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From: Karl on 22 Apr 2010 17:56
y'=y - (x^2)/y

What wouold be the easier way to solve this? Consider it a Bernoulli
equation?
From: James Burns on 22 Apr 2010 18:18
Karl wrote:
> y'=y - (x^2)/y
>
> What wouold be the easier way to solve this?
> Consider it a Bernoulli equation?

Multiply both sides by y(x), then substitute
z = y^2. Then solve

1/2*z' - z = -(x^2)



Jim Burns

From: Pfsszxt on 23 Apr 2010 09:41
On Thu, 22 Apr 2010 23:56:47 +0200, Karl <948mechastraisand(a)yahoo.com>
wrote:

>y'=y - (x^2)/y
>
>What wouold be the easier way to solve this? Consider it a Bernoulli
>equation?

y'y - y^2 = -x^2
y^2 = U

U'/2 - U = -x^2
From: Dan Cass on 23 Apr 2010 05:49
> y'=y - (x^2)/y
>
> What wouold be the easier way to solve this? Consider
> it a Bernoulli
> equation?
By far the easiest way is to put it in Maple
or other CAS...

dsolve( diff(y(x),x)=y(x)-x^2/y(x),y(x));
2 1/2
y(x) = 1/2 (2 + 4 x + 4 x + 4 exp(2 x) _C1) ,

2 1/2
y(x) = -1/2 (2 + 4 x + 4 x + 4 exp(2 x) _C1)
From: Dan Cass on 23 Apr 2010 05:50
> > y'=y - (x^2)/y
> >
> > What wouold be the easier way to solve this?
> Consider
> > it a Bernoulli
> > equation?
> By far the easiest way is to put it in Maple
> or other CAS...
>
> dsolve( diff(y(x),x)=y(x)-x^2/y(x),y(x));
> 2 1/2
> y(x) = 1/2 (2 + 4 x + 4 x + 4 exp(2 x) _C1) ,
>
> 2
> 2
> 2 1/2
> y(x) = -1/2 (2 + 4 x + 4 x + 4 exp(2 x) _C1)

This makes no sense unless you view as plain text...
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