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Homework Help: Finding the intergral function (dy/dt = a(q-y))

  1. Apr 27, 2012 #1
    1. The problem statement, all variables and given/known data
    I was working on an assignment and when I got my draft back, my teacher said I've made some errors working this out, however I'm not sure what I did wrong...

    Find the intergral function of:
    dy/dt=a(q-y) where t ≥0, y(0)=0 a and q are constants.

    2. Relevant equations

    3. The attempt at a solution




    [itex]\int\frac{dy}{(q-y)}[/itex]=[itex]\int adt[/itex]


    q-y=eat+ec where: ec is a constant so let it = A



    [itex]\frac{-y}{-1}[/itex]= -(Aeat) [itex]\frac{ (-q)}{-1}[/itex]

    y= -Aeat+ q

    ∴ y= -Aeat+ q

    Can anyone see any mistakes?
  2. jcsd
  3. Apr 27, 2012 #2
    Your mistake was in converting from the first to the next statement. Try and think where the mistake is.

    Hint: What is the derivative of ln[f(x)]?
  4. Apr 27, 2012 #3
    I know the derivative of ln(x) to be

    Last edited: Apr 27, 2012
  5. Apr 27, 2012 #4
    The derivative of ln[f(x)] is f'(x)/f(x), where f'(x) is the derivative of f(x) with respect to x.

    Substitute f(x) = x into this formula to convince yourself that the formula I wrote is a generalisation of the derivative of ln(x).

    Then try to work out the derivative of ln(q-y), where q is a constant.
  6. Apr 27, 2012 #5
    wow... why must I divide the derivative function by the original function?
  7. Apr 27, 2012 #6
    You mean 'why must I divide the derivative function by the original function to obtain the derivative of ln[f(x)]?

    I think you are asking why f'(x)/f(x) has to be the derivative of ln[f(x)]. This is a very theoretical question, and as such it is unlikely to come up in your tests. But just so you know, that formula can be obtained by considering the mother of all formulas you have ever studied in differentiation. It is this:

    [itex]\frac{dg}{dx} = \underbrace{lim}_{h \rightarrow 0} \frac{g(x+h) - g(x)}{(x+h) - (x)}[/itex], where [itex]\frac{dg}{dx}[/itex] is the derivative of a function g(x) with respect to x.

    This is the fundamental definition of the differentiation of a function and all other formulas that you have learnt can be obtained by using this mother of all formulas. For instance, to obtain the derivative of ln[f(x)], substitute g(x) = ln[f(x)] in that fundamental definition and work out the result. The result has to be f'(x)/f(x).

    On a more practical note, d/dx { ln[f(x)] } = f'(x)/f(x) means that d/dx { ln(x) } = ????

    Try to work out the answer and then find d/dy { ln(q-y)}.
  8. Apr 27, 2012 #7
    I've always had trouble deriving by first principles, could I use the chain rule?

    let ln(q-y)=ln u?

    where u = (q-y) where q is constant


    [itex]\frac{d}{dy}[/itex] = [itex]\frac{d}{du}[/itex] x [itex]\frac{du}{dy}[/itex] ?
    Last edited: Apr 27, 2012
  9. Apr 27, 2012 #8
    Oh no! Don't try to find d/dy { ln(q-y) } using first principles. I elaborated on first principles just to answer your question on why you must divide the derivative function by the original function to obtain the derivative of ln[f(x)].

    Anyway, by the looks of your last post, you appear comfortable with the topic of differentiation, so try the chain rule.

    And you can check that you will get the same answer if you use d/dx { ln[f(x)] } = f'(x)/f(x).
  10. Apr 27, 2012 #9

    d/du = 1/u

    du/dy = -1?

    d/dy = 1/(q-y) x -1

    d/dy = -1/q-y

    errr... this can't be right, can it?
  11. Apr 27, 2012 #10
    Do you know how to do integration by substitution?
  12. Apr 27, 2012 #11
    that would not be ln(q-y). it will be -
  13. Apr 27, 2012 #12
    Nope, I've only learn the basic rules for intergrating trigonometric and basic polynomials (i.e. f(x)). Also the basic rule for intergrating 1/x which equals lnx

    Also the method that I used in my opening post... which I guess hasn't gotten me anywhere =D
  14. Apr 27, 2012 #13
    [tex]\int \frac{1}{q-y}dy[/tex]
    [tex]u = q-y[/tex]
    [tex]\frac{du}{dy} = -1\rightarrow dy = -du[/tex]

    We then substitute using q-y = u and dy = -du
    [tex]\int \frac{1}{u} \left (-du \right )=-\int \frac{1}{u} du[/tex]
    Once you accomplish the integration, treating u just like you'd treat x, you replace any u in your answer with q-x.
  15. Apr 27, 2012 #14
    Hmmm, why in this case are we deriving du/dy? instead of dy/du?
  16. Apr 27, 2012 #15
    It doesn't matter which one you do (if you can do both easily). When you do a U-substitution, though, you will write

    u = f(y)

    So it makes sense to differentiate with respect to y. You then get du/dy = f'(y).
  17. Apr 28, 2012 #16
    Ok mate, you've lost me
    [tex]\int \frac{1}{q-y}dy[/tex]
    [tex]u = q-y[/tex]


    [tex]\int \frac{1}{u}dy[/tex] right?

    So why doesn't...

    d = lnu y

    because [itex]\int dy[/itex] = y and [itex]\int\frac{1}{u}[/itex] = ln u ?

    Where is my mistake?
  18. Apr 28, 2012 #17


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    Other than your abuse of notation, what you wrote is correct. So if f(y) = ln (q-y), you have that f'(y) = -1/(q-y). You can use this information to solve
    $$\int\frac{1}{q-y}\,dy = \int -f'(y)\,dy = \ ?$$
  19. Apr 28, 2012 #18
    You can't integrate u with respect to y like that. You have to also substitute out the dy so that it is in terms of du. When you do this substitution, it introduces the negative sign.

    When you solve for this substitution, you write

    [tex]u = f(y)[/tex]
    [tex]\frac{du}{dy} = f'(y)[/tex]

    Now, at this point, it is important to note we are treating du/dy like division of two regular variables. In the case where f'(y) is simply a constant (here, it is -1), it is simplest to just multiply dy over and divide the constant over, resulting in
    [tex]dy = \frac{du}{f'(y)}= \frac{du}{-1} = -du[/tex]

    We then replace the "dy" in the original equation with "-du" since we established this equivalence. You can then integrate your 1/u with respect to u, which is important.

    Maybe another example will help you see the process:
    [tex] \int xe^{x^2} dx[/tex]

    So the trick with u substitution is you can take out an entire f'(x)dx and replace it with a single du.

    [tex]u = x^2[/tex]
    [tex]\frac{du}{dx} = 2x \rightarrow du = 2xdx[/tex]

    We then proceed with the substitution. Let's do it slowly so you can see each effect separately. Let's put in the u first:
    [tex]\int e^u xdx[/tex]

    Now, if only there were a 2 in there, the 2xdx would equal du. We can do the good old fashioned constant/constant = 1 trick:

    [tex]\int e^u \frac{2}{2}xdx = \frac{1}{2} \int e^u 2xdx[/tex]

    We now see we have the 2xdx = du. Let's replace that now

    [tex]\frac{1}{2}\int e^u du = \frac{1}{2} e^u[/tex]

    Finally, we replace u = x^2. That is our original substitution.

    [tex]\int xe^{x^2} dx=\frac{1}{2} e^{x^2}[/tex]

    As an exercise, you could try differentiating that final equation to see it equals the original integrand.

    Maybe you could try (using u substitution)

    [tex]\int \frac{x^2}{x^3-1}dx[/tex]
    Last edited: Apr 28, 2012
  20. Apr 29, 2012 #19
    Alright, I'll have a crack at this one...

    [tex]\int \frac{x^2}{x^3-1}dx[/tex]

    Let u = x3-1 So:

    [tex]\int \frac{x^2}{u}dx[/tex]


    [itex]\int x^2 × \frac{1}{u}[/itex]

    [itex] \frac{x^3}{3} × lnu [/itex]

    Sub in [itex]u = x^3 -1 [/itex]


    [itex]\int \frac{dy}{dx} = \frac{x^3}{3}ln x^3 -1 [/itex]
    Last edited: Apr 29, 2012
  21. Apr 29, 2012 #20
    It will help to look more clinically on what's on your paper:

    [itex] \int \frac{x^2}{x^3-1} dx = \int \frac{x^2 dx}{x^3-1}[/itex]

    Now sub [itex] x^3 -1 [/itex] with [itex]u[/itex].

    This means you got [itex] u = x^3 -1[/itex] and also [itex] du = 3x^2dx [/itex] wich means that [itex] \frac{du}{3} = \frac{1}{3} du =x^2 dx [/itex]

    Now look at your integrand, and become excited, since the right side of the equation over looks much like something in your integrand, now sub the [itex]x^2 dx[/itex] with [itex] \frac{1}{3}du[/itex] and you got:

    (I'm sure you can figure out the rest from here, but I will put the rest in spoiler)

    Moderator note: Even though this is just a practice problem, I removed the solution. It's against the forum rules to provide complete solutions.
    Last edited by a moderator: Apr 29, 2012
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