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From: Blue Venom on 4 May 2010 12:10
y'=SQRT(1+x^2+y^2)
y(0)=0

Assuming there exists a local and unique solution defined on (-d,d), prove:
that the solution y is an odd function of x
that y(x) >= sinh(x) for all x >=0.

Any hint?
From: Blue Venom on 4 May 2010 12:15
> y'=SQRT(1+x^2+y^2)
> y(0)=0
>
> Assuming there exists a local and unique solution defined on (-d,d), prove:
> that the solution y is an odd function of x
> that y(x) >= sinh(x) for all x >=0.

Minor correction: the first question asks to prove that the solution is
odd for every x in (-d,d).

From: Robert Israel on 4 May 2010 13:52
Blue Venom <mandalayray1985(a)gmail.com> writes:

> > y'=SQRT(1+x^2+y^2)
> > y(0)=0
> >
> > Assuming there exists a local and unique solution defined on (-d,d),
> > prove:
> > that the solution y is an odd function of x
> > that y(x) >= sinh(x) for all x >=0.
>
> Minor correction: the first question asks to prove that the solution is
> odd for every x in (-d,d).
>

No, you were right the first time. The solution is an odd function of x.

Hint: what does the change of variables X = -x, Y = -y do to the differential
equation and initial condition?
--
Robert Israel israel(a)math.MyUniversitysInitials.ca
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia Vancouver, BC, Canada
From: Blue Venom on 4 May 2010 14:10
04/05/2010 19.52, Robert Israel:

> No, you were right the first time. The solution is an odd function of x.
>
> Hint: what does the change of variables X = -x, Y = -y do to the differential
> equation and initial condition?

I did it. The second question still troubles me, though. I must show
that the solution y is defined on all of R and that it is s.t.
y(x)>= senh(x) for all x>=0
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