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Differential Equation using Separation of Variable Watch

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    (Original post by davros)
    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

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

    then I would simply write down something like this:

    Integrating both sides with respect to x, we obtain:

    y = \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) \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:

    \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

    I think we are of one mind regarding the horrendous multiply by dx concept

    I agree that I would not see the need to write Integral 4 dy in this case
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    (Original post by Chlorophile)
    Firstly, simplify the x terms on the LHS to get 2e^{-x}-5x=4\frac{dy}{dx}. You then want to multiply both sides by dx, giving you (2e^{-x}-5x) dx=4 dy and then you can integrate both sides.

    thank you for your help I appreciate it a lot


    I done the question now with the following answer is this correct?

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    1)collect the x^2 together 1st and divide by 4 - then try again : c= \frac{3}{2}
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    (Original post by Mo-Student)
    thank you for your help I appreciate it a lot


    I done the question now with the following answer is this correct?

    In addition to Hasufel's comment, don't put the 'therefore' symbol on both sides of an equation - it's meaningless! Remember how therefore is used in English - it introduces a statement e.g. "Therefore x is equal to y". You don't say "therefore x = therefore y"
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    (Original post by Hasufel)
    1)collect the x^2 together 1st and divide by 4 - then try again : c= \frac{3}{2}
    Do you mean in
    2e^-x -4x -x +c=4y

    placing the x's together to form the -5x?
    2e^-x -5x +c=4y

    (2e^-x) -(2.5x^2) +c= 4y <--- i think this is the same as

    (2e^-x) - (2x^2) - (0.5x^2) +C = 4y

    I tryed by plugging in the y=1 and x=0 and had the same answers for both methods as having -5x or leaving -4x -x after integrating them

    Can you elaborate a little bit more please
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    on your 3rd line down, divide by 4 (or you can do this before you integrate) so you get:

    \displaystyle  - \frac{1}{2e^{x}}- \frac{5x^{2}}{8}+k=y where k=c/4 then:

    \displaystyle  - \frac{1}{2e^{0}}+k=1....
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    (Original post by Mo-Student)
    thank you for your help I appreciate it a lot


    I done the question now with the following answer is this correct?

    You integral signs are in the wrong place .... They should be on the previous line

    When you integrate 2e^-x you will get -2x^-x

    I would suggest collecting the -4x -x together prior to integrating but that for ease and neatness rather than correctness


    You should get c=6
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    (Original post by Hasufel)
    on your 3rd line down, divide by 4 (or you can do this before you integrate) so you get:

    \displaystyle  - \frac{1}{2e^{x}}- \frac{5x^{2}}{8}+k=y where k=c/4 then:

    \displaystyle  - \frac{1}{2e^{0}}+k=1....

    this kind of what dont like magical numbers appearing (sorry thats how it looks to me) why should I divide by 4? then again where did you get the 5x^2/8 where did you get the 8 from?

    Then there is -1/2e^x isnt this 2e^-x?
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    (Original post by TenOfThem)
    You integral signs are in the wrong place .... They should be on the previous line

    When you integrate 2e^-x you will get -2x^-x

    I would suggest collecting the -4x -x together prior to integrating but that for ease and neatness rather than correctness


    You should get c=6

    When you integrate 2e^-x you will get -2x^-x

    what happened to the e there? should it be -2e^-x?
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    (Original post by Mo-Student)
    When you integrate 2e^-x you will get -2x^-x

    what happened to the e there? should it be -2e^-x?
    Yes, of course ... Mistype
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    (Original post by TenOfThem)
    Yes, of course ... Mistype

    Ok if I do that I will get C=6

    but there is another problem

    If keep my constant from the RHS and I do not use the other constant from the LHS then the answer would be C=-6

    So if I use the answer as above with LHS constant as for my image and have -2e^-x then this answer is C=6

    why is that?
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    (Original post by Mo-Student)
    Ok if I do that I will get C=6

    but there is another problem

    If keep my constant from the RHS and I do not use the other constant from the LHS then the answer would be C=-6

    So if I use the answer as above with LHS constant as for my image and have -2e^-x then this answer is C=6

    why is that?
    Because

    x+6=y is the same as x=y-6
 
 
 
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