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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).

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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.

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