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    Could someone help me with this please?

    x\frac{dy}{dx}+3y=\frac{sinx}{x^  2}

    Thanks.
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    (Original post by RamocitoMorales)
    Could someone help me with this please?

    x\frac{dy}{dx}+3y=\frac{sinx}{x^  2}

    Thanks.
    If you divide both sides through by x, you'll have a first order de in the form of:

    \dfrac{dy}{dx} +Py =Q, where P and Q are functions of x. You can then use an integrating factor.
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    Seriously? You're supposed to be a university maths student and you need help with this?

    EDIT: Yes, I originally counted wrong. But that doesn't detract from my point.
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    (Original post by davidmarsh01)
    Seriously? You're supposed to be a second year maths student and you need help with this?
    Seriously? You're supposed to be an A level maths student and you cant count?
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    (Original post by rbnphlp)
    Seriously? You're supposed to be an A level maths student and you cant count?
    Seriously? Can't you read my signature? I do Advanced Highers, don't just assume :rolleyes:
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    (Original post by davidmarsh01)
    Seriously? Can't you read my signature? I do Advanced Highers, don't just assume :rolleyes:
    So do I :borat:
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    (Original post by davidmarsh01)
    Seriously? Can't you read my signature? I do Advanced Highers, don't just assume :rolleyes:
    Seriously? It still doesnt take away the fact that you cant count ..
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    (Original post by rbnphlp)
    Seriously? It still doesnt take away the fact that you cant count ..
    I made a mistake, get over it. I'd like to see you devise an education system that eliminates all errors :rolleyes:
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    (Original post by davidmarsh01)
    Seriously? You're supposed to be a university maths student and you need help with this?

    EDIT: Yes, I originally counted wrong. But that doesn't detract from my point.
    I think it's important to remember that this is at the upper end of Further Maths (for A-Level students) and a lot of university courses tend to spend a sizeable chunk of the first year reinforcing this knowledge so it's not so strange that he's stuck on a question which seems to be of A-Level/HL difficulty to you. He may not have fully understood it when he originally learnt it, or he may have possibly forgotten how to do it all together!
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    (Original post by Farhan.Hanif93)
    I think it's important to remember that this is at the upper end of Further Maths (for A-Level students) and a lot of university courses tend to spend a sizeable chunk of the first year reinforcing this knowledge so it's not so strange that he's stuck on a question which seems to be of A-Level/HL difficulty to you. He may not have fully understood it when he originally learnt it, or he may have possibly forgotten how to do it all together!
    Really? This type of question is seen as quite a basic question in Advanced Higher Maths, it just surprised me that someone in university didn't know something like this.
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    I've been given this exact same question for an assesment for 1st year maths. I still don't get the maths behind it and before anyone goes being a **** saying why don't you know the fact is people come on here for help not abuse
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    Ok I got some help withis in school today.

    x(dy/dx) + 3y = Sin(x)/(x^2) (*1/x)
    dy/dx + 3y/x = Sin(x)/(x^3)

    The next bit I don't really understand the principle behind however it seems to work

    Take 3/x as and integration factor integrate it to get 3ln(x). The raise e to the power of 3ln(x) to get x^3. Now multiply the whole thing by x^3

    (dy/dx)x^3 +3yx^2 = Sin(x) (*dx)
    dyx^3 + dx3x^2 = dxSin(x)

    Now integrate the above to find

    yx^3 = -Cos(x) +c
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    If the reason someone has just given me negative rating is because they think its wrong please say whats up with it you'd be helping me
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    I'm a 4th year university student and I have forgotten pretty much everything about differential equations. I intend to re-teach myself sometime. Haven't done them since second year.
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    Nice what sort of stuff are you doing at the moment then.
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    I've actually finished all my modules now, just my dissertation to finish and exams in the summer term. But the two maths modules I did last term (I'm on a joint course) were semigroup theory and algebraic geometry. I've been focusing on the algebra related modules, so calculus and other areas of maths have been neglected.
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    (Original post by Cameron F)
    Ok I got some help withis in school today.

    x(dy/dx) + 3y = Sin(x)/(x^2) (*1/x)
    dy/dx + 3y/x = Sin(x)/(x^3)

    The next bit I don't really understand the principle behind however it seems to work

    Take 3/x as and integration factor integrate it to get 3ln(x). The raise e to the power of 3ln(x) to get x^3. Now multiply the whole thing by x^3

    (dy/dx)x^3 +3yx^2 = Sin(x) (*dx)
    dyx^3 + dx3x^2 = dxSin(x)

    Now integrate the above to find

    yx^3 = -Cos(x) +c
    What's the point of multiplying it by x^3 again if you divide the expression by x at first. Can you have coefficients with dy/dx to carry out integration?
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    (Original post by gunmetalpanda)
    What's the point of multiplying it by x^3 again if you divide the expression by x at first. Can you have coefficients with dy/dx to carry out integration?
    If you've had enough practice you may realise that multiplying by x^2 in the first place gives x^3\frac{dy}{dx}+3x^2y which is \frac{d(x^3y)}{dx}.

    If you can do that then you probably don't need to be doing these questions.

    The point of dividing by x in the first place is to get the DE into standard form. From there the integrating factor method can be followed.
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    (Original post by Cameron F)
    The next bit I don't really understand the principle behind however it seems to work
    This is exactly where you can have a bit of fun with maths.

    Think about it. Keep it general.

    dy/dx + y p(x) = q(x)

    You can figure out why it works.
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    When you put it like that it does make sense
 
 
 
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