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    Solve the differential equation
    dy/dx = 3y^2/(2x+1)^3
    given that y=2 when x=1
    give your answer in the formofy=f(x)

    Help. I seem to get the wrong answer each time.

    int 1/3y^2 dy+ int 1/(2x+1)^3
    =ln l3y^2l = ln l(2x+1)^3l + ln lkl
    =ln l(3y^2)/(2x+1)^3l = ln lkl
    =ln l12/27l= ln lkl
    thefore
    ln l3y^2l = ln l(2x+1)^3l + ln l12/27l
    3y^2= (12(2x+1)^3)/27
    y^2= 4(2x+1)^3/27

    help!!!
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    dy/dx = 3y^2/(2x+1)^3
    int y-2 dy = int 3(2x+1)-3 dx
    -y-1 = -3/8 (2x+1)-2 + C
    ....


    Edit: int f'(x)/f(x) dx = ln f(x) + C
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    you cant intergrate (u)^-3 by whacking it in an ln||
    you need to intergrate it properly
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    I get

    y = [ 4(2x+3)2] / [ 3 + (19/50)(2x+3)2]

    Looks horrible, is it correct? If not, whats the answer?
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    seems easiest to start by saying

    dy/3y^2 = dx/(2x+1)^3

    1/3 y^-2 dy = (2x+1)^-3 dx
    then integrate
    -1/3y = (2x+1)^-2 / -2 x 2 + C
    then sub in values of x and y to find C
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    (Original post by BeitHaller)
    seems easiest to start by saying

    dy/3y^2 = dx/(2x+1)^3

    3y^-2 dy = (2x+1)^-3 dx
    then integrate
    -3/y = (2x+1)^-2 / -2 x 2 + C
    then sub in values of x and y to find C
    it's 1/3 . y^-2 not 3y^-2
    and it doesnt matter where u put the 3...
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    yup, obv 1/3. easier to do maths when using paper than on a screen lol. and i know perfectly well it doesn't matter where it goes.
 
 
 

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