Differential Eqns: particular soln to xy' + 4y = 15xcox(x^5)  TOPIC_SOLVED

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Differential Eqns: particular soln to xy' + 4y = 15xcox(x^5)

Postby doggy on Thu Apr 16, 2009 12:21 am

I'm having a bit of trouble figuring out this differential equation. The question is:

Find the particular solution of the differential equation:
xy ' +4y = 15xcos(x^5) sastisfying the initial condition y(pi)=0.
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Re: Differential Equations

Postby Martingale on Thu Apr 16, 2009 12:35 am

doggy wrote:I'm having a bit of trouble figuring out this differential equation. The question is Find the particular solution of the differential equation:
xy ' +4y = 15xcos(x^5) sastisfying the initial condition y(pi)=0.



Use an Integrating factor.
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Re: Differential Equations

Postby doggy on Thu Apr 16, 2009 2:36 am

I have tried using an integrating factor of x^4 but I'm still not getting the correct answer
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Re: Differential Equations

Postby Martingale on Thu Apr 16, 2009 2:46 am

doggy wrote:I have tried using an integrating factor of x^4 but I'm still not getting the correct answer











then

so

now take care of the initial condition
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Re: Differential Equations

Postby doggy on Thu Apr 16, 2009 4:06 am

I have gotten up to that point as well but I'm having trouble with the conditions when I sub in pi I got:
3 cos(pi)^5/(pi)^4 + c/(pi/4) =0
thus c= -3 cos((pi)^5)

So the final answer should be :
3 cos (x^5) - 3 cos(pi)^5 but that's not correct
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Re: Differential Equations

Postby Martingale on Thu Apr 16, 2009 4:11 am

doggy wrote:I have gotten up to that point as well but I'm having trouble with the conditions when I sub in pi I got:
3 cos(pi)^5/(pi)^4 + c/(pi/4) =0
thus c= -3 cos((pi)^5)

So the final answer should be :
3 cos (x^5) - 3 cos(pi)^5 but that's not correct



Why do you have ?
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Re: Differential Eqns: particular soln to xy' + 4y = 15xcox(x^5)

Postby doggy on Fri Apr 17, 2009 12:31 am

Don't I need the cos (pi)^5?
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Re: Differential Eqns: particular soln to xy' + 4y = 15xcox(x^5)

Postby Martingale on Fri Apr 17, 2009 12:34 am

doggy wrote:Don't I need the cos (pi)^5?


no.

Martingale wrote:so

now take care of the initial condition




solve for
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Re: Differential Eqns: particular soln to xy' + 4y = 15xcox(x^5)

Postby doggy on Fri Apr 17, 2009 1:45 am

So then the final answer would be c=0

and y= 3 sin(x^5)/x^4?
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Re: Differential Eqns: particular soln to xy' + 4y = 15xcox(x^5)

Postby Martingale on Fri Apr 17, 2009 1:46 am

doggy wrote:So then the final answer would be c=0

and y= 3 sin(x^5)/x^4?


no
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