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    Find the general solution of the differential equation 4 + x(dy/dx) = y^2

    So far I've separated the variables; 4 + dy/y^2 = dx/x . Next I would normally integrate both sides however what variable do I integrate 4 with respect to?
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    You haven't separated those variables correctly.
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    (Original post by Mr M)
    You haven't separated those variables correctly.
    Yep I've seen where I've done wrong. Cheers!!
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    (Original post by Mr M)
    You haven't separated those variables correctly.
    I've done:
    x(dy/dx) = y^2 - 4
    dy/(y^2 - 4) = dy/x
    But can't figure out how to integrate 1/(y^2 -4). I tried rewriting using difference of two squares and tried splitting using partial fractions. Is there a substitution method I can make?
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    (Original post by Freerider101)
    I've done:
    x(dy/dx) = y^2 - 4
    dy/(y^2 - 4) = dy/x
    But can't figure out how to integrate 1/(y^2 -4). I tried rewriting using difference of two squares and tried splitting using partial fractions. Is there a substitution method I can make?
    That was right. What did you get as the partial fractions?
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    You could also look in your formula book at the standard integrals.
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    (Original post by Mr M)
    That was right. What did you get as the partial fractions?
    I couldn't seem to split it.

    1/(y+2)(y-2) = A(y+2) + B(y-2)

    (y+2)(y-2) = B(y+2) + A(y-2)

    Then let y = 2
    B = 0
    let y = -2
    A = 0

    So it didn't seem to work out.
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    (Original post by Freerider101)
    I couldn't seem to split it.

    1/(y+2)(y-2) = A/(y+2) + B/(y-2)

    (y+2)(y-2) = B(y+2) + A(y-2)

    Then let y = 2
    B = 0
    let y = -2
    A = 0

    So it didn't seem to work out.
    I'm not liking that second line one bit. I added a couple of division symbols in your first line that were missing.
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    (Original post by Mr M)
    I'm not liking that second line one bit. I added a couple of division symbols in your first line that were missing.
    Okay I've got it. So its just 1/(y-2)(y+2) =1/4(y-2) -1/4(y+2)
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    (Original post by Freerider101)
    Okay I've got it. So its just 1/(y-2)(y+2) =1/4(y-2) -1/4(y+2)
    Yes. The answer (once integrated) appears in the formula book too so you could have just looked it up.
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    For those that want to know:

    \int\displaystyle\dfrac{1}{a^2 + x^2}dx =



\dfrac{1}{a}tan^{-1}(\dfrac{x}{a}) + C

    _Kar.
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    (Original post by Mr M)
    Yes. The answer (once integrated) appears in the formula book too so you could have just looked it up.
    Ahh sweet good revision of partial fractions I guess. Also I was told that all the integration formulae were for FP2 etc but I guess this isn't the case since this came from a C4 book.
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    (Original post by Kareir)
    For those that want to know:

    \int\displaystyle\dfrac{1}{a^2 + x^2}dx =



\dfrac{1}{a}tan^{-1}(\dfrac{x}{a}) + C

    _Kar.
    That was random. This question involved the difference of two squares not the sum of them.
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    (Original post by Freerider101)
    Ahh sweet good revision of partial fractions I guess. Also I was told that all the integration formulae were for FP2 etc but I guess this isn't the case since this came from a C4 book.
    Well, as you discovered, you can do it without reference to the formula book.
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    I know this is irrelevant but I REALLY need an answer (mock tomorrow)!

    What is 2root-3 as a complext number?
    Thats 2 x (-3)^0.5

    thanks.
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    (Original post by Glee)
    I know this is irrelevant but I REALLY need an answer (mock tomorrow)!

    What is 2root-3 as a complext number?
    Thats 2x-3^0.5

    thanks.
    2 \sqrt{-3} = 2 \times \sqrt{3} \times \sqrt{-1}

    Does that help?
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    (Original post by Mr M)
    2 \sqrt{-3} = 2 \times \sqrt{3} \times \sqrt{-1}

    Does that help?
    Yes a bit thanks, but in terms of 'i'?
    I know i^2=-1 ...
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    Oh sorry, so its 2 root3i?
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    (Original post by Glee)
    Oh sorry, so its 2 root3i?
    Yes.
 
 
 
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