Can't figure this out: $dy/dx = \cos(y)\sin(x)$ Problem: Find the solution and domain of validity for the following differential equation:
$\dfrac{dy}{dx} = \cos(y) \cdot \sin(x)$
What I tried: I noticed that this is a separable equation, so I wrote it in the form: $\dfrac{1}{\cos(y)} \cdot dy = \sin(x) \cdot dx$
Next, I found out that if $\cos(y) = 0$ then $\dfrac{1}{\cos(y)}$ is undefined. So I separate into two following cases;
Case1: $\cos(y) \neq 0$;
Then $\int\dfrac{dy}{\cos(y)} = \int\sin(x)dx$. By solving these integrals I get that $\tan(y) \cdot \sec(y)+ c_0 = -\cos(x)$ which implies that
$\arccos(-\tan(y) \cdot \sec(y)+ c) = x$,   for $c=-c_0$ which is an arbitrary constant.
I wonder if this is a general solution.
Case2: $\cos(y)=0$;
This implies that $y\in \{ \mathbb{R} -\dfrac{(2k+1)\pi}2:k\in \mathbb{Z}\}$
Pick an arbitrary $y_0\in \{ \mathbb{R} -\dfrac{(2k+1)\pi}2:k\in \mathbb{Z}\}$, substituting this into the differential equation gives that;
$\dfrac {dy_0}{dx} = \cos(y_0)\cdot \sin(x)$.
I am pretty sure that $\cos(y_0) = 0$. But I do not know how can I show that $\dfrac {dy_0}{dx} = 0$ to find out if there is a singular/lost solution or not.
Moreover, I have been taught that $dx$ means derivative, for instance $x^2\cdot dx = 2x$. Then why can't we solve a separable differential equation just by taking derivative of both sides for instance in $\dfrac{1}{\cos(y)} \cdot dy = \sin(x) \cdot dx$
 A: $$\frac{dy}{dx}=\sin x \cos y$$
Separate variables
$$\frac{dy}{\cos y}=\sin x$$
Integrate both sides
$$\int \frac{dy}{\cos y}=\int \sin x \,dx$$
Set $$\cos y=\frac{1-t^2}{1+t^2};\;t=\tan\frac{y}{2};\;y=2\arctan t;dy=\frac{2dt}{1+t^2}$$
$$\int \frac{dy}{\cos y}=\int \frac{2dt}{1+t^2}\cdot \frac{1+t^2}{1-t^2}=\int \frac{2dt}{1-t^2}=2 \text{arctanh}\, t+C=\\=
2 \text{arctanh}\, \tan\frac{y}{2}+C$$
So the solution can be written as
$$2 \text{arctanh}\, \tan\frac{y}{2}=-\cos x + C\\
\text{arctanh}\, \tan\frac{y}{2}=-\frac{\cos x}{2} + \frac{C}{2}\\
\tan\frac{y}{2}=\tanh\left(\frac{\frac{C}{2}-\cos x}{2}  \right)\\
y=2\arctan\left(\tanh \frac{1}{2}\left(c-\cos x \right)\right) $$
A: $$\int \frac{dy}{\cos y}=\int \sin x \,dx=-\cos x +C $$
Then for the integral on the LHS:
$$I=\int \frac{dy}{\cos y}=\int \frac{d \sin y}{\cos^2 y}$$
$$I=\int \frac{d \sin y}{1-\sin^2 y}=\int \frac{d u}{1-u^2}$$
$$I=\dfrac 12 \left (\int \frac{du }{u+1}-\int \frac{du}{u-1}\right )$$
$$ I=\dfrac 12 \ln \left |\dfrac {u+1}{u-1}\right |+C= \dfrac 12\ln \left |\dfrac {(u+1)^2}{u^2-1}\right |+C$$
Where $u =\sin y$.
$$ I= \ln \left |\dfrac {1+\sin y}{\cos y}\right |+C$$
You can also use this result:
$$ I=\dfrac 12 \ln \left |\dfrac {u+1}{u-1}\right |+C=\tanh^{-1} u+C$$
$$ \implies I=\tanh^{-1} ( \sin y)+C$$
So that we have:
$$\tanh^{-1} ( \sin y)=(C-\cos x)$$
$$ y(x)=\arcsin (\tanh(C-\cos x))$$
A: $$\int\frac{dy}{\cos(y)}=\int\frac{\cos(y)\,dy}{1-\sin^2(y)}=\text{artanh}(\sin(y))=\int\sin (x)\,dx$$
gives
$$y=\arcsin(\tanh(c-\cos(x)))$$ and periodic replicas.
$$y=\pm\dfrac\pi2$$ and replicas are also (degenerate) solutions ($y'=0$, corresponding to $c=\pm\infty$).

