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C4 intergrationby parts help!! watch

1. Solve the differential equation
dy/dx = 3y^2/(2x+1)^3
given that y=2 when x=1

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!!!
2. 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
3. you cant intergrate (u)^-3 by whacking it in an ln||
you need to intergrate it properly
4. I get

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

Looks horrible, is it correct? If not, whats the answer?
5. 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
6. (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...
7. 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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Updated: June 13, 2006
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