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The problem is to solve the separable differential equation
$sqrt{xy}~~frac{dy}{dx}=sqrt{4-x}$.

My work thus far:
$ sqrt{xy}~~ {dy}=sqrt{4-x}~~dx
\ sqrt{y}~~ {dy}=sqrt{frac{4-x}{x}}~~dx
\ int sqrt{y}~~ {dy}= int sqrt{frac{4-x}{x}}~~d x
\ frac{2}{3} y^{3/2} =int sqrt{frac{4-x}{x}}~~dx
\ frac{2}{3} y^{3/2} =int sqrt{frac 4 x - 1 }~~dx
$

The right side is a difficult integral to evaluate.

Is there an easier approach I have missed, or should I persist in trying to solve the integral.

Hint: Apply the substitution $u=frac{4}{x}$, then integrate by parts and then again look for a clever substitution to get to the form $intfrac{1}{t^2+1}dt=arctan(t)$.

Hint Sometimes an integral $int sqrt{f(x)} , dx$ can be resolved by rearranging $u = sqrt{f(x)}$ to give a reverse substitution (where we may need to restrict the domain of $f$).

In our case, setting $u = sqrtfrac{4 - x}{x}$ and rearranging and differentiating yields the substitution
$$x = frac{4}{u^2 + 1}, qquad dx = -frac{8 u, du}{(u^2 + 1)^2},$$
which gives
$$int sqrtfrac{4 - x}{x} dx = -8 int frac{u^2 ,du}{(u^2 + 1)^2} .$$
The latter integrand is rational and so can be solved with standard techniques.

The occurrence of the form $u^2 + 1$ in the denominator suggests the substitution $u = tan theta$, $du = sec^2 theta, dtheta$.

Make the substitution $t = sqrt{x}$, then

$$ int 2sqrt{4-x} frac{dx}{2sqrt{x}} = int 2sqrt{4-t^2} dt $$

This is a standard trig substitution. Let $t = 2sin theta$

begin{align}
int 2sqrt{4-t^2}dt &= int 2cdot 2costhetacdot 2costheta dtheta \
&= int 8 cos^2theta dtheta \
&= int 4(1+cos2theta) dtheta \
&= 4theta + 2sin2theta + C
end{align}

Going backwards

begin{align}
4theta + 2sin2theta + C &= 4theta + 4sinthetacos theta + C \
&= 4arctanfrac{t}{2} + tsqrt{4-t^2} + C \
&= 4arcsinfrac{sqrt{x}}{2} + sqrt{x}sqrt{4-x} + C
end{align}

Thankyou to the tips given. I will attempt to answer my question.
$$ sqrt{xy}~~frac{dy}{dx}=sqrt{4-x}
\ sqrt{xy}~~ {dy}=sqrt{4-x}~~dx
\ sqrt{y}~~ {dy}=sqrt{frac{4-x}{x}}~~dx
\ int sqrt{y}~~ {dy}= int frac{ sqrt {4-x}}{sqrt x}~~dx
\ frac{2}{3} y^{3/2} =int frac{ sqrt {4-x}}{sqrt x}~~dx
\ text{ Make the substitution u = √x }
\ frac{2}{3} y^{3/2} = 2int sqrt { 4- u^2} du
\ \ text{ Substitute w = u/2 }
\ frac{2}{3} y^{3/2} = 8 int sqrt{ 1 - w ^2} ~dw
\ text{ using trig substitution}
\ frac{2}{3} y^{3/2} = 4 w sqrt {1 -w^2}+4 arcsin left( w right) + C
\ text{back substituting }
\ frac{2}{3} y^{3/2} = 4 frac u 2 sqrt {1 -left( frac u 2 right) ^2}+4 arcsin left( frac u 2 right) + C
\ frac{2}{3} y^{3/2} = 2 sqrt x sqrt {1 -left( frac {sqrt x }{ 2} right) ^2}+4 arcsin left( frac {sqrt x }{ 2} right) + C
\ frac{2}{3} y^{3/2} = 2 sqrt x sqrt {1 -left( frac {x}{4} right)}+4 arcsin left( frac {sqrt x }{ 2} right) + C
\ frac{2}{3} y^{3/2} = 2 sqrt x frac{ sqrt {4 -x }}{ sqrt{4} } +4 arcsin left( frac {sqrt x }{ 2} right) + C
\ frac{2}{3} y^{3/2} = sqrt x sqrt {4 -x } +4 arcsin left( frac {sqrt x }{ 2} right) + C
\ y^{3/2} =frac{3}{2} left( sqrt x sqrt {4 -x } +4 arcsin left( frac {sqrt x }{ 2} right) + C right)
\ y^{3/2} = left( frac{3}{2} sqrt x sqrt {4 -x } + 6 arcsin left( frac {sqrt x }{ 2} right) + frac 3 2 C right)
\ y = left( frac{3}{2} sqrt x sqrt {4 -x } + 6 arcsin left( frac {sqrt x }{ 2} right) + C right) ^{2/3}$$