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Topic: Differential Equation (Read 684 times) |
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ThudnBlunder
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Differential Equation
« on: Dec 13th, 2006, 4:38pm » |
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Find y(x) such that y''(x) = y'(x)*y(x)
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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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Icarus
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Re: Differential Equation
« Reply #1 on: Dec 13th, 2006, 6:31pm » |
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y=-2/x is one solution. (Go for the easy one first is my motto!)
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"Pi goes on and on and on ...
And e is just as cursed.
I wonder: Which is larger
When their digits are reversed? " - Anonymous
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Michael Dagg
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Re: Differential Equation
« Reply #2 on: Dec 13th, 2006, 7:49pm » |
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A couple others are
y(x) = 0
and
y(x) = sqrt(2) tan(sqrt(2)/2 x)
In general
y(x) = sqrt(2) tan(sqrt(2)(x + c1)/(2c2) )/c2, for c2 =/= 0
are solutions as well.
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| « Last Edit: Dec 13th, 2006, 8:21pm by Michael Dagg » |
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Regards,
Michael Dagg
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Icarus
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Re: Differential Equation
« Reply #3 on: Dec 14th, 2006, 6:10am » |
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For a derivation: note that y'y = (1/2)(d/dx)y2, so y" = (d/dx) (y2/2), and so y' = (y2 + c2)/2 for some c2.
If c2 = 0, then y'/y2 = 1/2, -1/y = x/2 + c1/2, or y = -2/(x + c1).
If c2 =/= 0, then y'/(y2 + c2) = 1/2.
If c2 > 0, let s2 = sqrt(c2). Then Arctan (y/s2)/s2 = (x+c1)/2, or y = s2*tan(s2(x+c1)/2).
If c2 < 0, let s2 = sqrt(-c2). Then tanh-1(-y/s2)/s2 = (x+c1)/2, or y = -s2*tanh(s2(x+c1)/2).
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"Pi goes on and on and on ...
And e is just as cursed.
I wonder: Which is larger
When their digits are reversed? " - Anonymous
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balakrishnan
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Re: Differential Equation
« Reply #4 on: Jan 6th, 2007, 12:20pm » |
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If we put
a=y''(x),v=y'(x)
the problem is same as
a=y*v
or
vdv/dy=y*v
which gives
v=y^2/2+C
or
dy/dx=(y^2+K)/2
which gives
y=C*tan(Cx/2+B)
where B,C are constants(including complex)
Sorry.I did not look at Icarus's post
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| « Last Edit: Jan 6th, 2007, 12:23pm by balakrishnan » |
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Icarus
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Re: Differential Equation
« Reply #5 on: Jan 7th, 2007, 3:27pm » |
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It's alright. I'm glad you had the satisfaction of solving it yourself!
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"Pi goes on and on and on ...
And e is just as cursed.
I wonder: Which is larger
When their digits are reversed? " - Anonymous
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