<s>[INST] Locate my error for this initial value separable differential equation? The problem is to solve $ sin\,2x\,dx + cos\,3y\,dy = 0, \;\;\;\;y({\pi\over 2}) = {\pi\over 3}$
Here are my steps:
$$ cos \,3y \,dy = -sin \,2x \,dx $$
$$\int cos\,3y\,dy = \int -sin\,2x\,dx$$
$$ {1\over 3}\,sin\,3y = {1\over 2}\,cos\,2x\,+C $$
$$ 2\,sin\,3y = 3\,cos\,2x+C$$
Using the initial value then.. $$ 2\,sin({3\pi\over 3}) = 3\,cos({2\pi\over 2}) + C $$
$\qquad\qquad\qquad\qquad\qquad\qquad 0 = -3+C $ and it follows that $ C = 3 $$$$$Then, 
$$ 2\,sin\,3y = 3\,cos\,2x+3 $$
$$ sin\,3y = {3\over 2}cos\,2x+{3\over 2} = 3({1\over 2}cos\,2x+{1\over 2})$$
Using the trig. identity, $$ sin\,3y = 3\,cos^2x$$
$$ arcsin(sin\,3y) = arcsin(3\,cos^2x) $$
$$ 3y = arcsin(3\,cos^2x)$$
$$ y = {arcsin(3\,cos^2x)\over 3} $$
However, the solution is stated as: $$ y = {\pi - arcsin(3\,cos^2x)\over 3} $$
Could someone point out what I missed?
 [/INST] The equation
\begin{align}
- \sin(2x) \ dx =  \cos(3y) \ dy
\end{align}
can be seen to have the solution 
\begin{align}
\frac{1}{2} \cos(2x) = \frac{1}{3} \sin(3y) + c.
\end{align}
This can be seen in the form
\begin{align}
\sin(3y) = \frac{3}{2} \left( \cos(2x) - 2 c \right)
\end{align}
or
\begin{align}
y(x) = \frac{1}{3} \sin^{-1}\left( \frac{3}{2}\left[ \cos(2x) - 2c \right]\right).
\end{align}
From the "general solutions" section of the Wiki page this can be placed in the form
\begin{align}
y(x) = \frac{1}{3}\left\{\pi -  \sin^{-1}\left( \frac{3}{2}\left[ \cos(2x) - 2c \right]\right) \right\}.
\end{align}
Now for the condition $y(\pi/2) = \pi/3$ this becomes
\begin{align}
\frac{\pi}{3} =\frac{\pi}{3} - \frac{1}{3} \sin^{-1}\left( \frac{3}{2}\left[ \cos(\pi) - 2c \right]\right)
\end{align}
or $c = -1/2$. It can now be seen that
\begin{align}
y(x) = \frac{1}{3}\left\{\pi -  \sin^{-1}\left(3 \cos^{2}(x)\right) \right\}.
\end{align}
</s>