How to solve the initial value problem $y'(x)=\lambda \sin(x+y(x))$, $y(0)=1$. For $\lambda \in \mathbb{R}$, consider the initial value problem $y'(x)=\lambda \sin(x+y(x))$, $y(0)=1$. Then this initial value problem has
  
  
  no solution in any neighbourhood of $0$.
  a solution in $\mathbb{R}$ if $|\lambda|&lt;1$
  a solution in a neighbourhood of $0$.
  a solution in $\mathbb{R}$ only if $|\lambda|&gt;1$.
  


This is the first time I have encountered this kind of IVP, and have no idea to proceed. The entanglement of $x$ and $y(x)$ in $\sin(x+y(x))$ is what causing me the trouble to make any headway. So please help me to solve this. Thanks.
Hint: Observe we have
\begin{align}
y' = F(x, y)
\end{align}
where $F$ is Lipschitz in $y$ variable since
\begin{align}
|F(x, y_1) -F(x, y_2)| = |\lambda| |\sin(x+y_1)-\sin(x+y_2)| \leq |\lambda||y_1-y_2|.
\end{align}
Note we have used the fact $|\sin u-\sin v| \leq |u-v|$. 

Now, by Picard-Lindelof theorem, one can guarantee local existence, i.e. (3) holds if we do not know anything about $\lambda$. 

Moreover, one can use a Banach fixed point argument to show that the ode has a global solution, i.e. solution on all of $\mathbb{R}$ if $|\lambda|&lt;1$, which means (2) holds.
You first make a string of numbers for the range of $x$. Let's say $x\in[0,1]$. You could select $x=0, 0.1 , 0.2, ..., 1.0$.

Next, substitute $x=0$ to the equation to get:
$y^\prime(0) = \lambda \text{sin}(0+y(0)) = 0$.

Then use:

$y(0.1) = y(0) + 0.1 y^\prime(0) = 0$.

Next, substitute $x=0.1$ in the equation:

$y^\prime(0.1) = \lambda \text{sin}(0.1+y(0.1)) = \lambda \text{sin}(0.1)$

You continue the iteration up to $x=1.0$. The finer the meshing, the more precise the answer.