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November 14th, 2010, 09:59 AM   #1
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Differential Equations

I need assistance with part 1, please. For part 2, I just need it to be checked.









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[color=#0080FF][2][/color]
[color=#0080FF][1][/color]

[color=#0080FF][1] + [2][/color]



[color=#0080FF][1][/color]




















































































when 0 = A



when t = 0





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November 14th, 2010, 12:30 PM   #2
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Re: Differential Equations

No help?

Is it too long?
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November 14th, 2010, 01:31 PM   #3
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Re: Differential Equations

Today is my sabbath, so I don't help anyone except myself!
But seriously... The first one in the second part is correct. That's all I had time to look at.
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November 14th, 2010, 03:46 PM   #4
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Re: Differential Equations

Quote:
Originally Posted by The Chaz
Today is my sabbath, so I don't help anyone except myself!
But seriously... The first one in the second part is correct. That's all I had tine to look at.
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November 14th, 2010, 06:11 PM   #5
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(1) dy/dx + y/x = y

(i) Multiplying by the integrating factor -2/(xy) gives -2(x(dy/dx) + y)/(xy) = -2/x,
which can be integrated to give 1/(xy) = 2/x + C, etc. What was that about not being able to use an integrating factor?

(ii) Differentiating 1/y = -2z w.r.t. x gives -(2/y)(dy/dx) = -2dz/dx,
so -(2/y)(y - y/x) = -2dz/dx, i.e., -2(1 - 1/(xy)) = -2dz/dx, so dz/dx - 2z/x = 1.

(iii) Let 1/y = -2z, so that dz/dx - 2z/x = 1 and z = -1/8 when x = 1.
Dividing by x and integrating gives z/x = -1/x + 7/8,
so -1/(xy) = -2/x + 7/4. In hindsight (to avoid division by zero), I'll assume x > 0.
By rearranging, one gets y = 2/?(8x - 7x), so I'll also assume x < 8/7.
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November 14th, 2010, 06:15 PM   #6
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Re: Differential Equations

Quote:
Originally Posted by MathematicallyObtuse
I need assistance with part 1, please. For part 2, I just need it to be checked.





ii)



, do some substitutions.....



iii)

So, solve at first.







Conditions, ,

so, ,

solution is or
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November 14th, 2010, 06:31 PM   #7
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Re:

Quote:
Originally Posted by skipjack
(1) dy/dx + y/x = y


(iii) Let 1/y = -2z, so that dz/dx - 2z/x = 1 and z = -1/8 when x = 1.
Dividing by x and integrating gives z/x = -1/x + 7/8,
so -1/(xy) = -2/x + 7/4. In hindsight (to avoid division by zero), I'll assume x > 0.
By rearranging, one gets y = 2/?(8x - 7x), so I'll also assume x < 8/7.
If z/x = x + C,

dz/dx-2z/x=x not the const "1"
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November 14th, 2010, 07:21 PM   #8
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Re: Differential Equations

Quote:
Originally Posted by MathematicallyObtuse




General solution,



Conditions,

yields

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November 14th, 2010, 07:23 PM   #9
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Oops, I integrated 1 instead of 1/x. I've amended my working above to change the x to -1/x and recalculated the constant.
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November 14th, 2010, 07:30 PM   #10
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Re:

Quote:
Originally Posted by skipjack
Oops, I integrated 1 instead of 1/x. I've amended my working above to change the x to -1/x and recalculated the constant.
It's so glad our results agree
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