Functions I have some familiarity with look so, $y^\prime(x) = \tan(x+2)$: straightforward expression of the first derivative of y as a function of x.
But say I have a function, $y^\prime(x) = \cos{(y)}$? I'm not sure what 'y' is supposed to signify when it's being called recursively like this: $y(x), \ y^\prime(x), \ $ or something else.
If it's not saying, $y^\prime(x)= \cos{\left(\int y^\prime(x) \ \text{d}x \right)}$, what is it saying? Any insights would be greatly appreciated...
On
$$ y'=\cos y\Longrightarrow \int\frac{dy}{\cos y}=\int dx=x+A. $$ Therefore, setting $$ t=\tan\frac{y}{2}, $$ we have: \begin{eqnarray} x+A&=&\int\frac{1}{\cos y}\,dy=\int\frac{\cos^2\frac{y}{2}+\sin^2\frac{y}{2}}{\cos^2\frac{y}{2}-\sin^2\frac{y}{2}}\,dy=\int\frac{1+t^2}{1-t^2}\frac{2}{1+t^2}\,dt\\ &=&2\int\frac{1}{1-t^2}\,dt=\int\left(\frac{1}{1+t}+\frac{1}{1-t}\right)\,dt=\ln\left|\frac{1+t}{1-t}\right|. \end{eqnarray} It follows that $$ \frac{1+t}{1-t}=\underbrace{\pm e^A}_{B}e^x \iff 1+t=(1-t)Be^x \iff t=\frac{Be^x-1}{Be^x+1}, $$ hence $$ y(x)=2\arctan\left(\frac{Be^x-1}{Be^x+1}\right), $$ where $B$ is a real constant.
You have: $\dfrac{dy}{dx} = \cos y \to \dfrac{dy}{\cos y} = dx \to x = \displaystyle \int \sec ydy$. Can you continue? more generally, you have a separable ODE: $y' = f(y)$, then the way to solve it is "separate" the $x$ and the $y$. like the one I did.