# Need help with exact solution and proof of well-posedness for linear first order IVP

Apologies for any mistakes or inefficiencies in my submission and wording of the question. It is my first time using stackexchange.

The IVP is as follows:

$$u' = \cos(u(t))$$, subject to the initial conditions $$u(0) = 0$$, where $$0\le t\le 1$$.

I separated the variables, solved for the general solution, and applied the initial conditions giving me the solution $$u(t) = 2\arctan(e^{t + \log(\tan(\pi))}) - \frac{\pi}{2}$$

According to wolfram alpha, however, the correct solution is $$u(t) = 2\arctan(\tanh(\frac{t}{2}))$$.

I would like to see where my error is as well as how to show well-posedeness of the problem. I have already shown the solution to exist and be unique, however, I don't quite know how to show that it's behaviour changes continuously with respect to the initial conditions(stability).

It is difficult to find where your error is if you do not show your work. However, something is definitely wrong, since $$\log(\tan\pi)$$ is undefined.
Let's check now the well poshness. Let $$u(x,a)$$ be the solution with initial value $$u(0)=a$$. Then \begin{align} |u(x,a)-u(x,b)|&=\Bigl|a-b+\int_0^x\bigl(\cos(u(t,a))-\cos(u(t,b))\bigr)\,dt\Bigr|\\ &\le|a-b|+\int_0^x|\cos(u(t,a))-\cos(u(t,b))|\,dt\\ &\le|a-b|+\int_0^x|u(t,a)-u(t,b)|\,dt. \end{align} Now use Gronwall's inequality.