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?

DrHuang.com

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

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wolfram gives
(c1-x)^2/x
is it right? did it seem wrong? what is yours?

DrHuang.com

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On Saturday, 9 July 2022 at 17:04:16 UTC+10, Nasser M. Abbasi 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?

DrHuang.com

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

thanks
DrHuang.com

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Nasser M. Abbasi <[email protected]> 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?

DrHuang.com

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]

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

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[email protected] schrieb:

Nasser M. Abbasi <[email protected]> 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?

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]

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.

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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