Diferential equation solution satisfying $y(0) = \pi$ I have the following question from a past exam:

Show that the differential equation $\frac{dy}{dx} = \cfrac{e^x + x}{\sin y + 2}$ has a solution satisfying $y(0) = \pi$

What I have done:
$$\int (\sin y + 2) \; \mathrm{d}y = \int (e^x + x) \; \mathrm{d}x$$
$$-\cos y + 2y = e^x + \frac{x^2}2 + C$$
Sub in $y = \pi, x = 0$
$$-\cos {\pi} + 2\pi = e^0 + 0 + C$$
$$1+2\pi - 1 = C, C = 2\pi$$
$$-\cos y + 2y = e^x + \frac{x^2}2 + 2\pi$$
Now I am not sure what this question actually wants me to do. Any ideas how I can 'show that the differential equation has a solution satisfying$\dots$'?
Note: I am apparently meant to solve it implicitly using an 'appropriate' theorem. Perhaps there is a real-analysis way of solving this?(This was for an Analysis past exam, and I seem to have solved it using only calculus)
 A: The question asks for the existence of a solution. You have derived a candidate for a solution using justifiable but ultimately sketchy techniques (or as you said, 'calculus' methods). 
Method One: Using the work you've done so far.
In precalculus we would say that you've done the proof "backwards".
For a proof that the equation has a solution, you have to go forwards. So just take the derivative of the solution, show it is what you wanted it to be, and you're done. 
[Note: It is not really true that you have done it backwards: each step in your solution can be fully formalized into applications of theorem statements. But it is tedious and nobody would do this, since the easier method suffices.]
However, there is a  complication is that the solution you derived is not written in functional form; indeed it is not obvious that $y$ is a function of $x$ at all. To fully formalize the proof, you will need to show that $f(y)=2y-\cos(y)$ is a bijection; equivalently that it is unbounded and monotonic. The first of these is trivial; the second is clear by noticing that it is differentiable everywhere and considering the derivative.
Method Two: Using (insufficiently) heavy machinery.
The theorem that you were looking for in your edit is called the Peano Existence Theorem. Sometimes theoretical courses don't cover PET and instead opt for the stronger Picard-Lindelöf Theorem, better known to differential equations students as the "existence and uniqueness theorem".
The usual statement of the P-LT is

Consider the initial value problem $y'=f(t,y)$ with $y(t_0)=y_0$. If $f$ is Litschitz continuous in $y$ for all $t$, and continuous in $t$ for all $y$, then there exists an $\varepsilon>0$ and a function $y$ satisfying the IVP for all $t\in[t_0-\varepsilon, t_0+\varepsilon]$.

(the usual statement of PET is similar)
Unfortunately, it may be insufficient to consider this theorem as a black box to get the existence of a solution. It seems likely that the questioner had in mind the existence of a solution $\Bbb R\to \Bbb R$, which the P-LT does not give. If this is the case, to complete the solution one must understand a method by which one can derive a particular $\varepsilon$. Unless you discussed methods of doing this, it means you must know the proof well enough to provide some (derivation of an) estimate. If you're lucky, the proof will in this instance allow you to choose this $\varepsilon$ arbitrarily large. 
However, it is also quite possible that you can only bound the maximal choice of $\varepsilon$ from below, at which point you must "piece together" solutions inductively. Note that the lower bound on maximal $\varepsilon$ is a crucial piece of this method: if you do not have it, then it is possible your intervals will shrink so quickly that the solutions do not piece together to all of $\mathbb R$.

It probably goes without saying that I don't recommend method two on an exam, but it is at least good to know that something like it is possible. I think "on an exam" is an important qualifier here. What's causing you all this trouble is that you need to show how you get the estimate for $\varepsilon$. In a more relaxed setting, you could probably just say "It follows from the proof of PLT that $\varepsilon$ may be chosen..." and then state the estimate; the proof is then quite short.
