On the Bernoulli differential equation $ y'(x) = \dfrac{y}{x + \sqrt{xy}} $ $$ y'(x) = \frac{y}{x + \sqrt{xy}} $$
I have attempted to solve it. However, I am not able to arrive at the exact expression of the final answer. Maybe I am making a mistake in a step? Can you please help?
The answer should be
$$ \sqrt{x} = \sqrt{y}(\log\sqrt{y} + \mbox{Constant}) $$
PS: This is my first question on this site, I will be more than willing to provide my own attempts at this, if it is required. Thanks.
Here are my workings,
$$ \frac{1}{\sqrt{x}}x'(y)-\frac{\sqrt{x}}{y}=\frac{1}{\sqrt{y}} $$
$$ Taking, \sqrt{x}=z $$
Differentiating  wrt  to x and dividing by dy$$ ,\frac{1}{2\sqrt{x}}x'(y)=z'(y) $$
Substituting, $$ z'(y)-\frac{z}{2y}=\frac{1}{2\sqrt{y}} $$
Now taking integrating factor $$ e^{\int_{}{}-\frac{1}{2y} dy}\ $$
$$ =e^{-\frac{1}{2}\log y } $$
Multiplying both sides by the integrating factor, $$ z'(y)e^{-\frac{1}{2}\log y }-\frac{z}{2y}e^{-\frac{1}{2}\log y }=\frac{1}{2\sqrt{y}}e^{-\frac{1}{2}\log y } $$
$$ (ze^{-\frac{1}{2}\log y })'(y)=\frac{1}{2\sqrt{y}}e^{-\frac{1}{2}\log y } $$
$$ ze^{-\frac{1}{2}\log y }=\int_{}{}\frac{1}{2y} dy\ $$
$$ze^{-\frac{1}{2}\log y }=\frac{1}{2}log(y)+Constant $$
$$ \frac{\sqrt{x}}{\sqrt{y}}=\frac{1}{2}log(y)+Constant $$
There you go, did I mess up some logarithmic property? or is it something with the integration I am missing? The answer is close, but not it.

$$y'(x) = \frac{y}{x + \sqrt{xy}}$$
It's Bernoulli' s differential equation only if you consider $x'$ instead  of $y'$:
$$\dfrac 1 {x'(y)} = \frac{y}{x + \sqrt{xy}}$$
$$yx'-x =  \sqrt{xy}$$

Your answer is correct. Note that you have the log property: $n\log y=\log(  y^n)$
here $n=\dfrac 12$:
$$\frac{\sqrt{x}}{\sqrt{y}}=\frac{1}{2}\log(y)+C$$
$$\frac{\sqrt{x}}{\sqrt{y}}=\log(\sqrt y)+C$$
$$\sqrt{x}={\sqrt{y}}(\log(\sqrt y)+C) $$
