Re: Help with solutions to two DEs

Ο <carlos5100@xxxxxxxxx> έγραψε στο μήνυμα
news:1119639900.697947.159340@xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx
>
> Hello,
>
> 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
>
> Moving things around gives us
>
> dy/((y^2+1)dx)-x/(x^2+1)=0
>
> Integrating gives us, after moving things around again.
>
> InverseTan(y)=c+ln(x^2+1)/2
>
> Thus,
>
> y=tan(c+ln(x^2+1)/2)
>
> 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?

Yes.

> 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?

Now apply exp(x) to both sides of your last equation and solve for y. You
will get two constraints on the x in terms of C, which will come from the
denominators of the two resultant equations, when you solve for y.

> Thanks a lot for your help guys.
>
> Sam J.
--
I. N. Galidakis
http://users.forthnet.gr/ath/jgal/
Eventually, _everything_ is understandable

.

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