$$u_t+\frac{x}{t}u_x=0,\hspace{20mm}t>1,-\infty<x<\infty,$$ where the initial data is given at $t=t_0$ at $t=1$ and we have $u(x,t=1)=u_{in}(x)=e^{-x^2}$.
I used method of characteristics. I find that $\frac{dx}{dt}=\frac{x}{t}\implies x=ct$. Now, I have $c=\frac{x}{t}$, so my solution would be $u(x,t)=u_{in}(\frac{x}{t})=e^{-(\frac{x}{t})^2}$. Could someone please tell me if I am correct?
Why asking if your solution is correct ? You can check yourself : $$u(x,t)=e^{-(x/t)^2}$$ $$u_x=\frac{-2x}{t^2}e^{-(x/t)^2}$$ $$u_t=\frac{2x^2}{t^3}e^{-(x/t)^2}$$ $$u_t+\frac{x}{t}u_x=\left(\frac{2x^2}{t^3}e^{-(x/t)^2}\right)+\frac{x}{t}\left(\frac{-2x}{t^2}e^{-(x/t)^2}\right)=0$$ Thus $u=e^{-(x/t)^2}$ is solution of the PDE. $$u(x,1)=e^{-(x/1)^2}=e^{-x^2}$$ Thus $u=e^{-(x/t)^2}$ satisfies the condition.