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Differential Equation using Separation of Variable

How would you solve this Differential Equation using separation of variable.


y=1 when x=0

solving

(2/e^x) - 4x = x + (4 dy/dx)


(4 dy/dx) is this one 4y?

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Original post by Mo-Student
How would you solve this Differential Equation using separation of variable.


y=1 when x=0

solving

(2/e^x) - 4x = x + (4 dy/dx)


(4 dy/dx) is this one 4y?


What do you mean by that? Before you can integrate the expression you need to make sure the side with the x terms is multiplied by dx and the side with the y terms (or in this case, the constant 4) multiplied by dy. You don't want dy/dx on one side.
Reply 2
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Mo-Student
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Original post by Chlorophile
What do you mean by that? Before you can integrate the expression you need to make sure the side with the x terms is multiplied by dx and the side with the y terms (or in this case, the constant 4) multiplied by dy. You don't want dy/dx on one side.




from the rules I understand that separation variables says x should be on one side of the equal sign and all the y's on the other one but from the question I can really see a y term

hope this attach file work

20150120_141035-1.jpg194.7KB
Original post by Mo-Student
from the rules I understand that separation variables says x should be on one side of the equal sign and all the y's on the other one but from the question I can really see a y term

hope this attach file work



You don't need a y term to integrate with respect to y, all you need is a constant. For instance, it's perfectly possible to integrate 1 with respect to x - you'll just get x. What you want to do is rearrange the equation so that you have all the x terms on one side and in this case, a constant on the right side. Then you integrate. Does that make sense?
Reply 4
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Mo-Student
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Original post by Chlorophile
You don't need a y term to integrate with respect to y, all you need is a constant. For instance, it's perfectly possible to integrate 1 with respect to x - you'll just get x. What you want to do is rearrange the equation so that you have all the x terms on one side and in this case, a constant on the right side. Then you integrate. Does that make sense?



4dy/dx is the only constant? but as for the second step straight after the question aren't the -4x and the x as well going to be constants because would be differentiating at this point?
later after this stage I would be starting with integration and as you said place all x's on one side and all the constants on the other side and integrate?
Original post by Mo-Student
4dy/dx is the only constant? but as for the second step straight after the question aren't the -4x and the x as well going to be constants because would be differentiating at this point?
later after this stage I would be starting with integration and as you said place all x's on one side and all the constants on the other side and integrate?


I'm not entirely sure I understand what you're asking. -4x and x are not constants because they have x terms in them.
Reply 6
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Mo-Student
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Original post by Chlorophile
I'm not entirely sure I understand what you're asking. -4x and x are not constants because they have x terms in them.



when you differentiate -4x is going to be just 4 and you get rid of x that way? same with x.

As in an example 3x^2 would be 6x because (3)(2) = 6 and x is still there. so 3x^1 is same as 3x when it is differentiated it would be (3)(1) = 3 and x is gone because x is the ^1 which comes down?
Original post by Mo-Student
when you differentiate -4x is going to be just 4 and you get rid of x that way? same with x.

As in an example 3x^2 would be 6x because (3)(2) = 6 and x is still there. so 3x^1 is same as 3x when it is differentiated it would be (3)(1) = 3 and x is gone because x is the ^1 which comes down?


Why are you differentiating? If you're trying to solve the differential equation you should be integrating...
Reply 8
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Mo-Student
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Original post by Chlorophile
Why are you differentiating? If you're trying to solve the differential equation you should be integrating...



ok nice one thanks mate just the wording got me confused :smile: they should call it something else like integral equation or something the wording sends you to the wrong direction :/

so to beautify it down i should integrate this?

(2e^-x) -4x= x + 4dy/dx?

integrate straight way?
Original post by Mo-Student
ok nice one thanks mate just the wording got me confused :smile: they should call it something else like integral equation or something the wording sends you to the wrong direction :/

so to beautify it down i should integrate this?

(2e^-x) -4x= x + 4dy/dx?

integrate straight way?


Yeah I know, it's a bit confusing!

You can't integrate the expression you've given above. Firstly, you've still got x terms on both sides. Secondly, you've still got a dy/dx on one side. One side (with the x terms) needs to have dx, the other dy.
Reply 10
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Mo-Student
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Original post by Chlorophile
Yeah I know, it's a bit confusing!

You can't integrate the expression you've given above. Firstly, you've still got x terms on both sides. Secondly, you've still got a dy/dx on one side. One side (with the x terms) needs to have dx, the other dy.




This is something I haven't done before still struggling with breaking the dy/dx

this is how my separation for this expression I know it is wrong how should the correct way be?
2e^-x -4x -x = 4 dy/dx
Original post by Mo-Student
This is something I haven't done before still struggling with breaking the dy/dx

this is how my separation for this expression I know it is wrong how should the correct way be?
2e^-x -4x -x = 4 dy/dx


Firstly, simplify the x terms on the LHS to get 2e−x−5x=4dydx2e^{-x}-5x=4\frac{dy}{dx}. You then want to multiply both sides by dx, giving you (2e−x−5x)dx=4dy(2e^{-x}-5x) dx=4 dy and then you can integrate both sides.
(edited 9 years ago)
Reply 12
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davros
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Original post by Chlorophile
Firstly, simplify the x terms on the LHS to get 2e−x−5x=4dydx2e^{-x}-5x=4\frac{dy}{dx}. You then want to multiply both sides by dx, giving you (2e−x−5x)dx=4dy(2e^{-x}-5x) dx=4 dy and then you can integrate both sides.


I may be being dumb, but why not leave the dy/dx as it is and just integrate it to get y - there's no actual need to split it up in this example :smile:
Original post by davros
I may be being dumb, but why not leave the dy/dx as it is and just integrate it to get y - there's no actual need to split it up in this example :smile:


:confused::confused::confused:

How would you integrate without a dx
(edited 9 years ago)
Reply 14
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davros
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Original post by TenOfThem
:confused::confused::confused:

How would you integrate without a dx


Because you basically have:

dy/dx = f(x)

so you can integrate both sides directly with respect to x

(I think we're slightly talking semantics here - yes. when you write the integral out there is a 'dx' in the notation but the question as stated is a poor example of 'separation of variables' IMHO)
Original post by TenOfThem
:confused::confused::confused:

How would you integrate without a dx


I think he means integrating both sides with respect to x (leaving the dy/dx on the RHS which integrates to y)? That should work, I'd have thought you'd get to the same result.
Original post by Chlorophile
I think he means integrating both sides with respect to x (leaving the dy/dx on the RHS which integrates to y)? That should work, I'd have thought you'd get to the same result.


Yes ∫dydxdx=∫1dy\int \frac{dy}{dx}dx = \int 1 dy
Original post by davros
Because you basically have:

dy/dx = f(x)

so you can integrate both sides directly with respect to x

(I think we're slightly talking semantics here - yes. when you write the integral out there is a 'dx' in the notation but the question as stated is a poor example of 'separation of variables' IMHO)


Perhaps we are talking semantics but I would not do this without correct notation any more than I would say "multiply by dx"
Original post by Chlorophile
I think he means integrating both sides with respect to x (leaving the dy/dx on the RHS which integrates to y)? That should work, I'd have thought you'd get to the same result.


That is what we do when we use separation of variables ... We integrate both sides wrt x
Reply 19
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davros
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Original post by TenOfThem
Perhaps we are talking semantics but I would not do this without correct notation any more than I would say "multiply by dx"


Sorry, I got called away from the computer so my last post was a bit rushed.

What I was getting at was this - if I were given the equation

dydx=f(x)\dfrac{dy}{dx} = f(x)

then I would simply write down something like this:

Integrating both sides with respect to x, we obtain:

y=∫f(x)dxy = \int f(x) dx

because we know that integration reverses differentiation so we don't even need to think about manipulating the LHS,

What I would NOT do is write down an intermediate step like this:

dy = f(x) dx

- which is what seemed to be being suggested here -
and then write down the next step with an integral sign in front of each side, because dy/dx is not a fraction and quantities like 'dy' and 'dx' don't have any meaning on their own.

In a "more sensible" example of a separation of variables question you would have something like this:

g(y)dydx=f(x)g(y) \dfrac{dy}{dx} = f(x)

and it is common to see textbooks writing things like

g(y)dy = f(x)dx

although I try to avoid this sort of abuse of notation myself and go straight to the "final version" with integral signs on both sides:

∫g(y)dy=∫f(x)dx\int g(y) dy = \int f(x) dx

The end result is the same - I just try not to give people the impression that you can just move dx's and dy's around as if they were standalone algebraic quantities.

Hope that explains better what I was trying to say :smile:

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