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    Q. By eliminating the arbitrary constant A find the first order differential equation that is equivalent to;

     x^2 + y^2 = A

    My answer;
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    d/dx(constant) = 0
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    (Original post by ckfeister)
    Q. By eliminating the arbitrary constant A find the first order differential equation that is equivalent to;

     x^2 + y^2 = A

    My answer;
    You forgot to differentiate the constant A in the second line!
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    (Original post by NotNotBatman)
    d/dx(constant) = 0
    (Original post by SlashaRussia)
    You forgot to differentiate the constant A in the second line!
    dy/dx = y/x (can't use latex as the / (oppose side one) is broken for some annoying reason)

    (1/y) dy = (1/x) dx
    ln(y) = ln(x) + c
    y = x + e^c
    :. y = x + A

    The answer says it is
    y = Ax
    ???
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    (Original post by ckfeister)
    ln(y) = ln(x) + c
    y = x + e^c
    This step is wrong.
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    (Original post by RDKGames)
    This step is wrong.
    Ive bee trying to see where for a few hours with other questions having the same issue, isn't that anti-log? I'm going what I know from textbook only as I'm self-studying.
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    (Original post by ckfeister)
    dy/dx = y/x (can't use latex as the / (oppose side one) is broken for some annoying reason)

    (1/y) dy = (1/x) dx
    ln(y) = ln(x) + c
    y = x + e^c
    :. y = x + A

    The answer says it is
    y = Ax
    ???
    e^{lnx + c} = e^{lnx} \cdot e^c
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    (Original post by NotNotBatman)
    e^{lnx + c} = e^{lnx} \cdot e^c
    Well thats new...
    So
    ln[y] = ln[x] + c
    y = e^(ln[x] + c)
    y = Ax (as e^ln[x] = x)
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    (Original post by ckfeister)
    Well thats new...
    So
    ln[y] = ln[x] + c
    y = e^(ln[x] + c)
    y = Ax (as e^ln[x] = x)
    Bit worrying if that's new!

    a^{b+c} = a^b \cdot a^c is something Y10s work with.
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    (Original post by RDKGames)
    Bit worrying if that's new!

    a^{b+c} = a^b \cdot a^c is something Y10s work with.
    In year 10 I got E/Fs which was normal in my school as it is one of the worst as well as the one of the most disadvantaged area in South East England...
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    (Original post by ckfeister)
    Well thats new...
    So
    ln[y] = ln[x] + c
    y = e^(ln[x] + c)
    y = Ax (as e^ln[x] = x)
    This is a rule you should know, if it's new, go over any other little bits of algebra, to make sure that you can do these type of questions.
 
 
 
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