How can I prove
$$ u(t,x)=\frac{1}{2}(\varphi(x+t)+\varphi(x-t))+\frac{1}{2}\int_{x-t}^{x+t}\psi(y)dy $$ satisfies the following Cauchy problem:
$$\begin{cases} \partial_{t}^{2}u-\partial_{x}^{2}u=0\\ u(0,x)=\varphi(x)\\ \partial_{t}u(0,x)=\psi(x) \end{cases} $$
Here are my results (so far):
$$ \begin{split} \partial_{t}u &= \frac{1}{2}\varphi_{t}^{,}(x+t)-\frac{1}{2}\varphi_{t}^{,}(x-t)+\frac{1}{2}\psi(x+t)+\frac{1}{2}\psi(x-t)\\ \partial_{t}^{2}u &= \frac{1}{2}\varphi_{tt}^{,,}(x+t)-\frac{1}{2}\varphi_{tt}^{,,}(x-t)+\frac{1}{2}\psi_{t}^{,}(x+t)-\frac{1}{2}\psi_{t}^{,}(x-t)\\ \partial_{x}u &= \frac{1}{2}\varphi_{x}^{,}(x+t)+\frac{1}{2}\varphi_{x}^{,}(x-t)+\frac{1}{2}\psi(x+t)-\frac{1}{2}\psi(x-t)\\ \partial_{x}^{2}u &= \frac{1}{2}\varphi_{xx}^{,,}(x+t)+\frac{1}{2}\varphi_{xx}^{,,}(x-t)+\frac{1}{2}\psi_{x}^{,}(x+t)-\frac{1}{2}\psi_{x}^{,}(x-t) \end{split} $$ Substituting into $\partial_{t}^{2}u-\partial_{x}^{2}u=0\\$, I got $\varphi_{tt}^{,,}-\varphi_{xx}^{,,}=0$.
What did I missed?
Substituting $t$ into the derivatives I also got the initial values therefore it must be correct, but how do I know $\varphi_{tt}^{,,}=\varphi_{xx}^{,,}$?
I tried to do something with the operators: $\partial_{t}^{2}-\partial_{x}^{2}=(\partial_{t}-\partial_{x})(\partial_{t}+\partial_{x})$ but it didnt help.
Sorry if the question is obvious but I really can't solve it.
The $-\frac{1}{2}\varphi_{tt}^{,,}(x-t)$ term should be positive. You forgot to multiply by negative one.
Also, the notation $\varphi_{tt}^{,,}(x-t)$ is slightly confusing as you didn't specify that it means to take two partial derivatives with respect to $t$. It might be best to stick with $\varphi_{tt}$ or $\varphi''$.
We need to show that d'Alembert's formula
$$ u(t,x)=\frac{1}{2}\big(\varphi(x+t)+\varphi(x-t)\big)+\frac{1}{2}\int_{x-t}^{x+t}\psi(y)dy $$
satisfies the Cauchy problem
$$\begin{cases} \partial_{t}^{2}u-\partial_{x}^{2}u=0\\ u(0,x)=\varphi(x)\\ \partial_{t}u(0,x)=\psi(x) \end{cases} $$
Let's start with the second condition. We are given
$$ u(0,x)=\frac{1}{2}\big(\varphi(x)+\varphi(x)\big)+\frac{1}{2}\int_{x}^{x}\psi(y)dy=\varphi(x) $$
so this condition is satisfied. Next, for the third condition we apply the fundamental theorem of calculus
$$\partial_{t}u(t,x)=\frac{1}{2}\big(\varphi'(x+t)-\varphi'(x-t)\big)+\frac{1}{2}\big(\psi(x+t)+\psi(x-t)\big)$$
therefore
$$\partial_{t}u(0,x)=\frac{1}{2}\big(\varphi'(x)-\varphi'(x)\big)+\frac{1}{2}\big(\psi(x)+\psi(x)\big)=\psi(x)$$
which satisfies the third condition. To verify the first condition, we calculate three more partial derivatives
$$\partial_{t}^{2}u(t,x)=\frac{1}{2}\big(\varphi''(x+t)+\varphi''(x-t)\big)+\frac{1}{2}\big(\psi'(x+t)-\psi'(x-t)\big)$$ $$\partial_{x}u(t,x)=\frac{1}{2}\big(\varphi'(x+t)+\varphi'(x-t)\big)+\frac{1}{2}\big(\psi(x+t)-\psi(x-t)\big)$$ $$\partial_{x}^{2}u(t,x)=\frac{1}{2}\big(\varphi''(x+t)+\varphi''(x-t)\big)+\frac{1}{2}\big(\psi'(x+t)-\psi'(x-t)\big)$$
from which we see that
$$\partial_{t}^{2}u(t,x)-\partial_{x}^{2}u(t,x)=0$$
as required.