let $b \in \mathbb{R} $be constant .Solve:

$$\frac{\partial u}{\partial t} (x,t) + b \frac{ \partial u}{\partial x}(x,t) =0 , \quad x \in \mathbb{R}, \quad t >0$$

$$ u(x,0) = x^2$$

My attempt : $\frac{\partial}{\partial t} u (x,t) =- b \frac{ \partial u}{\partial x}(x,t) $ , Now again differenting both l.h.S and r.h.s , we get $$\frac{\partial^2}{\partial t^2} (x,t) =- b \frac{ \partial^2 u}{\partial x^2}(x,t) $$

after that applying the D alembert formula we have $$u(x,t) = \frac{(x+ bt)^2 +(x-bt)^2}{2} + \frac{1}{2b}\int\_{x-bt}^{x+bt}0\,{\rm d}s.$$

Is its true ?