Locate my error for this initial value separable differential equation?

Question

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?

Answer 1

Note that your answer does not satisfy the initial condition. The mistake lies in concluding from $\sin z=w$ that $z=\arcsin w$.

If we know that $\sin z=w$, then $z=2\pi n+\arcsin w$ or $z=(2n+1)\pi -\arcsin w$, for some integer $n$. We need to choose the "branch" that is correct at the initial position, that is, the branch that goes through the point $(\pi/2,\pi/3)$.

Answer 2

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}