Inverse Trigonometric functions - Boyce & Diprima 2.2.19 The problem asks for a solution to the initial value problem:
\begin{align}
&\sin(2x)dx+\cos(3y)dy=0\\
&y\left(\frac{\pi}{2}\right)=\frac{\pi}{3}
\end{align}
The problem is separable and I arrive at the following implicit solution:
\begin{align}
\sin(3y)-3\cos^2(x)=0
\end{align}
If I now take an $\arcsin$ of both sides and solve for $y$ I would naively have:
\begin{align}
y(x)=\frac{1}{3}\arcsin\left(3\cos^2(x)\right)
\end{align}
However, when $x=\frac{\pi}{2}$ we have LHS $=\frac{\pi}{3}$ and RHS $=0 $ which seems to be a problem. The solution provided in the book shifts the RHS by $\frac{\pi}{3}$. Could someone explain what is really going on here? thx
The author's given solution is:
\begin{align}
y(x)=\frac{1}{3}\left(\pi-\arcsin\left(3\cos^2(x)\right)\right)
\end{align}

After integrating you have:

$$ - \frac{1}{2} \cos{2x} + \frac{1}{3} \sin{3y} = K, \quad K \in \mathbb{R}, $$

which is as nice as an explicit solution. Substitute the initial data to find the value of the constant of integration:

$$ - \frac{1}{2} \cos{\pi} + \frac{1}{3} \sin{\pi} = \frac{1}{2} = K, $$

and you are done. I don't think finding an explicit solution for $y(x)$ before is a good idea.
Cheers!
