Eq (2.5.7) and Eq (2.5.12) of Sakurai say

[itex]\psi(x'',t) = \int d^3x' K(x'',t;x',t_0)\psi(x',t_0)[/itex]

and

[itex]\left(H-i\hbar\frac{\partial}{\partial t}\right)K(x'',t,x',t_0) = -i\hbar\delta^3(x''-x')\delta(t-t_0)[/itex]

We know from the basic Schrodinger equation

[itex]\left(H-i\hbar\frac{\partial}{\partial t}\right)\psi(x,t) = 0 [/itex]

So, I applied the differential operator to Eq (2.5.7) and use Eq(2.5.12). But, I couldn't get the right Schrodinger equation like this.

[itex]\left ( H - i \hbar \frac{\partial}{\partial t}\right ) \psi (x'',t) = \left ( H - i \hbar \frac{\partial}{\partial t}\right ) \int dx' K(x'',t;x',t_0) \psi(x',t_0) = \int \left [ \left ( H - i \hbar \frac{\partial}{\partial t}\right ) K(x'',t;x',t_0) \right ] \psi(x',t_0) dx' = -i \hbar \int \psi(x', t_0) \delta(x''-x') \delta(t-t_0)dx'[/itex]

[itex]=-i \hbar \psi(x'',t_0) \delta(t-t_0)[/itex]

which is non zero at t=t_0

What is the point that I am missing?