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solution of x* y'+y=2* (x* y)^0.5 ?

15 views

Dr Huang

Jul 9, 2022, 2:43:50 AM7/9/22
to
wolfram gives
(c1-x)^2/x
is it right? did it seem wrong? what is yours?

DrHuang.com

Nasser M. Abbasi

Jul 9, 2022, 3:04:16 AM7/9/22
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Mathematica 13.1

ClearAll[x, y]
ode = x*y'[x] + y[x] == 2*(x*y[x])^(1/2);
sol = DSolve[ode, y[x], x, IncludeSingularSolutions -> True]

gives

{{y[x] -> (E^(C[1]/2) + x)^2/x}, {y[x] -> x}}

The first is the general solution and the second is singular solution.

The solution you give is a particular solution which satisfies the ode
under the condition x>=C[1] only. You can see this by doing

sol = y -> Function[{x}, (C[1] - x)^2/x]
result = ode /. sol // FullSimplify

Sqrt[(x - C[1])^2] + C[1] == x

The above becomes identity when x >= C[1]

ps. I do not use Wolfram Alpha.

--Nasser

Dr Huang

Jul 9, 2022, 5:59:11 AM7/9/22
to
thanks
DrHuang.com

anti...@math.uni.wroc.pl

Jul 9, 2022, 5:47:42 PM7/9/22
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Well, it depends on what Sqrt (or ^(1/2)). In general there
are two square roots and in complex domain you can not
really separate them, one will analytically continue to
the other.

AFAICS the (C[1] - x)^2/x is better. With complex C[1]
writing E^(C[1]/2) is just obscure way of saying that we
have arbitrary nonzero constant. This is even more obscure
since replacing E^(C[1]/2) leads to valid solution. With
real C[1] writing constat as E^(C[1]/2) excludes negative
values and conseqently some solutions. And if you insist
that argument to Sqrt must be positive, then you get
restrictions on x anyway.

BTW: AFAICS 0 is singular solution, not covered by
formulas above. You can glue it with other solutions
at point where other solution is 0, so starting from
0 initial condition you get infintely many solutions.

--
Waldek Hebisch

nob...@nowhere.invalid

Jul 10, 2022, 1:28:19 PM7/10/22
to

anti...@math.uni.wroc.pl schrieb:
>
> Nasser M. Abbasi <n...@12000.org> wrote:
> > On 7/9/2022 1:43 AM, Dr Huang wrote:
> > > wolfram gives
> > > (c1-x)^2/x
> > > is it right? did it seem wrong? what is yours?
> > >
> >
Using Derive 6.10 assuming default real variables:

DSOLVE1_GEN(y - 2*SQRT(x*y), x, x, y, c)

LN(SQRT(y/x) - 1) = - LN(x) - c

Exponentiation gives:

EXP(LN(SQRT(y/x) - 1)) = EXP(- LN(x) - c)

SQRT(y/x) - 1 = EXP(-c)/x

Manipulating for x >= 0 to recover the original SQRT(x*y):

SQRT(x*y) = x + EXP(-c)

where EXP(-c) = d can now be negative, but the real solution is
obviously limited x >= -d, which may now also be negative. Checking the
implicit solution:

x*IMP_DIF(SQRT(x*y) - x - d, x, y) + y = 2*SQRT(x*y)

true

Alright. Finally solving for y is trivial:

y = (x + d)^2/x

but plugging this y(x) into the original ODE confirms that the solution
is indeed limited to x >= -d.

Martin.
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