We have $$\frac{dt}{t^{2}+1}=\frac{dx}{1}=\frac{du}{u}$$

Xi Yue is right until the last line. Instead we need to proceed as follows: $$\frac{du}{u}=dx\Rightarrow \ln(u)=x+d\Rightarrow u=De^{x}=\phi(x-\arctan(t))e^{x}$$

So for part c): $$u(x,0)=\phi(x)e^x=\cos(x)\Rightarrow \phi(x)=\cos(x)e^{-x}$$ since $\arctan(0)=0$. Therefore, the solution to the IVP is:

$$u(x,t)=\cos(x-\arctan(t))e^{\arctan(t)-x}e^{x}=\cos(x-\arctan(t))e^{\arctan(t)}$$

The solution is fully determined because the original equation was first order in time, with no restrictions on x. Therefore one initial condition uniquely defines a solution.