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    OK ive done something so wrong here bit i cant see it:

    Find a particular solution:
    (1 - x^2)dy/dx = xy + y (when x=0.5, y=6)

    So i did:
    dy/dx = (1 + x)y/(1 - x^2)
    INT 1/y dy = INT (1+x)/(1+x)(1-x) dx
    INT 1/y dy = INT 1/(1-x) dx
    ln|y| = ln|1-x| + ln|k|
    ln|y| = ln|k(1-x)|
    y = k(1 - x)

    6 = 0.5k
    k = 12

    y = 12(1-x)

    What did i do wrong? (coz that is not the right answer by a long way)
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    (Original post by joker_900)
    OK ive done something so wrong here bit i cant see it:

    Find a particular solution:
    (1 - x^2)dy/dx = xy + y (when x=0.5, y=6)

    So i did:
    dy/dx = (1 + x)y/(1 - x^2)
    INT 1/y dy = INT (1+x)/(1+x)(1-x) dx
    INT 1/y dy = INT 1/(1-x) dx
    ln|y| = ln|1-x| + ln|k|
    ln|y| = ln|k(1-x)|
    y = k(1 - x)

    6 = 0.5k
    k = 12

    y = 12(1-x)

    What did i do wrong? (coz that is not the right answer by a long way)
    the same mistakes someone else had done in aonther thread:
    int 1/(1-x) dx = - ln(1-x) not ln(1-x)
    as d(1-x)/dx = -1 not 1
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    that would make the answer
    y = k/(1-x)
    find k... and u should get that answer u have..
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    ahhhhhhh thx a lot at least it wasnt some fundamental flaw

    cheers
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    (Original post by joker_900)
    ahhhhhhh thx a lot at least it wasnt some fundamental flaw

    cheers
    well you should always (in such cases) chech for the SIGN of x too!
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    (Original post by joker_900)
    OK ive done something so wrong here bit i cant see it:

    Find a particular solution:
    (1 - x^2)dy/dx = xy + y (when x=0.5, y=6)

    So i did:
    dy/dx = (1 + x)y/(1 - x^2)
    INT 1/y dy = INT (1+x)/(1+x)(1-x) dx
    INT 1/y dy = INT 1/(1-x) dx
    ln|y| = ln|1-x| + ln|k|
    ln|y| = ln|k(1-x)|
    y = k(1 - x)

    6 = 0.5k
    k = 12

    y = 12(1-x)

    What did i do wrong? (coz that is not the right answer by a long way)

    Why did you do + ln(k) ??

    isnt it just + k ?
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    (Original post by __WiZaRD__)
    Why did you do + ln(k) ??

    isnt it just + k ?
    well lnk is a constant too.. but there is a ln in the integral... it's easier to make it lnk so as to put the k in the other ln.

    ln|y| = ln|1-x| + ln|k|
    ln|y| = ln|k(1-x)|

    if it was
    ln|y| = ln|1-x| + k
    you will have to find k in the ln form. so putting the constant as lnk is easier to deal with...
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    but surely it would give you a completely different answer if youdo this? how can you just put it in ln form?
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    k is a constant that can be anything, so putting it as lnk doesnt make any difference, it just changes the value of k. it makes life much easier since your dealing with logs.

    if you left it as
    ln|y| = ln|1-x| + k:

    'e' both sides
    y= (1-x)*e^k
    y= e^k - xe^k
    = constant - (constant)x

    compare this with
    ln|y| = ln|1-x| + lnk = ln|k(1-x)|
    y= k(1-x) = (constant) - (constant)x which is the same thing and arrived at much quicker.
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    (Original post by __WiZaRD__)
    but surely it would give you a completely different answer if youdo this? how can you just put it in ln form?
    no it would give u the same answer!
    try it!

    ln|y| = -ln|1-x| + ln|k|
    y = k/(1-x)
    6 = k/(0.5)
    k = 3
    y = 3/(1-x)


    and

    ln|y| = -ln|1-x| + C
    ln6 = ln(1/0.5) + C
    ln6 = ln2 + C
    ln6 - ln2 = C
    C = ln3

    lny = ln(1/1-x) + ln3
    lny = ln(3/(1-x))
    y = 3/(1-x)


    same answer... but more work
 
 
 
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