1.solve the differential equation
dy/dx = 8y^3 sin^2 x
given that y = 1/2 at x = pi/4
2.1. Use the substitution u = x^3 + 1 to evaluate
x^2/(x^3 + 1)^.5 dx between x =2 and x= 0
2.2 Using integration by parts, or otherwise, find
{ x(3x +1)^-.5 dx
Hence evaluate { x(3x +1)^-.5 dx between x=5 and x = 1.
Thanks.
x
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jonnymcc2003
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- 06-06-2004 19:03
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- 06-06-2004 19:10
(Original post by jonnymcc2003)
1.solve the differential equation
dy/dx = 8y^3 sin^2 x
given that y = 1/2 at x = pi/4
dy/8y^3 = sin^2 x dx
Integrate the RHS using cos 2x = 1 - 2sin^2 x and then put your x and y values in to work out the constant. -
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- 06-06-2004 19:59
hey johnny, do u have the answers?? thanx
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- 06-06-2004 20:04
(Original post by jonnymcc2003)
2.1. Use the substitution u = x^3 + 1 to evaluate
x^2/(x^3 + 1)^.5 dx between x =2 and x= 0
du/dx = 3x^2
so dx = du/3x^2
x=2, u=8
x=0, u=0
so it now becomes, INT [x^2/(u^3 +1)^0.5](du/3x^2)
the x^2 cancels, so you just get 1/3[INT (u+1)^-0.5 du] , which can be done using p1 methods... -
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- 06-06-2004 20:09
(Original post by mockel)
u=x^3
du/dx = 3x^2
so dx = du/3x^2
x=2, u=8
x=0, u=0
so it now becomes, INT [x^2/(u^3 +1)^0.5](du/3x^2)
the x^2 cancels, so you just get 1/3[INT (u+1)^-0.5 du] , which can be done using p1 methods...
u = x^3 + 1
du/dx = 3x^2
du = 3x^2 dx
INT [1/3u^5]du -
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- 06-06-2004 20:14
(Original post by wonkey)
hmmm.....i've done it differently :
u = x^3 + 1
du/dx = 3x^2
du = 3x^2 dx
INT [1/3u^5]du
my way would still be correct, although for this one, the way you did it is the only way, since it says "use the substitution u= x^3 + 1" -
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- 06-06-2004 20:17
(Original post by mockel)
oops, misread the substitution....i thought it said u=x^3
my way would still be correct, although for this one, the way you did it is the only way, since it says "use the substitution u= x^3 + 1"
INT ]1/3u^5]du = INT [3u^-5] du.....so i end up with [3u^-4/-4]....is this the point where i substitute 'x^3 +1' back where 'u' appears??
or do i just change the 'x' limits into 'u'? thanx -
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- 06-06-2004 20:20
(Original post by wonkey)
cool........so i've done what i've donw up to the point where i end up with :
INT ]1/3u^5]du = INT [3u^-5] du.....so i end up with [3u^-4/-4]....is this the point where i substitute 'x^3 +1' back where 'u' appears??
or do i just change the 'x' limits into 'u'? thanx
if its not a definitee integral simply subsitute and write the expression -
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- 06-06-2004 20:24
johnny, i really need some answers pls, its driving me up the wall! lol
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jonnymcc2003
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- 06-06-2004 21:14
Can anyone do 2.2 ?
The answers for 1. is 2 tan(x/2) - x + c, and 2.1 is 4/3. -
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- 06-06-2004 22:37
i think...
INT x(3x+1)^-5 dx
u = 3x+1
du = 3dx
dx = du/3
x = (u-1)/3
therefore: INT 1/9.(u-1).u^-5 du
which gives: 1/9 INT u^-4-u^-5 du with limits which are u = 16 and u = 4
which gives 1/9[(-1/3u^-3) - (-1/4u^-4)] = 1/9[(-1/3u^-3) + (1/4u^-4)]
put that into ur calculator you 4.615... x 10^-4 -
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- 06-06-2004 22:40
(Original post by jonnymcc2003)
Can anyone do 2.2 ?
The answers for 1. is 2 tan(x/2) - x + c, and 2.1 is 4/3.
can u show me how to do 1 pls. thx -
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- 07-06-2004 07:12
(Original post by wonkey)
can u show me how to do 1 pls. thx
Not sure...
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