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From: Ash on 10 Jun 2010 12:28
Hi,

I have the differential equation

y'(x) = sqrt[ (B-y)/y) ]

Is there a general method, or infact any method I can use to solve
this?

Thank you,

Ashley
From: Alois Steindl on 10 Jun 2010 12:31
Ash <asriwi(a)googlemail.com> writes:

> Hi,
>
> I have the differential equation
>
> y'(x) = sqrt[ (B-y)/y) ]
>
> Is there a general method, or infact any method I can use to solve
> this?
>
> Thank you,
>
> Ashley
Hello,
yes, any reasonable introductory book about ODEs will have some remark
about separable differential equations.

Alois
From: Ray Vickson on 10 Jun 2010 12:36
On Jun 10, 9:28 am, Ash <asr...(a)googlemail.com> wrote:
> Hi,
>
> I have the differential equation
>
> y'(x) = sqrt[ (B-y)/y) ]
>
> Is there a general method, or infact any method I can use to solve
> this?
>
> Thank you,
>
> Ashley

If you adopt sloppy notation you get sqrt(y/(B-y)) dy = dt, hence t =
const + integral(sqrt(y/(B-y)) dy). Now that you have this you can
justify it rigorously.

R.G. Vickson
From: Ron on 10 Jun 2010 12:36
On Jun 10, 12:28 pm, Ash <asr...(a)googlemail.com> wrote:
> Hi,
>
> I have the differential equation
>
> y'(x) = sqrt[ (B-y)/y) ]
>
> Is there a general method, or infact any method I can use to solve
> this?
>
> Thank you,
>
> Ashley

This differential equation is separable. dy/dx=sqrt((B-y)/y). Thus you
have sqrt(y/(B-y)) dy = dx. Integrate both sides.
From: Ash on 10 Jun 2010 12:41
Sorry, I wasn't completely clear.

I am having trouble with solving the integral itself. Any hints would
be appreciated, thanks.
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