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differential equation and initial value

 
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Sep23-09, 10:48 PM   #1
 

differential equation and initial value


1. The problem statement, all variables and given/known data

[img=http://img3.imageshack.us/img3/3417/questionec.th.jpg]


2. Relevant equations



3. The attempt at a solution

I tried dividing both sides by (x^5+1), however the integration becomes really complex... can someone give me suggestions on how to do this?
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Sep23-09, 10:54 PM   #2
 
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Quote by -EquinoX- View Post

I tried dividing both sides by (x^5+1), however the integration becomes really complex... can someone give me suggestions on how to do this?
put the -10x4y on the left side then divide by tr x5+1. Integrating factor it.
Sep24-09, 12:04 AM   #3
 
that's what I did and then I need to integrate -10x^4/(x^5+1) dx right and then then to get the integrating factor is just e to the power of whatever the result of the integration is... however the integration is quite hard...
Sep24-09, 03:06 PM   #4
 
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differential equation and initial value


Quote by -EquinoX- View Post
that's what I did and then I need to integrate -10x^4/(x^5+1) dx right and then then to get the integrating factor is just e to the power of whatever the result of the integration is... however the integration is quite hard...
alright now, so


[tex]\int \frac{-10x^4}{x^5+1} dx[/tex]

see how d/dx(x5+1)=5x4 ?

Can you use a substitution to make this integral easier?
Sep24-09, 03:57 PM   #5
 
Quote by rock.freak667 View Post
alright now, so


[tex]\int \frac{-10x^4}{x^5+1} dx[/tex]

see how d/dx(x5+1)=5x4 ?

Can you use a substitution to make this integral easier?
Okay say I solve the integral and then the integrating factor would be e to the power of this resulting integral right? so then what do I need to do next in order to solve for this problem?
Sep24-09, 05:00 PM   #6
 
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Quote by -EquinoX- View Post
Okay say I solve the integral and then the integrating factor would be e to the power of this resulting integral right? so then what do I need to do next in order to solve for this problem?

right so for y'+P(x)y=Q(x), when you multiply by an integrating factor 'u', the left side becomes d/dx(uy) . That's why we multiply by u in the first place.


So you'll need to basically integrate uQ(x) w.r.t. x
Sep24-09, 06:23 PM   #7
 
and then divide that by the integrating factor right?
Sep24-09, 06:25 PM   #8
 
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Quote by -EquinoX- View Post
and then divide that by the integrating factor right?
yes you can if you feel the need to.
Sep24-09, 07:17 PM   #9
 
what do you mean by if I need to? doesn't it always works like that?
Sep24-09, 11:47 PM   #10
 
this now comes to:

[tex]\int \frac{x^2+5x-4}{x^5+1} e^{-2ln(x^5+1)} dx[/tex]

I guess this can be simplified to:

[tex]\int \frac{x^2+5x-4}{(x^5+1)^3} dx[/tex]

is this true?

how can I solve this such complex integration?
Sep25-09, 05:18 PM   #11
 
anyone please?
Sep25-09, 09:24 PM   #12
 
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Quote by -EquinoX- View Post
anyone please?
Re do your integrating factor as it was to integrate 10x4/(x5+1) not with the -ve sign.
Sep25-09, 11:34 PM   #13
 
Integrating [tex]\frac{10x^4}{(x^5+1)}[/tex] the result I got is [tex]2ln(x^5+1)[/tex] and so the integrating factor is [tex]e^{2ln(x^5+1)}[/tex] which simplifies to [tex](x^5+1)^2[/tex].

Then I do [tex] \int (x^2+5x-4)(x^5+1) [/tex] and the result of this integration I divide by [tex](x^5+1)^2[/tex] which is the integrating factor. Is this the correct step to find the solution?

Please correct me if I am wrong.
Sep25-09, 11:43 PM   #14
 
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Quote by -EquinoX- View Post
Integrating [tex]\frac{10x^4}{(x^5+1)}[/tex] the result I got is [tex]2ln(x^5+1)[/tex] and so the integrating factor is [tex]e^{2ln(x^5+1)}[/tex] which simplifies to [tex](x^5+1)^2[/tex].

Then I do [tex] \int (x^2+5x-4)(x^5+1) [/tex] and the result of this integration I divide by [tex](x^5+5)^2[/tex] which is the integrating factor. Is this the correct step to find the solution?

Please correct me if I am wrong.

yes just integrate and divide, also don't forget the constant of integration
Sep26-09, 12:28 AM   #15
 
Quote by rock.freak667 View Post
yes just integrate and divide, also don't forget the constant of integration
by the constant of integration you mean the C right? The terms after integration that I found is very long.....
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