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Advanced Higher Maths 2011-2012 : Discussion and Help

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    (Original post by Mr Dangermouse)
    If I post a list of questions could anyone help me? Wish I could offer more than rep
    I would help, although probably later on in the evening as I'm a bit busy atm.
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    (Original post by Mr Dangermouse)
    If I post a list of questions could anyone help me? Wish I could offer more than rep
    Just go for it.
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    (Original post by Mr Dangermouse)
    If I post a list of questions could anyone help me? Wish I could offer more than rep
    I'll try and help if I can. =)
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    (Original post by Mr Dangermouse)
    If I post a list of questions could anyone help me? Wish I could offer more than rep
    Mathematicians! Assemble!
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    I guess I'm in for some maths on a Friday night.
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    dy/dx + y/x = x


    Anybody?
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    (Original post by Mr Dangermouse)
    dy/dx + y/x = x


    Anybody?
    Need to solve it using the integrating factor method, where if you have dy/dx + P(x)y = Q(x) (which is purely a function of x).

    Then: e^\int{P(x)dx} is your integrating factor. Multiply both sides of the equation by this integrating factor and see if you can think where to go from there.

    The LaTex isn't clear but that's e raised to the integral of P(x).

    If you get stuck.
    Spoiler:
    Show
    P(x) = 1/x. So e raised to the integral of 1/x which is (as you know) lnx. e^lnx = x. Then multiply both sides by x.
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    So it's a first order linear differential equation of form:

    \frac{dy}{dx} + P(x)y = Q(x)

    What's P(x)? What's Q(x)?

    You can then use P(x) to find the integrating factor that's of the form e^{\int P(x)dx}, which you multiply the entire original equation by, surely you can remember where to go from here?

    My solution:
    Spoiler:
    Show
    P(x) = \frac{1}{x} \ Q(x) = x
     e^{\int \frac{1}{x} dx} = e^{lnx} \\ = x

    Multiply both sides of the original differential by the integrating factor.

     x\frac{dy}{dx} + y = x^2

    Now you have to integrate both sides of the equation to find a general solution in terms of y.

    I'll do the LHS first:
     \int (x\frac{dy}{dx} + y)dx which equals xy.
    Now to integrate the RHS:
     \int x^2 = \frac{x^3}{3} + c

    Set them equal to each other:
     xy = \frac{x^3}{3} + c
     y = \frac{x^2}{3} + c
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    Erm... wut?
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    I normally solve them by splitting the variables.
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    (Original post by Mr Dangermouse)
    I normally solve them by splitting the variables.
    You can't do that with this. Variables separable only applies to some first order differential equations. Integrating factor is the other method used to solve this type of first order differential equations.

    Steps for the Integrating factor method:

    1. Ensure the equation is in standard form - dy/dx + p(x)y = Q(x)

    2. Identify P(x) and \int P(x)dx

    3. Write out the integrating factor e^{\int P(x)dx}

    4.Multiply both sides by the integrating factor

    5. The LHS will be \frac{d}{dx}(I.F\cdot y)

    6. Integrate both sides.

    7. Substitute in initial conditions/rearrange to make y the subject.
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    (Original post by Hype en Ecosse)
    So it's a first order linear differential equation of form:

    \frac{dy}{dx} + P(x)y = Q(x)

    What's P(x)? What's Q(x)?

    You can then use P(x) to find the integrating factor that's of the form e^{\int P(x)dx}, which you multiply the entire original equation by, surely you can remember where to go from here?

    My solution:
    Spoiler:
    Show
    P(x) = \frac{1}{x} \ Q(x) = x
     e^{\int \frac{1}{x} dx} = e^{lnx} \\ = x

    Multiply both sides of the original differential by the integrating factor.

     x\frac{dy}{dx} + y = x^2

    Now you have to integrate both sides of the equation to find a general solution in terms of y.

    I'll do the LHS first:
     \int (x\frac{dy}{dx} + y)dx which equals xy.
    Now to integrate the RHS:
     \int x^2 = \frac{x^3}{3} + c

    Set them equal to each other:
     xy = \frac{x^3}{3} + c
     y = \frac{x^2}{3} + c
    Spoiler:
    Show
    Just semantics really, but you should divide the entire RHS by x.

    y = \frac{x^2}{3}+\frac{c}{x}

    But yep, exactly what I got/did.

    (Original post by Mr Dangermouse)
    I normally solve them by splitting the variables.
    Aren't able to split the variables here, have to use the integrating factor method. The separation of variables can only be done if you've got the derivative of a function equal to a function of x multiplied by a function of y.
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    (Original post by Mr Dangermouse)
    I normally solve them by splitting the variables.
    We don't have a differential equation where we can do that, I can't for the life of me remember what the first type that you're talking about are called, but I know what you mean. Where you'd have:

    dy/dx = xy

    And you'd make it dy/y = xdx and integrate. But you don't have that here, have you done first-order linear differential equations?


    (Original post by JordanR)
    Spoiler:
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    Just semantics really, but you should divide the entire RHS by x.

    y = \frac{x^2}{3}+\frac{c}{x}

    But yep, exactly what I got/did.
    c is a completely arbitrary constant, so I thought one didn't need to divide the c?
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    I don't know if they have a specific name actually. I don't think I've been told one if they do.

    But if they're like: \frac{d}{dx}P(x) = Q(x)R(y) then we can split the variables up.


    (Original post by Hype en Ecosse)

    c is a completely arbitrary constant, so I thought one didn't need to divide the c?
    Hence why it's semantics! Although it would become important if one was given an initial condition and one was also asked to find the value of one's constant, so it's definitely good practice for one to do so.
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    (Original post by JordanR)
    I don't know if they have a specific name actually. I don't think I've been told one if they do.

    But if they're like: \frac{d}{dx}P(x) = Q(x)R(y) then we can split the variables up.



    Hence why it's semantics! Although it would become important if one was given an initial condition and one was also asked to find the value of one's constant, so it's definitely good practice for one to do so.
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    (Original post by Hype en Ecosse)
    c is a completely arbitrary constant, so I thought one didn't need to divide the c?
    Yes, but they're not the same arbitrary constant, so you need to either show the division or rename it. Same reason x = k \not\rightarrow \tfrac{1}{2}x = k. It's not just semantics: your last step implies c = 0.
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    (Original post by TheUnbeliever)
    Yes, but they're not the same arbitrary constant, so you need to either show the division or rename it. Same reason x = k \not\rightarrow \tfrac{1}{2}x = k. It's not just semantics: your last step implies c = 0.
    When I first started doing differential equations, I always divided my c, too. The answer schemes never divided the c, and the teacher told me I didn't need to divide it through; I made the same argument as you, but the teacher was having none of it...

    This was shortly after I got a row about "remember that she's done a degree and you've not." So...
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    I hate it when teachers try and argue from authority without justifying it at all.
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    Hmm, this is quite a nice question. Quite a few different techniques required for it:

    Solve by use of standard integrals.

    \int \frac{2x+3}{x^2+4x+8}\ dx

    Also, I have a few questions:

    What does "find the term independent of x" mean and are we supposed to be able to do this? Is it just the term in the binomial expansion that has an x term that is to the power of 0?

    Are related rates examinable? Because I don't think we covered them in class, but I'm sure I seen them in an AH book (but not 100% sure).
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    (Original post by Quintro)
    Hmm, this is quite a nice question. Quite a few different techniques required for it:

    Solve by use of standard integrals.

    \int \frac{2x+3}{x^2+4x+8}\ dx

    Also, I have a few questions:

    What does "find the term independent of x" mean and are we supposed to be able to do this? Is it just the term in the binomial expansion that has an x term that is to the power of 0?

    Are related rates examinable? Because I don't think we covered them in class, but I'm sure I seen them in an AH book (but not 100% sure).
    Very nice indeed. On my phone but it seems like long division and then a couple of u-subs? I'll try it later.

    Yes, I'd think so. So if you had (x+y)^5 you'd find the term that only had y in it.

    Probably, but I've never seen them in any past papers. Not looked at many though.
Updated: August 12, 2012
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