[SOLVED] Integration of Inverse of f(x)

JayKo

[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

mjsd

yes, factorise and then partial fractions is ok

JayKo

wait a minutes,i show my partial fraction. please point out where gone wrong ;) thanks

mjsd

firstly you can further factorise y^2-6y+5

JayKo

[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

i know, is just that i m slow using teX, please gimme a min.thanks

mjsd

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

after i substitute A,B,C,and D.

i can get back my original 1/f(x)

mjsd

looks ok so far

JayKo

lol, i uses teX as its more tidy, i go n check the y^2 bit now.thanks.

mjsd

FYI i've got
A=6/25
B=1/5
C=-1/4
D=1/100

JayKo

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

OOops,gimme a minute,i recheck :)

JayKo

you are right! :D :D :D thanks a lot.for pointing it out !:D

JayKo

[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

anyone, can show me the right way?

JayKo

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.