
#1
Jan1308, 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 



#3
Jan1308, 05:28 AM

P: 128

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




#4
Jan1308, 05:32 AM

HW Helper
P: 858

[SOLVED] Integration of Inverse of f(x)
firstly you can further factorise y^26y+5




#5
Jan1308, 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}(y1)(y5)}[/tex] So, by Partial fraction, is this correct? [tex]\frac{1}{y^{2}(y1)(y5)}[/tex] =[tex]\frac{A}{y}[/tex]+[tex]\frac{B}{y^{2}}[/tex]+[tex]\frac{C}{y1}[/tex]+[tex]\frac{D}{y5}[/tex] and then, by comparing coefficient of y^3, y^2, y. (A+C+D){y^{3}}=0 (6A+B5CD){y^{2}}=0 (5A6B)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 



#6
Jan1308, 05:34 AM

P: 128





#7
Jan1308, 05:35 AM

HW Helper
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 



#8
Jan1308, 05:42 AM

P: 128

after i substitute A,B,C,and D.
i can get back my original 1/f(x) 



#10
Jan1308, 05:44 AM

P: 128





#12
Jan1308, 05:48 AM

P: 128

[tex]\frac{1}{y^{2}(y1)(y5)}[/tex]=[tex]\frac{6}{25y}[/tex]+[tex]\frac{1}{5y^{2}}[/tex]+[tex]\frac{3}{10(y1)}[/tex][tex]\frac{87}{50(y5}[/tex] 



#13
Jan1308, 05:50 AM

P: 128





#14
Jan1308, 05:59 AM

P: 128





#15
Jan1308, 06:08 AM

P: 128

[tex]\int\frac{1}{y^{4}6y^{3}+5y^{2}}[/tex] dy =
6/25(lny)1/4(lny1)+1/100(lny5)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(lny)1/4(lny1)+1/100(lny5)1/(5y) then i m stuck to make y the subject. anyone to point out the mistake.thanks 



#16
Jan1308, 06:14 AM

P: 128

anyone, can show me the right way?




#17
Jan1308, 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. 



#18
Jan1308, 11:36 AM

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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