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Quasi-homogeneous 1st Order ODEs

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    Wondering if anyone could explain this example to me, please?

    I'm looking at a worked example in our notes and the qu is:


    Solve dy/dx = (x + y + 3)/(x − y − 5).

    The solution starts:

    Put x = x0 + X, y = y0 + Y
    We require x0 + y0 + 3 = 0 and x0 − y0 − 5 = 0 which has solution x0 = 1 and y0 = −4.
    Then we have dY/dX = (X + Y)/(X − Y).


    I understand all of that.

    Then, it says:


    Put Y = uX to get X*du/dX + u = (1 + u)/(1 − u).

    I understand where the RHS came from... but how did they get the LHS of that?

    And what happens to the 1 and -4 we found earlier?

    Any help would be much appreciated. Thank you
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    differentiated uX with respect to X? (just a guess, i shouldnt really be looking at undergraduate)
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    (Original post by funkiichiicka)
    I understand where the RHS came from... but how did they get the LHS of that?
    The LHS is the product rule for differentiation. It is necessary to find \dfrac{\text{d}Y}{\text{d}X}= \dfrac{\text{d}(uX)}{\text{d}X}.

    (Original post by funkiichiicka)
    What happened to the 1 and -4 we found earlier?
    It was used to give us \dfrac{\text{d}Y}{\text{d}X}= \dfrac{X+Y}{X-Y}.

    I hope that helps.

    Darren
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    Yes, Y = uX, so \dfrac{d}{dX} Y = \dfrac{d}{dX}(uX).
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    (Original post by DPLSK)
    The LHS is the product rule for differentiation. It is necessary to find \dfrac{\text{d}Y}{\text{d}X}= \dfrac{\text{d}(uX)}{\text{d}X}.

    It was used to give us \dfrac{\text{d}Y}{\text{d}X}= \dfrac{X+Y}{X-Y}.

    I hope that helps.

    Darren

    Thank you.

    I thought it was something to do with product rule, but I can't figure out how exactly it was used

    Please can you explain :confused:
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    (Original post by funkiichiicka)
    Thank you.

    I thought it was something to do with product rule, but I can't figure out how exactly it was used

    Please can you explain :confused:
    You're welcome.

    We know the product rule for differentiation is the following.

    \dfrac{d}{dx}(u\cdot v)=u\cdot \dfrac{dv}{dx}+v\cdot \dfrac{du}{dx}

    Replace v with X and the required result follows.

    I hope that helps.

    Darren

    P.S.: The centre dot denotes product. It isn't necessary here, but some people choose to use it.
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    (Original post by funkiichiicka)

    Thank you.

    I thought it was something to do with product rule, but I can't figure out how exactly it was used

    Please can you explain :confused:
    uX is the function u(X) multiplied by the function X. We can just product rule it:

    \dfrac{d(uX)}{dX} = \dfrac{du}{dX} + \dfrac{dX}{dX}
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    (Original post by DPLSK)
    You're welcome.

    We know the product rule for differentiation is the following.

    \dfrac{d}{dx}(u\cdot v)=u\cdot \dfrac{dv}{dx}+v\cdot \dfrac{du}{dx}

    Replace v with X and the required result follows.

    I hope that helps.

    Darren

    P.S.: The centre dot denotes product. It isn't necessary here, but some people choose to use it.
    Aah!!! Got it! Thank you so much!

    Thank you for the help. Much appreciated.

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