I.V.P $y'=\sin(e^{y}), y(0)=a$ Is the I.V.P:
$$\begin{cases}
\dfrac{dy}{dx}=\sin(e^{y})\\[8pt]
y(0)=a
\end{cases}
\text{ where } a\in \mathbb{R}$$
a) Find the values ​​of $a$ for which $y(x, a)=0$
b) Prove that if $a=0$ then $0<y(x,a)<\ln(\pi)$,  (for $x>0$)

I try this:
$$y(x,a)=0=a+\int_{0}^{x}f(t,y(t))dt$$
$$a=-\int_{0}^{x}f(t,y(t))dt$$
Is this correct?
 A: Hint for (a): if $y$ is constant, $\dfrac{dy}{dx} = 0$.
EDIT: Now that you have (a), you know that one of the constant solutions is $y=\ln(\pi)$. 
Use the fact that two different solutions can't meet (because of the Existence and Uniqueness Theorem) to get one of the inequalities.  For the other one: what is the sign of $dy/dx$ at $x=0$?  How could $dy/dx$ ever change signs?
A: The problem describes an autonomous system $y'=f(y)$ with $y(0)=a$ and $f(y(x))=\sin(e^y)$.
The trivial solutions are immediately known because they are the constant solutions:
$$
y_k(x)=\ln(k\pi)\qquad k=1,2,\ldots
$$
To find the nonconstant solutions we may separate the variable and integrate:
$$
\int \frac{\operatorname{d}y}{f(y)}=\int \operatorname{d}x
$$
that is
$$
\int_{y_0}^y \frac{\operatorname{d}u}{\sin(e^u)}=x-x_0
$$
Note that the integration may be hard if not impossible to carry out. So it may not be possible to use the analytical technique. 
So we have to use other techniques to study the behaviour of the solutions.
The equilibria or constant solutions of this differential equation are the roots of the equation $f(y)=\sin(e^y)=0$ that is
$$
y(x)=\ln(k\pi)\qquad \forall x\in\Bbb{R},\;k=1,2,\ldots
$$
Using the existence and uniqueness theorem, these constant solutions will cut the entire plane (where the solutions live) into independent regions. This means, if an initial condition belongs to one of the regions, then the solution satisfying the initial condition will stay in that region all the time. 
The graph of $f(y)=\sin(e^y)$ is given below

We see that 


*

*if $y(x)$ is a solution satisfying $-\infty < y(0) < \ln(\pi)$, then $y(x)$ is always increasing; 

*if $y(x)$ is a solution satisfying $\ln(\pi) < y(0) < \ln(2\pi)$, then $y(x)$ is always decreasing;

*and so on.


Using the above results, we know that for the given initial condition, $y(x)$ is increasing if $-\infty < y(0) < \ln(\pi)$ and for every $x$ we have 
$$
-\infty < y(x) < \ln(\pi).
$$
Oberving that $\lim_{x\to-\infty}y(x)=-\infty$ and $\lim_{x\to +\infty}y(x)=\ln(\pi)$ and that $y(x)$ is increasing forall $x$, then $y(x,a)$ will have a unique root $y(x,a)=0$, for all $x$, if $a=y(0)$ is such that $-\infty < a< \ln(\pi)$. And if $a=y(0)=0$ obviously $0<y(x)<\ln(\pi)$ for $x\ge 0$.
If $y(x)$ is a solution satisfying $\ln(\pi) < y(0) < \ln(2\pi)$, then $y(x)$ is always decreasing and $\lim_{x\to-\infty}y(x)=\ln(2\pi)$ and $\lim_{x\to +\infty}y(x)=\ln(\pi)$, that is $\forall x\in\Bbb{R},\;\ln(\pi) < y(x) < \ln(2\pi)$.
And so on you can proceed for all the intervals determined by the critical points.
In the graph below you can see the behaviour of $y(x)$ for $x\ge0$ and for different values of $a=y(0)$.

A: This IVP enjoys existence and uniqueness of solutions. Also, its solution is global (i.e., defined in the whole of $\mathbb R$) as the flux function $f(y)=\sin(\mathrm{e}^y)$ is bounded. 
The solutions $\varphi=\varphi(x)$ of IVP of scalar autonomous ODEs (i.e., of the form $y'=f(y)$, $\,y(0)=y_0$) with smooth flux have the following properties:
a. If $f(y_0)=0$, then $\varphi(x)=y_0$, for all $x$. This is due to the fact that indeed that constant function is a solution, and the IVP enjoys uniqueness.
b. If $f(y_0)>0$ (corresp. $\,f(y_0)<0$), then $\varphi$ is everywhere strictly increasing (corresp. strictly decreasing), and in particularly $\varphi'(x)>0$ (corresp. $\,\,\varphi'(x)<0$ ) for all $x$, otherwise the derivative of $\varphi$ would vanish, say at $x_0$, which means that $f(\varphi(x_0))=0$, and hence $\varphi$ would also be a solution of the IVP
$$
y'=f(y), \quad y(x_0)=\varphi(x_0),
$$
which is satisfied by the constant function $\psi(x)=\varphi(x_0)$. Contradiction.
c. If $a<y_0<b$ and $f(a)=f(b)=0$, then $a<\varphi(x)<b$, for all $x$. Otherwise $\varphi$ would cross $a$ or $b$ and its derivative would vanish there.
In our case $f(y)=\sin(\mathrm{e}^y)$, the IVP with initial condition $y(0)=a$, has a constant solution iff $\sin(\mathrm{e}^a)=0$, equivalently $a=\log k\pi$, for some positive integer $k$. Otherwise the solution $\varphi$ satisfies the inequalities $y_1<\varphi(x)<y_2$, where $y_1$ and $y_2$ are consecutive roots of $f$, and $a$ lies in $(y_1,y_2)$.
