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Topic: differential equations (Read 705 times) |
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Nikki
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Consider the differential equation
2x.(dy/dx)+y3 = 0 x>0
a) Find all the solutions of this differential equation, carefully stating the largest open interval on which each particular solution is valid
b) Find the solution that satisfies y(1) = -1. On what interval is the solution valid?
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ThudnBlunder
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Re: differential equations
« Reply #1 on: May 26th, 2004, 4:26am » |
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Yes, I'm sure somebody will be happy to do all your homework for you. 
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THE MEEK SHALL INHERIT THE EARTH.....................................................................er, if that's all right with the rest of you.
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nikki
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Re: differential equations
« Reply #2 on: May 26th, 2004, 7:27am » |
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never mind,i think this is the answer
dy/dx + y3 = 0
dy/dx = -y3
dy/-y3 = dx
1/(2y2) = x + C
2y2 = 1/(x + C)
y2 = 1/[2(x + C)]
y = ± sqrt(1/[2(x + C)])
1/[2(x + C)] > 0
2(x + C) > 0
x+C > 0
x > -C and x > 0
1/(2*12) = -1 + C
1/2 = -1 + C
3/2 = C
1/(2y2) = x + 3/2
1/(2y2) = (2x + 3)/2
2y2 = 2/(2x + 3)
y2 = 1/(2x + 3)
y = ± sqrt(1/(2x + 3))
2x + 3 > 0
2x > -3
x > -3/2
But we have the condition that x > 0.
Hence x > 0 it our interval
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ThudnBlunder
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Re: differential equations
« Reply #3 on: May 26th, 2004, 8:57am » |
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Nikki, now that you have shown your working, I am even more sure you will get lots of help,
especially as Putnams are evidently getting rather easy these days.
(Instead of writing 'x2', you can write x2 by using the sup button above the edit window.)
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| « Last Edit: May 26th, 2004, 8:32pm by ThudnBlunder » |
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Sir Col
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Re: differential equations
« Reply #4 on: May 26th, 2004, 10:13am » |
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on May 26th, 2004, 3:58am, Nikki wrote:Consider the differential equation
2x.(dy/dx)+y3 = 0 x>0 |
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Nikki, you seem to have lost 2x in your solution, but you had the right idea. From your differential equation...
2x dy/dx = -y3
1/(2x) dx/dy = -1/y3
[int] 1/(2x) dx = [int] -1/y3 dy
(1/2)ln(2x) = 1/(2y2)+c
Multiply through by 2 and replace constant with -ln(k)
ln(2x) = 1/y2-ln(k)
ln(2x)+ln(k) = 1/y2
ln(2kx) = 1/y2
y = [pm]1/[sqrt]ln(2kx)
OR
2kx = exp(1/y2)
x = exp(1/y2)/(2k)
I'll leave you to figure the rest out.
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