While I know this is an extremely well known result, I was trying to derive the expression for $ \langle{x}|{p}\rangle$, as done in Sakurai. It is basically a two line equation, so our starting point is $$\langle x'|p|p' \rangle= -i \hbar \frac{\partial}{\partial x'}\langle x'|p' \rangle. $$ which then leads to $$ p'\langle x'|p' \rangle=-i \hbar \frac{\partial}{\partial x'}\langle x'|p' \rangle. $$ Now, the solution of this differential equation is given to be, $$ \langle x'|p' \rangle=N \text{exp}\Big(\frac {i p' x'}{\hbar}\Big).$$ While it's easy for me to understand the exponential bit of the solution on the RHS, as far as I remember the normalisation constant $N$ should be allowed to be a function of the momentum $p$, because the differential equation is basically of the form $$\frac{\partial f(x,p)}{\partial x}= c(p) f(x,p). $$ Did I get the math wrong or is there another reason for why $N$ is taken as a constant independent of $p$ and $x$?
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