Differential equation with separable, probably wrong answer in book I have a differential equation:

$$ \frac{dy}{dx} = y \log(y)\cot(x)$$

I'm trying solve that equation by separating variables and dividing by $y\log(y)$:

$$ dy = y \log(y) \cot(x) dx$$

$$ \frac{dy}{y \log(y)} = \cot(x) dx$$ 

$$   \cot(x) -  \frac{dy}{y \log(y)} = 0   $$  

Where of course $ y &gt; 0 $ regarding to division

Beacuse: 

$$ \int \frac{dy}{y \log(y)} = \ln | \ln(y) | +C $$

and:

$$ \int \cot(x) dx  = \ln| \sin(x) | + C$$

So: 

$$ \ln| \sin(x) | - \ln | \ln(y) | = C $$

$$ \ln \lvert\frac{\sin(x)} {\ln(y)} \rvert  = C $$

$$ \frac{\sin(x)}{\ln(y)} = \pm e^{C} $$

$d = \pm e^{C} $

$$ \sin(x) = d  \ln(y) $$

$$ \frac{\sin(x)}{d} = \ln(y)$$

$$ e^{\frac{\sin(x)}{d}} = y$$

This is my final answer. I have problem because in book from equation comes the answer to exercise is: 

$$ y = e^{c \sin(x)}$$  

Which one is correct?

I will be grateful for explaining Best regards
Note that
$$ \int \frac{du}{u} = \ln |u| + c_1 = \ln |u| + \ln |c| = \ln |cu|.$$

Starting from your result:

$$ \cot x\, dx =\frac{dy}{y \ln  y} $$
$$\implies \frac{d(\sin x)}{\sin x} = \frac{d(\ln y)}{y \ln  y}$$

we integrate both sides:

$$ \int \frac{d(\sin x)}{\sin x} \, dx = \int \frac{d(\ln y)}{\ln  y}\, dx$$
$$\implies \ln |a\sin x| = \ln |b\ln  y| $$
$$\implies a\sin x = b\ln  y, \quad(1)$$
$$\implies \ln  y = c\sin x$$
$$\implies y = e^{c\sin x}$$
where $a, b, c$ are arbitrary constants. Note that $(1)$ includes all possible solutions including the case $a=0$.
NOTE:
$$\ln|\frac{\sin x}{\ln y}|=C$$
Can also be written as
$$\ln|\frac{\sin x}{\ln y}|=\ln C$$ 
(Because both terms are constant.)

This shall give you $$\sin x =C \ln y$$
Which will give you:
$$y=e^{c \sin x}$$ where $c=\frac{1}{C}$. I will prefer this solution because in your answer d=0 won't be giving you any solution but Note that c=0 will do. {Credits to that comment}