There's a lot going on with this one. Three things:

(1) You can solve it directly by separation of variables as you mentioned. I get:

$\displaystyle 3\sqrt{y}+5\ln(5-3\sqrt{y})=-9/2(x+c)+5;\quad c=3\sqrt{2}+5\ln(5-3\sqrt{2}),\; 5-3\sqrt{y}>0$

(2) You can solve it numerically with Euler or other means. Not too hard to set up Euler right? Or if you've done a few Euler's, then you can set it up in Mathematica say for the interval (0,5):

Code:

sol = NDSolve[{Derivative[1][y][t] == 5 - 3*y[t]^(1/2),
y[0] == 2}, y, {t, 0, 5}]
p1 = Plot[Evaluate[y[t] /. sol], {t, 0, 5}]

Yeah, I know. It's almost a crime to do it that way.

(3) You can extract an explicit expression for y in terms of x in (1) using the Lambert $\displaystyle \textbf{W}$ function. Suppose:

$\displaystyle x=y^ne^y$

Then to solve for y in terms of x, it's:

$\displaystyle y=n\textbf{W}\left[\frac{1}{n}x^{1/n}\right]$

So if I have:

$\displaystyle 3\sqrt{y}+5\ln(5-3\sqrt{y})=-9/2(x+c)+5$

Can you show:

$\displaystyle y(x)=\Bigg(5/3+5/3\textbf{W}\bigg[-1/5 e^{-1/5(5-k)}\bigg]\Bigg)^2;\quad k=-9/2(x+c)+5$

However if you're like me you're skeptical about all this. So I plotted the NDSolve solution and the Lambert W solution and superimposed them. The plot is below. Works for me.