Consider the ODE:
$y'= x + \text{cos}(x+3y),\,\, y(0) = 1.$
Let $g(x,y) = x + \text{cos}(x+3y).$ Suppose $g$ is continuous on a neighbourhood of $(0,1)$ of the form $G = (-1,1) \times (0,2).$ I want to apply Picard-Lindelöf theorem and identify the set $X \subseteq C[-\delta,\delta]$ on which the contraction mapping is defined and in which the solution of the ODE lies.
$\mathbf{My\,\, attempt}:$
First, I want to find an upper-bound for $g(x,y).$ Let $ M = \sup_{(x,y) \in G} |g(x,y)|.$ Then, \begin{align} M &\leq \sup_{x} |x| + \sup_{x,y}|\text{cos}(x+3y)|\\ &\leq 1 + 1 = 2. \end{align}
Next, I want to show that $g$ is Lipschitz in the $y$-direction. I'm getting stuck here. How can I show that $|\text{cos}(x + 3y_1) - \text{cos}(x + 3y_2)| \leq K|y_1 - y_2|?$
Once, I'm able to find K, I can choose $\delta' < \delta, \frac{\varepsilon}{M}, \frac{1}{K}$ and find $X$ as follows:
$X = C(I',J)$ where $I' = [-\delta',\delta']$ and $J = [1-\varepsilon,1 + \varepsilon].$
By MVT $|\cos x -\cos y| =|x-y||\sin t|$ for some $t$ and $|\sin t | \leq 1$. So you can take $K=3$.