognize it =-)
At the core of this though is that I was taught (in physics mind you) to do a simple dy/dx = Cy equation by separation of variables and im wondering why. Thats where the confusion comes from.
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On November 24 2009 02:29 Sadist wrote:Show nested quote +On November 24 2009 02:23 ZBiR wrote:
you can multiply both sides by dy2 if you want
d2u=-C*dy2
d(du)=-C*dy*dy
then integrate both sides with indefinite boundaries (remember that du on the left side acts as a variable now, and integrated function =1):
du+D1=-C*y*dy+D2
since D1 and D2 are any numbers, these can be written as one symbol C1=D2-D1:
du=-C*y*dy+C1
again integrate with indefinite boundaries, this time left side's variable is u not du
u+D3=-C*y*y/2+C1*y+D4
then we take C2=D4-D3
u=-C*y*y/2+C1*y+C2
that's seroiusly one of the easiest diff eqs i've ever seen, what kind of equations did you have in those previous courses?
Obviously i misunderstood something. I know how to use bernoulli, laplace transforms, etc. It must be one of those that are so simple i didnt recognize it =-) At the core of this though is that I was taught (in physics mind you) to do a simple dy/dx = Cy equation by separation of variables and im wondering why. Thats where the confusion comes from.
Yeah, that must be the case 
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On November 24 2009 01:49 Sadist wrote:
obviously both sides were integrated twice to come up with the answer but I am curious as to what rule allows this to be done. I had messed this up on a quiz where we had to prove something the other day because i tried separation of variables which must not work for 2nd Order DE's or something.
Did the equation look something like this?
d2y/dx^2 + C dy/dx + y = 0 (a general 2nd order ODE)
You can't separate the variables here. Just try it. You'll get confused trying to separate them
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On November 24 2009 02:23 ZBiR wrote:
you can multiply both sides by dy2 if you want
d2u=-C*dy2
d(du)=-C*dy*dy
then integrate both sides with indefinite boundaries (remember that du on the left side acts as a variable now, and integrated function =1):
du+D1=-C*y*dy+D2
since D1 and D2 are any numbers, these can be written as one symbol C1=D2-D1:
du=-C*y*dy+C1
again integrate with indefinite boundaries, this time left side's variable is u not du
u+D3=-C*y*y/2+C1*y+D4
then we take C2=D4-D3
u=-C*y*y/2+C1*y+C2
that's seroiusly one of the easiest diff eqs i've ever seen, what kind of equations did you have in those previous courses?
I like this approach the most, since it doesn't skip on some parts (both sides get a constant) like they do in some books. But fro practical use, it's better to "just integrate and put a constant".
Well, to be precise (if I remember correctly), these actually are definite boundary integrals (as all indefinite integrals are), but they are from some constant/fixed value to eg. u or y. So one boundary is a definite constant (those are the constants that you introduce, ie. C1, D1, etc.) and the other is an indefinite variable (so the upper boundary is not fixed). I think that's how it got it's name.
PS: I think you can only separate functions that depend on 2 variables. This function that you wrote depends only on one. I think that what you're refering to is called differently (separating du and dy and all things depending on u and y on different sides). But I might be wrong, it's been a long time.
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On November 24 2009 02:41 Purind wrote:Show nested quote +On November 24 2009 01:49 Sadist wrote:
obviously both sides were integrated twice to come up with the answer but I am curious as to what rule allows this to be done. I had messed this up on a quiz where we had to prove something the other day because i tried separation of variables which must not work for 2nd Order DE's or something.
Did the equation look something like this? d2y/dx^2 + C dy/dx + y = 0 (a general 2nd order ODE) You can't separate the variables here. Just try it. You'll get confused trying to separate them
na I guess it wasnt a differential eq =-) I wouldnt do separation of variables on something like that.
Physics just fucked me up and had me using separation of variables when I didnt need to but it ended up working out in the end anyway which is odd.
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At the core of this though is that I was taught (in physics mind you) to do a simple dy/dx = Cy equation by separation of variables and im wondering why.
How else would you solve that equation?
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On November 24 2009 13:30 gyth wrote:Show nested quote +At the core of this though is that I was taught (in physics mind you) to do a simple dy/dx = Cy equation by separation of variables and im wondering why. How else would you solve that equation?
apparently just integrate both sides without dividing multiplying by dx? :d
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I'm not sure why you think of it that way.
You've got that -C is the second derivative of a function u(y) with respect to y.
Agree?
So by the fundamental theorem of calc (integration and derivative are inverse operations) to solve for u(y) just integrate it with respect to y twice.
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apparently just integrate both sides without dividing multiplying by dx? :d
You _do_ need to use separation of variables to solve exponential growth.
ddu/ddy = -C is already as separate as it needs to be.
If it had been x'' = -9.8m/s^2, could you have come up with x = x0 + v0t -4.9 t^s?
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EPT![[image loading]](http://img18.imageshack.us/img18/2795/integration.jpg)
