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    I'm stuck on part iii

    for ii) I got the partial fractions as 1/(x-1) - 1/(x+1)

    for part iii)
    the last line of the question is:
    "given that y=1 when x=3. Express y as a function of x" (photo cut it off)

    these are my workings so far:

    I integrated
    1/y dy = -[(x^2+1)/(x^2-1)] dx

    and got:
    ln y= integrate -1 + 2/x^-1 dx

    *2/x^-1 is from the part ii*

    ln y = integrate -1 + 1/(x-1) - 1/(x+1)

    ln y = -x + ln (1-x) - ln (x+1)
    I e'd both sides
    y= e^(-x) + [(1-x)/(x+1)]

    but the answer is supposed to be y= (x+1)/(2(x-1)) * e^(3-x)

    I have my fractions the wrong way up and I'm not sure how the 2 got there?
    and where the 3 came from?
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    (Original post by 0utdoorz)
    Name:  IMG_4511[1].jpg
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    I'm stuck on part iii

    for ii) I got the partial fractions as 1/(x-1) - 1/(x+1)

    for part iii)

    these are my workings so far:

    I integrated
    1/y dy = -[(x^2+1)/(x^2-1)] dx

    and got:
    ln y= integrate -1 + 2/x^-1 dx

    *2/x^-1 is from the part ii*

    ln y = integrate -1 + 1/(x-1) - 1/(x+1)

    ln y = -x + ln (1-x) - ln (x+1)
    I e'd both sides
    y= e^(-x) + [(1-x)/(x+1)]

    but the answer is supposed to be y= (x+1)/(2(x-1)) * e^(3-x)

    I have my fractions the wrong way up and I'm not sure how the 2 got there?
    and where the 3 came from?
    Where's your value for +C? Did they give any values of x or y to sub in to find this C?
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    you have forgotten that, when you expanded the RHS to partial fractions, it was multipled by -1 (you haven`t done that)

    (I,e the signs of your partial fractions for part 3 are not correct)
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    (Original post by 0utdoorz)
    Name:  IMG_4511[1].jpg
Views: 87
Size:  488.1 KB

    I'm stuck on part iii

    for ii) I got the partial fractions as 1/(x-1) - 1/(x+1)

    for part iii)

    these are my workings so far:

    I integrated
    1/y dy = -[(x^2+1)/(x^2-1)] dx

    and got:
    ln y= integrate -1 + 2/x^-1 dx

    *2/x^-1 is from the part ii*

    ln y = integrate -1 + 1/(x-1) - 1/(x+1)

    ln y = -x + ln (1-x) - ln (x+1)
    I e'd both sides
    y= e^(-x) + [(1-x)/(x+1)]

    but the answer is supposed to be y= (x+1)/(2(x-1)) * e^(3-x)

    I have my fractions the wrong way up and I'm not sure how the 2 got there?
    and where the 3 came from?
    How confident are you that the answer is correct, because I'm not getting it!
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    (Original post by metaltron)
    How confident are you that the answer is correct, because I'm not getting it!
    pretty sure, the back of the book says it is?

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    (Original post by 0utdoorz)
    pretty sure, the back of the book says it is?

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    How far have you got now?
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    still stuck

    I tried substituting in the values y=1 and x=3 into ln y=-x-ln(1-x)+ln(x+1) +c

    I got
    0 = -3-ln (2)+ln(4)+c
    ln (2/4) = -3 +c
    1/2 = -3+c
    7/2 = c

    but I don't think that's right?
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    (Original post by 0utdoorz)
    still stuck

    I tried substituting in the values y=1 and x=3 into ln y=-x-ln(1-x)+ln(x+1) +c

    I got
    0 = -3-ln (2)+ln(4)+c
    ln (2/4) = -3 +c
    1/2 = -3+c
    7/2 = c

    but I don't think that's right?
    Did you get told it goes through the points (1,3) somewhere. Maybe I'm being blind!
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    (Original post by metaltron)
    Did you get told it goes through the points (1,3) somewhere. Maybe I'm being blind!
    yeah! sorry!! I'm such a fail, I just realized my photo cut off the last line!

    "given that y=1 when x=3. Express y as a function of x"
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    (Original post by 0utdoorz)
    still stuck

    I tried substituting in the values y=1 and x=3 into ln y=-x-ln(1-x)+ln(x+1) +c

    I got
    0 = -3-ln (2)+ln(4)+c
    ln (2/4) = -3 +c
    1/2 = -3+c
    7/2 = c

    but I don't think that's right?
    How about you try getting it in terms of y before subbing in for the c value. So start with:

     lny = ln(x+1) - ln(x-1) - x + c

    and you should get to (don't look until you've done it yourself please):

    Spoiler:
    Show


     y = \frac{A(x+1)}{e^x(x-1)}



    Then find a and you're home.
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    (Original post by 0utdoorz)
    yeah! sorry!! I'm such a fail, I just realized my photo cut off the last line!

    "given that y=1 when x=3. Express y as a function of x"
    OK don't worry, see my post above. Finally, my answer agrees with your book. Phew!
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    (Original post by metaltron)
    How about you try getting it in terms of y before subbing in for the c value. So start with:

     lny = ln(x+1) - ln(x-1) - x + c

    and you should get to (don't look until you've done it yourself please):

    Spoiler:
    Show


     y = \frac{A(x+1)}{e^x(x-1)}



    Then find a and you're home.
    Found the answer!! thanks!
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    (Original post by 0utdoorz)
    I got

    ln y= (ln x+1/ln x-1) -x +c

    e both sides: y=(x+1/x-1) *e^-x *A

    e^-x = 1/e^x

    so
    y= (x+1)A/e^x(x-1)

    I substituted x=3 and y=1

    1= 4A/2e^3
    2e^3 = 4A

    A= e^3/2 ???

    :eek:
    This is right Just sub back into the y equation.
 
 
 
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