Solving and sketching the differential equation $\cos y \,dy=\sin x\,dx$ Consider the differential equation
$$\cos y \,dy=\sin x\,dx$$
I'm not sure how this is supposed to end up to be $\sin y +\cos x=C$.  How did the position of the $y$ and $x$ change? I see an integral occurs, but not sure how or where.
Also, I'm told the sketch is supposed to be the straight line $x+y=\frac{π}{2}$ as the trajectory passing through $(\frac{π}{2}, 0)$.  Where am I supposed to get $x+y=\frac{π}{2}$ from?  I feel like I'm missing some very simple things, but it's not clicking for me.  Any help is appreciated!
 A: When you solve the Differential equation
$$\cos y \,dy=\sin x\,dx$$
You take the integral of Both sides
$$\int \cos y \,dy=\int \sin x\,dx$$
$$\sin y +C_1=-\cos x +C_2$$
Now throught some manipulation
$$\sin y + \cos x =C$$
Which is the answer you got
A: You want to solve the differential equation. Derive $\sin(y)+\cos(x)=C$ implicityly, you get:
$$\cos(y)\frac{d y}{d x} - \sin(x)=0$$
"Multiplying" by $dx$ you obtain the expression you are looking for (you are actually not multypling but working with differential forms notation. But for practicity think of multiplying). To come up with that expression yourself since it is a separable differential equation you just had to integrate both sides.
With respect to your second question, consider the case $C=0$ (this is because $\sin(0)=\cos(\pi/2)=0$) and the trigonometric identity $ \sin(y)=-\cos(\pi/2+y)$. Then 
$$\sin(y)+\cos(x)=C=0=-\cos(\pi/2+y)+\cos(x)$$
which implies after solving and taking $\cos^{-1}$ in both sides
$$ \pi/2 + y = x. $$
Are you totally sure about your last expression?
A: Since the form is separated before, integrating both sides will result in $siny=-cosx+c$ which gives $siny+cosx=c$. 
A: A method for solving a differential equation is to "invert" the differential, by integration or anti-differentiation.
For variety, I'll toss out a different way to run the calculation (although most of the methods you see are essentially the same):


*

*$\cos y \, dy = \sin x \, dx$

*$d(\sin y) = d(-\cos x)$

*$d(\sin y + \cos x) = 0$


and we know the only things whose differential is zero are the constants. (or locally constant functions, if you're in a disconnected setting)
Thus, $\sin y + \cos x = C$ for some value of $C$.
Another method is 
