|
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?
|
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.
|
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?
|
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.
|
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)
|
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.
|
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")
|
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.
|
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.
|
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?
|
LOL that isn't even a differential equation. It's just basic calculus.
|
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.
|
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 
|
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
|
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.
|
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.
|
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?
|
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
|
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.
|
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?
|
On November 24 2009 02:24 Luddite wrote:
LOL that isn't even a differential equation. It's just basic calculus.
ROFL
two simple indefinite integrations haha
|
On November 24 2009 02:24 Luddite wrote:
LOL that isn't even a differential equation. It's just basic calculus.
What do you think "basic calculus" is?
"Basic calculus" classes are just teaching you to solve basic integral and differential equations and understand what they mean at an intuitive level.
This problem is exactly an "ordinary differential equation." It just so happens to be one you should be able to figure out with Calc 1 techniques alone.
|
|
|
|
|
|
EPT![[image loading]](http://img18.imageshack.us/img18/2795/integration.jpg)
