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Math Help - Anti-Differentiation and Differential Equations

  1. #1
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    Anti-Differentiation and Differential Equations

    Okay Update,

    1.
    Find the area enclosed by the curves:
    y = 4 + x and x + 2y = 4

    I worked it out myself, so dont worry




    2.
    Fine the volume of the solid of revolution generated by rotating about the y-axis the region enclosed by the y-axis and the curves y = 2x^2 - 1 and y =
    √(x)

    V = integral 1 to 0 of ( pi (√(x) - 2x^2 - 1)^2 )

    So i expanded (√(x) - 2x^2 - 1)^2

    to get (simplified): 4x^4 - 4x^2.5 - 4x^2 + 2√(x) + x + 1

    then got the integral is:

    pi((4/5)x^5 - (4/3.5)x^3.5 - (4/3)x^3 + (2/1.5)x^1.5 + (x^2)/2 + x) between 0.42 and 0

    so subbing in those values: pi((81/70) - 0)

    = 3.64 units^3

    Correct?



    3.
    Use the method of cylindrical shells to find the volume of the solid of revolution generated by rotating about the y-axis the region enclosed by y = √(x+1), y = 0, x=0 and x=4

    V = integral_{4 to 0} of [ 2pi x ( √(x)+1 ) ] dx

    = (2pi)integral_{4 to 0} of (x^1.5 + x) dx

    = (2pi)[(x^2.5)/2.5 + (x^2)/2] between 4 and 0

    = (2pi)(12.8 + 8)

    = 41.6 pi units^3

    Correct?


    4.
    Solve
    dy/dx = (x(y+3)/y)

    (y/(y^2+3)) dy/dx = x

    so: integral (y/(y^2+3)) dy = integral x dx + C

    so the integral is: (1/2)ln(y^2+3) = x^2/2 + C

    then, we say ln(y^2+3) = x^2 + A
    where A = 2C

    so, y^2 = = e^(x^2 + A) - 3

    y = √(e^(x^2 + A) - 3)

    Correct?



    5.
    Solve
    dy/dx = e^(x-2y), y(1) = 0

    ok.. so im pretty much stumped after this step

    ln(dy) - ln(dx) = x - 2y

    ln(dy) + 2y = ln(dx) + x

    Help?



    6.
    dy/dx = xy-2e^(x), y(1) = -1
    Use Euler method with 4 steps to find and approx value for y(2)

    The formula for calculating succcessive values of x and y is :
    y_n+1 = y_n + 0.25[(x_n^2)(y_n)-2e^(x_n)]

    (_n+1 means subscript n+1 etc)

    y_1 = -1 + 0.25[(1^2)(-1)-2e^(1)] = -2.6091
    y_2 = -2.6091 + 0.25[(1.25^2)(-2.6091)-2e^(1.25)] = -5.3735
    y_3 = -5.3735 + 0.25[(1.50^2)(-5.3735)-2e^(1.50)] = -10.6369
    y_4 = -10.6369 + 0.25[(1.75^2)(-10.6369)-2e^(1.75)] = -21.6581


    So the value at y(2) = -21.6581

    Yes? Correct?
    Last edited by Jozsa; May 13th 2007 at 08:08 AM.
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  2. #2
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    Quote Originally Posted by Jozsa View Post
    Hey, theres a few questions i need help with urgently, would apreciate any help on as many of them as fast as possible. Thanks in advance.

    1.
    Find the area enclosed by the curves:
    y = 4 + x and x + 2y = 4
    First sketch the curves, see attachment.

    Now to find this area it is easier to integrate in the y-direction, so write
    the curves in the form x=f(y), to get:

    a) x = y^2-4
    b) x = 4 - 2y

    Then the area is:

    A = integral_{y=-4 to 2} (4 - 2y)-(y^2-4) dy

    which is elementary and you should be able to do it yourself
    (its 36, which is what I get by counting squares ).

    RonL
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  3. #3
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    Quote Originally Posted by Jozsa View Post
    4.
    Solve
    dy/dx = (x(y+3)/y
    This is of variables seperable type, so we have:

    [y/(y^2+3)] dy/dx = x

    so:

    integral [y/(y^2+3)] dy = integral x dx + C

    or:

    (1/2)ln(y^2+3) = x^2/2 + C

    and I will leave you to simplify this.

    RonL
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  4. #4
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    Thanks RonL for the help so far!

    I actually managed to work out the first one myself, turns out i just made a simple mistake the first time i tried to do it, but it was good to see that your answer backed up mine.


    as for question 4, im having some trouble simplifying.

    (1/2)ln(y^2+3) = x^2/2 + C


    I think i may have it

    Firstly, we say ln(y^2+3) = x^2 + A
    Where A = 2C

    Second, y^2 = = e^(x^2 + A) - 3

    Finally, y = √(e^(x^2 + A) - 3)

    Is this correct?

    Would appreciate some help on the other questions too!

    For Question 2, i got 0.35 units^3
    Feedback?
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  5. #5
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    Quote Originally Posted by CaptainBlack View Post
    integral [y/(y^2+3)] dy = integral x dx + C

    or:

    (1/2)ln(y^2+3) = x^2/2 + C
    I may be wrong... but isnt (1/2)ln(y^2+3) the integral of 1/(y^2+3)?
    The question stated that it is y/(y^2+3) not 1/(y^2+3)

    I could be, and probably am wrong but just making sure.
    Last edited by Jozsa; May 13th 2007 at 06:40 AM.
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  6. #6
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    Quote Originally Posted by Jozsa View Post
    I may be wrong... but isnt (1/2)ln(y^2+3) the integral of 1/(y^2+3)?
    The question stated that it is y/(y^2+3) not 1/(y^2+3)

    I could be, and probably am wrong but just making sure.
    d/dy[1/(y^2+3)] = -2y/(y^2+3)^2
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  7. #7
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    could you help me out with 5 and 6, im totally stuck.

    and let me know if ive gone about the other questions correctly.

    thanks!
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  8. #8
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    I think i worked out Q6, still stuck on 5 though, REALLY need help.
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  9. #9
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    Quote Originally Posted by Jozsa View Post
    5.
    Solve
    dy/dx = e^(x-2y), y(1) = 0
    dy/dx = e^x e^{-2y)

    so:

    e^{2y}dy/dx = e^x

    hence:

    integral e^{2y} dy = integal e^x dx + C

    (1/2) e^{2y} = e^x + C

    RonL
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  10. #10
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    Quote Originally Posted by CaptainBlack View Post
    dy/dx = e^x e^{-2y)

    so:

    e^{2y}dy/dx = e^x

    hence:

    integral e^{2y} dy = integal e^x dx + C

    (1/2) e^{2y} = e^x + C

    RonL

    Youre my hero, thank you so much friend.
    Sorry for suddenly joining and being a pest
    But i really honestly do appreciate it.
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  11. #11
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    Also, could you verify that my solutions to the other problems are correct?

    Thank you.
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