Differential equation and initial value

-EquinoX-
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Homework Statement



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


Homework Equations





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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-EquinoX- said:
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.
 
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...
 
-EquinoX- said:
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


\int \frac{-10x^4}{x^5+1} dx

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

Can you use a substitution to make this integral easier?
 
rock.freak667 said:
alright now, so


\int \frac{-10x^4}{x^5+1} dx

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?
 
-EquinoX- said:
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
 
and then divide that by the integrating factor right?
 
-EquinoX- said:
and then divide that by the integrating factor right?

yes you can if you feel the need to.
 
what do you mean by if I need to? doesn't it always works like that?
 
  • #10
this now comes to:

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

I guess this can be simplified to:

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

is this true?

how can I solve this such complex integration?
 
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  • #11
anyone please?
 
  • #12
-EquinoX- said:
anyone please?

Re do your integrating factor as it was to integrate 10x4/(x5+1) not with the -ve sign.
 
  • #13
Integrating \frac{10x^4}{(x^5+1)} the result I got is 2ln(x^5+1) and so the integrating factor is e^{2ln(x^5+1)} which simplifies to (x^5+1)^2.

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

Please correct me if I am wrong.
 
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  • #14
-EquinoX- said:
Integrating \frac{10x^4}{(x^5+1)} the result I got is 2ln(x^5+1) and so the integrating factor is e^{2ln(x^5+1)} which simplifies to (x^5+1)^2.

Then I do \int (x^2+5x-4)(x^5+1) and the result of this integration I divide by (x^5+5)^2 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
 
  • #15
rock.freak667 said:
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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