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Integration of Inverse of f(x)

by JayKo
Tags: integration, inverse, solved
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JayKo
#1
Jan13-08, 05:11 AM
P: 128
[tex]\int\frac{1}{y^{4}-6y^{3}+5y^{2}}[/tex] dy =

[tex]\frac{1}{4y^{3}-18y^{2}+10y}[/tex](ln|[tex]{y^{4}-6y^{3}+5y^{2}}[/tex]|)


is this correct way or doing it? thanks. i was a bit of confuse, if i should do it the above or partial fraction the denominator it? thanks
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mjsd
#2
Jan13-08, 05:26 AM
HW Helper
mjsd's Avatar
P: 858
yes, factorise and then partial fractions is ok
JayKo
#3
Jan13-08, 05:28 AM
P: 128
wait a minutes,i show my partial fraction. please point out where gone wrong ;) thanks

mjsd
#4
Jan13-08, 05:32 AM
HW Helper
mjsd's Avatar
P: 858
Integration of Inverse of f(x)

firstly you can further factorise y^2-6y+5
JayKo
#5
Jan13-08, 05:33 AM
P: 128
[tex]\frac{1}{y^{4}-6y^{3}+5y^{2}}[/tex]

=[tex]\frac{1}{y^{2}(y^{2}-6y+5)}[/tex]

=[tex]\frac{1}{y^{2}(y-1)(y-5)}[/tex]

So, by Partial fraction, is this correct?

[tex]\frac{1}{y^{2}(y-1)(y-5)}[/tex]
=[tex]\frac{A}{y}[/tex]+[tex]\frac{B}{y^{2}}[/tex]+[tex]\frac{C}{y-1}[/tex]+[tex]\frac{D}{y-5}[/tex]

and then, by comparing coefficient of y^3, y^2, y.

(A+C+D){y^{3}}=0
(-6A+B-5C-D){y^{2}}=0
(5A-6B)y=0
answer is A=6/25, B=1/5, C=3/10. D=-87/50.

but doubt my answer. anyone can point out?thanks
JayKo
#6
Jan13-08, 05:34 AM
P: 128
Quote Quote by mjsd View Post
firstly you can further factorise y^2-6y+5
i know, is just that i m slow using teX, please gimme a min.thanks
mjsd
#7
Jan13-08, 05:35 AM
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P: 858
I can tell you the answer....


no need use latex for such simple expression

most likely you have treated the y^2 bit incorrectly
JayKo
#8
Jan13-08, 05:42 AM
P: 128
after i substitute A,B,C,and D.

i can get back my original 1/f(x)
mjsd
#9
Jan13-08, 05:43 AM
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P: 858
looks ok so far
JayKo
#10
Jan13-08, 05:44 AM
P: 128
Quote Quote by mjsd View Post
I can tell you the answer....


no need use latex for such simple expression

most likely you have treated the y^2 bit incorrectly
lol, i uses teX as its more tidy, i go n check the y^2 bit now.thanks.
mjsd
#11
Jan13-08, 05:48 AM
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P: 858
FYI i've got
A=6/25
B=1/5
C=-1/4
D=1/100
JayKo
#12
Jan13-08, 05:48 AM
P: 128
Quote Quote by mjsd View Post
looks ok so far
if so, then
[tex]\frac{1}{y^{2}(y-1)(y-5)}[/tex]=[tex]\frac{6}{25y}[/tex]+[tex]\frac{1}{5y^{2}}[/tex]+[tex]\frac{3}{10(y-1)}[/tex]-[tex]\frac{87}{50(y-5}[/tex]
JayKo
#13
Jan13-08, 05:50 AM
P: 128
Quote Quote by mjsd View Post
FYI i've got
A=6/25
B=1/5
C=-1/4
D=1/100
OOops,gimme a minute,i recheck :)
JayKo
#14
Jan13-08, 05:59 AM
P: 128
Quote Quote by mjsd View Post
FYI i've got
A=6/25
B=1/5
C=-1/4
D=1/100
you are right! :D :D :D thanks a lot.for pointing it out !:D
JayKo
#15
Jan13-08, 06:08 AM
P: 128
[tex]\int\frac{1}{y^{4}-6y^{3}+5y^{2}}[/tex] dy =

6/25(ln|y|)-1/4(ln|y-1|)+1/100(ln|y-5|)-1/(5y). right?

but this problem is actually make y the subject in this equation, how to go about it further?
[tex]\int\frac{1}{y^{4}-6y^{3}+5y^{2}}[/tex] dy = [tex]\int\[/tex] dt


the original question is [tex]\frac{dy}{dt}[/tex]=[tex]y^{4}-6y^{3}+5y^{2}[/tex] find y(t).


if [tex]\int\frac{1}{y^{4}-6y^{3}+5y^{2}}[/tex] dy =

6/25(ln|y|)-1/4(ln|y-1|)+1/100(ln|y-5|)-1/(5y)
then i m stuck to make y the subject. anyone to point out the mistake.thanks
JayKo
#16
Jan13-08, 06:14 AM
P: 128
anyone, can show me the right way?
JayKo
#17
Jan13-08, 06:32 AM
P: 128
another differential eq. question.

y'=1+xy find y. how to go about it. i m clueless
my method of multiply dx to (1+xy) doesn't work in this case.
rohanprabhu
#18
Jan13-08, 11:36 AM
rohanprabhu's Avatar
P: 414
You can solve this by the solution for Linear Differential equation. The equation you have is:

[tex]
\frac{dy}{dx} = 1 + xy
[/tex]

which can be written as:

[tex]
\frac{dy}{dx} + y(-x) = 1
[/tex]

which is similar to:

[tex]
\frac{dy}{dx} + yP = Q
[/tex]

Here, [itex]P = (-x); Q = 1[/itex]

So, you have:

[tex]
I.F = e^{\int (-x)dx}
[/tex]
[tex]
I.F = e^{-\frac{x^2}{2}}
[/tex]

And hence, your solution is given by:

[tex]
ye^{\frac{-x^2}{2}} = \int (1)e^{\frac{-x^2}{2}}dx + C
[/tex]

But, it's gonna be a real pain with this:

[tex]
\int e^{\frac{-x^2}{2}}dx
[/tex]

a method to which i can't think of right now. Maybe there's some other way..


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