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Help with Integration general solution question watch

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    Hi

    If anyone can advise on this one it would be much appreciated:

    So far I have:
     u \frac {du}{dv} = v +2

    (I don't fully understand this next part I'm just following the pattern from my examples, if anyone could explain that would be a big help) I know you have to separate the variables but I'm not 100% on it yet.

    \int u \:du = \int v+2 dv
     \frac {u^2}{2} = \frac {v^2}{2} + 2v + A

    Not sure what to do next? and also whether that is correct or not?

    thank you!
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    What are you trying to do? You have integrated correctly. If you want to make v the subject multiply by 2 and complete the square.
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    (Original post by Mr M)
    What are you trying to do? You have integrated correctly. If you want to make v the subject multiply by 2 and complete the square.
    It asks to find the general solution so I think that is make v the subject?
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    I've got a answer on the sheet that is
     u^2 = v^2 + uv + B

    where has the u on the right come from?
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    (Original post by mh1985)
    I've got a answer on the sheet that is
     u^2 = v^2 + uv + B

    where has the u on the right come from?
    Seemingly out of nowhere, but you might have gone wrong before you got to the stage of writing u\frac{du}{dx} = v+2. What was the original question?
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    (Original post by nuodai)
    Seemingly out of nowhere, but you might have gone wrong before you got to the stage of writing u\frac{du}{dx} = v+2. What was the original question?
    u\frac{du}{dx} = v+2 was the original question :confused:
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    (Original post by mh1985)
    u\frac{du}{dx} = v+2 was the original question :confused:
    Ah right; in that case, the u shouldn't be there. [In fact, it should be a 4.] I think it's just a typo; they typed "u" instead of "4"... easy enough to do :p:
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    (Original post by nuodai)
    Ah right; in that case, the u shouldn't be there. [In fact, it should be a 4.] I think it's just a typo; they typed "u" instead of "4"... easy enough to do :p:
    ah good I've got the right answer in that case!
 
 
 
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