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    Okay question:

    Given that k is an arbitrary positive constant, show that y^2+kx^2=9k is a general solution of the differential equation \frac{dy}{dx}=\frac{-xy}{9-x^2}, |x|\leq 3.

    I have been able to do this by making a substitution, but trying to do it using partial fractions doesn't work. Can someone tell me what I've done wrong?

    \int\frac{1}{y}dy=\int\frac{x}{x  ^-9}dx

    \int\frac{1}{y}dy=\frac{1}{2} \int \frac{1}{x+3} +\frac{1}{2} \int \frac{1}{x-3}

    lny^2=ln|(x+3)(x-3)k|

    y^2=kx^2-9k^2


    Which is not what I'm trying to achieve.
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    I've got it, how do you write maths on here like that?
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    (Original post by ViralRiver)
    Okay question:

    Given that k is an arbitrary positive constant, show that y^2+kx^2=9k is a general solution of the differential equation \frac{dy}{dx}=\frac{-xy}{9-x^2}, |x|\leq 3.

    I have been able to do this by making a substitution, but trying to do it using partial fractions doesn't work. Can someone tell me what I've done wrong?

    \int\frac{1}{y}dy=\int\frac{x}{x  ^-9}dx

    \int\frac{1}{y}dy=\frac{1}{2} \int \frac{1}{x+3} +\frac{1}{2} \int \frac{1}{x-3}

    lny^2=ln|(x+3)(x-3)k|

    y^2=kx^2-9k^2


    Which is not what I'm trying to achieve.
    As |x|<3
    ln\left (|(x+3)(x-3)|\right )+lnk=ln\left (|x+3||x-3|\right )+lnk=ln(3+x)(3-x)k
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    (Original post by ztibor)
    As |x|<3
    ln\left (|(x+3)(x-3)|\right )+lnk=ln\left (|x+3||x-3|\right )+lnk=ln(3+x)(3-x)k
    I don't understand, why doesn't my method work, without messing around with mod signs?
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    Well, I could tell you the answer but like I said earlier, I don't know how to write the maths on here. All I can do is x^2 etc
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    (Original post by r_mar)
    Well, I could tell you the answer but like I said earlier, I don't know how to write the maths on here. All I can do is x^2 etc
    http://www.thestudentroom.co.uk/showthread.php?t=403989

    open the spoiler for brief introduction to LaTeX
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    (Original post by ViralRiver)
    Okay question:

    Given that k is an arbitrary positive constant, show that y^2+kx^2=9k is a general solution of the differential equation \frac{dy}{dx}=\frac{-xy}{9-x^2}, |x|\leq 3.

    I have been able to do this by making a substitution, but trying to do it using partial fractions doesn't work. Can someone tell me what I've done wrong?

    \int\frac{1}{y}dy=\int\frac{x}{x  ^-9}dx
    how did you get to that first step? surely

    \int\frac{1}{y}dy=-\int\frac{x}{9-x^2}dx
    \int\frac{1}{y}dy=\int\frac{x}{x  ^2-9}dx

    although you have worked on from that, so maybe a typo?
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    (Original post by ViralRiver)
    lny^2=ln|(x+3)(x-3)k|

    y^2=kx^2-9k^2


    Which is not what I'm trying to achieve.

    Last step really, why have you got k^2

    and you have to use the modulus part of the natural log to make it agree with the answer.
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    Cheers!

     \frac{1}{y}\frac{dy}{dx}=\frac{-x}{9-x^2}



/int\frac{1}{y}dy=\frac{1}{2}/int\frac{-2}{9-x^2}dx



ln|y|=\frac{1}{2}ln|9-x^2|+c



lny^2=ln|9-x^2|+c



y^2=9-x^2+e^c



y^2=k(9-x^2)



y^2+kx^2=9k

    Gave it a go haha.
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    Couple typos, should be k at the end, and x^2 at the start, sorry.

    (Original post by r_mar)
    Cheers!

     \frac{1}{y}\frac{dy}{dx}=\frac{-x}{9-x^2}



/int\frac{1}{y}dy=\frac{1}{2}/int\frac{-2}{9-x^2}dx



ln|y|=\frac{1}{2}ln|9-x^2|+c



lny^2=ln|9-x^2|+c



y^2=9-x^2+e^c



y^2=k(9-x^2)



y^2+kx^2=9k

    Gave it a go haha.
    r_mar, sorry but I don't understand what you're trying to do =\ .

    (Original post by 2^1/2)
    Last step really, why have you got k^2

    and you have to use the modulus part of the natural log to make it agree with the answer.
    Hmm, can you explain the modulus thing please? I've never really understood it properly, and I really need to now apparently =\ .
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    (Original post by ViralRiver)
    Couple typos, should be k at the end, and x^2 at the start, sorry.



    r_mar, sorry but I don't understand what you're trying to do =\ .



    Hmm, can you explain the modulus thing please? I've never really understood it properly, and I really need to now apparently =\ .
    when you integrate 1/x you get ln(x),

    though you write it like that but you really mean ln|x| (where |x| means only the positive value of x) as ln(negative number) doesn't exist.

    so modulus, for example |-3| = 3
    or |x-1|=|1-x| <- try with some examples (eg. |2-1|=1=|1-2|=|-1|)

    and in this case you have |(x+3)(x-3)|=|x^2-9|=|9-x^2|

    make sense?
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    Your 1st answer is mostly right. The k^2 term should just be k. Remember k is an arbitrary constant. That means you can just replace it with -k, or in fact whatever you like. Replacing with -k sets you straight
 
 
 
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