Re: Help with solutions to two DEs
- From: israel@xxxxxxxxxxx (Robert Israel)
- Date: 24 Jun 2005 20:27:41 GMT
In article <1119639900.697947.159340@xxxxxxxxxxxxxxxxxxxxxxxxxxxx>,
<carlos5100@xxxxxxxxx> wrote:
>I think I have solved these two DEs below correctly, but I am not
>really sure about my concussion on there interval of existence and was
>wondering if someone could tell me if I am right or wrong. If I am
>wrong, I would greatly appetite an explanation why.
>First DE
>(x^2+1)dy/dx=xy^2+x
....
>y=tan(c+ln(x^2+1)/2)
correct
>Now since InverseTan is defined by, y=InverseTan(x) if and only if
>tan(y)=x and -Pi/2<y<Pi/2, we have that the solution is valid for the
>interval Sqrt[e^(-Pi-C)-1]<x< Sqrt[e^(Pi-C)-1], correct?
You're using x and y for two different purposes here, which makes
this a bit confusing.
You want (n-1/2) Pi < c + ln(x^2+1)/2 < (n+1/2) Pi
where n is an integer, to make the solution continuous.
Of course if you add a multiple of Pi to c the solution is unchanged,
so you may as well take n=0 as you'll still get all solutions.
The inequalities are then equivalent to
exp(-2c-Pi) < x^2+1 < exp(-2c+Pi)
There are now three cases:
1) if c >= Pi/2, exp(-2c+Pi) <= 1 so this does not correspond to real
solutions
2) if -Pi/2 < c < Pi/2, exp(-2c-Pi) < 1 < exp(-2c+Pi) and the solution
is valid for - sqrt(exp(-2c+Pi)-1) < x < sqrt(exp(-2c+Pi)-1).
3) if c < -Pi/2, 1 < exp(-2c-Pi) and the solution is valid for
sqrt(exp(-2c-Pi)-1) < x < sqrt(exp(-2c+Pi)-1) or
-sqrt(exp(-2c+Pi)-1) < x < -sqrt(exp(-2c-Pi)-1).
>Second DE
>
>e^x(y^2-4y)dx+4dy=0
>
>Moving things around gives us
>
>dy/((y^2-4y)dx)+e^x/4=0
>
>integrating, we find.
>
>ln(|y-4|/|y|)/4+e^x/4=C
>
>moving things around
>
>ln(|y-4|/|y|)=-e^x+C
>
>and since we cannot not solve for y we cannot determine the interval
>the solution is valid for, correct?
No, you can solve for y. Start by taking exp of both sides.
Robert Israel israel@xxxxxxxxxxx
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia Vancouver, BC, Canada
.
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