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c4 differential equations and binomial

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    hey guys, stuck on q. 8(ii)(iii) and 9(ii) on this paper

    for 8(ii), I simplified my partial fractions expression into:

    (1+x)^-1 - (2+x)^-1 + 2(2+x)^-2
    = (1+x)^-1 - 2(1+0.5x)^-1 + 4(1+0.5x)^-2
    = 1-x+x^2 - 2[(1-0.5x+0.25x^2)] + 4[(1-x+0.75x^2)]
    = 1-x+x^2 - (2-x+0.5x^2) + (4-4x+3x^2)
    = 3.5x^2 - 4x + 3.

    However the mark scheme says the answer is 1 - 1.25x + 1.25x^2, where have U gone wrong? also, how do I approach part (iii) on this q?

    and for 9(ii), i've re-arranged and integrated and have:
    -kt + c = ln (theta - 20)
    When t = 0, theta is 100 so sub them in and get c = ln80
    Sub in other data they tell u:
    -5k + ln80 = ln48
    k = -1/5 ln(3/5) therefore -(1/5 ln 3/5)t + ln80 = ln (theta - 20)

    However the mark scheme says k = -1/5 ln(5/3), any ideas where I have gone wrong? and also how would you convert this equation into the way they want it, i.e. in terms of e?

    cheers!
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    (Original post by Next Level)
    hey guys, stuck on q. 8(ii)(iii) and 9(ii) on this paper

    for 8(ii), I simplified my partial fractions expression into:

    (1+x)^-1 - (2+x)^-1 + 2(2+x)^-2
    = (1+x)^-1 - 2(1+0.5x)^-1 + 4(1+0.5x)^-2
    = 1-x+x^2 - 2[(1-0.5x+0.25x^2)] + 4[(1-x+0.75x^2)]
    = 1-x+x^2 - (2-x+0.5x^2) + (4-4x+3x^2)
    = 3.5x^2 - 4x + 3.

    However the mark scheme says the answer is 1 - 1.25x + 1.25x^2, where have U gone wrong? also, how do I approach part (iii) on this q?

    and for 9(ii), i've re-arranged and integrated and have:
    -kt + c = ln (theta - 20)
    When t = 0, theta is 100 so sub them in and get c = ln80
    Sub in other data they tell u:
    -5k + ln80 = ln48
    k = -1/5 ln(3/5) therefore -(1/5 ln 3/5)t + ln80 = ln (theta - 20)

    However the mark scheme says k = -1/5 ln(5/3), any ideas where I have gone wrong? and also how would you convert this equation into the way they want it, i.e. in terms of e?

    cheers!
    For the first part; your error sits at taking a factor of 2 out of (2+x)^-1 and (2+x)^-2

    (2+x)^{-1} = 2^{-1}(1+0.5x)^{-1}

    I'll take a look at the second part now.

    EDIT: For the second part, the mark scheme says k = (1/5)ln(5/3)

    You had k = -\frac{1}{5}\ln(\frac{3}{5}) = \frac{1}{5}\ln((\frac{3}{5})^{-1})
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    (Original post by hassi94)
    For the first part; your error sits at taking a factor of 2 out of (2+x)^-1 and (2+x)^-2

    (2+x)^{-1} = 2^{-1}(1+0.5x)^{-1}

    I'll take a look at the second part now.

    EDIT: For the second part, the mark scheme says k = (1/5)ln(5/3)

    You had k = -\frac{1}{5}\ln(\frac{3}{5}) = \frac{1}{5}\ln((\frac{3}{5})^{-1})
    cheers for the feedback! I get my mistake for 8(ii), any ideas on how to do 8(iii)?

    and oh yeah my bad, so I have:

    1/5 ln (5/3)t + ln80 = ln(theta - 20)
    how do I simplify this to: theta = 20 + 80e^-(1/5 ln 5/3)t as the q wants it? cheers!
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    (Original post by Next Level)
    cheers for the feedback! I get my mistake for 8(ii), any ideas on how to do 8(iii)?

    and oh yeah my bad, so I have:

    1/5 ln (5/3)t + ln80 = ln(theta - 20)
    how do I simplify this to: theta = 20 + 80e^-(1/5 ln 5/3)t as the q wants it? cheers!
    For 8iii), note that the expansion for (1+x)^n (where n is non-natural) is only valid for -1 \leq x \leq 1


    And for 9. Make sure not to miss out any minuses; you have -1/5 ln (5/3)t + ln80 = ln(theta - 20)

    Take the exponent of both sides. So e^{ln80 - 1/5ln(5/3)t} = \theta - 20 then just mess around with the left hand side using laws of indices and that e^{ln(x)} = x
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    ive had sex twice in the last hour, jelly bro?
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    (Original post by hassi94)
    For 8iii), note that the expansion for (1+x)^n (where n is non-natural) is only valid for -1 \leq x \leq 1


    And for 9. Make sure not to miss out any minuses; you have -1/5 ln (5/3)t + ln80 = ln(theta - 20)

    Take the exponent of both sides. So e^{ln80 - 1/5ln(5/3)t} = \theta - 20 then just mess around with the left hand side using laws of indices and that e^{ln(x)} = x
    for 8(iii) why is this? where did u get that from? I don't follow why x has to be bounded by this inequality

    and for 9 I thought the minus is 'taken out' by using it to invert the 3/5 if u get what I mean?
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    (Original post by Next Level)
    for 8(iii) why is this? where did u get that from? I don't follow why x has to be bounded by this inequality

    and for 9 I thought the minus is 'taken out' by using it to invert the 3/5 if u get what I mean?
    It's just a fact for all expansions of that kind. It will say so in your formula booklet and your text book.

    And for 9 - yes we got rid of the minus IN k; but there was a minus before k anyway:

    You got k = 1/5ln(5/3) right?

    And you also have -kt + ln80 = ln(theta-20)

    Subbing k into that equation; we have -1/5ln(5/3)t + ln80 = ln(theta-20)
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    (Original post by hassi94)
    For 8iii), note that the expansion for (1+x)^n (where n is non-natural) is only valid for -1 \leq x \leq 1


    And for 9. Make sure not to miss out any minuses; you have -1/5 ln (5/3)t + ln80 = ln(theta - 20)

    Take the exponent of both sides. So e^{ln80 - 1/5ln(5/3)t} = \theta - 20 then just mess around with the left hand side using laws of indices and that e^{ln(x)} = x
    ah sorry I forgot the minus; but I don't see how you got it from that stage to the exponent one, what do you mean take the exponent of both sides? how do you do this with all terms in the expression as ln terms?
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    (Original post by Next Level)
    ah sorry I forgot the minus; but I don't see how you got it from that stage to the exponent one, what do you mean take the exponent of both sides? how do you do this with all terms in the expression as ln terms?
    Take the exponent just means do e^ each side.
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    (Original post by hassi94)
    Take the exponent just means do e^ each side.
    so why is the last term not e^theta - 20?
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    (Original post by Next Level)
    so why is the last term not e^theta - 20?
    Because its e^ln(theta - 20) and e^lnx = x

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