Suppose we have a time dependent Hamiltonian $H(t)$ such that $H(t)=H^{(0)}+\delta H(t)$.
$H^{(0)}$ is a known Hamiltonian and is time independent.
Now define $|\tilde\psi(t)\rangle$ as
$|\tilde\psi(t)\rangle=exp\Big(\frac{i}{\hbar}H^{(0)}t\Big)|\psi(t)\rangle \tag{1}$
Now,
$i\hbar\frac{d}{dt}|\tilde\psi(t)\rangle=-H^{(0)}|\tilde\psi(t)\rangle+exp\Big(\frac{i}{\hbar}H^{(0)}t\Big)(H^{(0)}+\delta H(t))|\psi(t)\rangle$
Solving this, we get
$i\hbar\frac{d}{dt}|\tilde\psi(t)\rangle=exp\Big(\frac{i}{\hbar}H^{(0)}t\Big)\delta H(t) exp\Big(\frac{-i}{\hbar}H^{(0)}t\Big)|\tilde\psi(t)\rangle$
So,
$\boxed{i\hbar\frac{d}{dt}|\tilde\psi(t)\rangle=\tilde{\delta H(t)}|\tilde\psi(t)\rangle} \tag{2}$
I have a doubt with the expression $(2)$
We can write $\psi(t)$ as
$|\psi(t)\rangle=exp\Big(-\frac{i}{\hbar} H^{(0)}t\Big)exp\Big(-\frac{i}{\hbar}\int_0^t\delta H(t')dt'\Big)|\psi(0)\rangle$
From $(2)$,
$|\tilde\psi(t)\rangle=exp\Big(-\frac{i}{\hbar}\int_0^t\delta H(t')dt'\Big)|\psi(0)\rangle \tag{3}$
So, $|\tilde\psi(t)\rangle$ is the wave function which is acted upon by only $\delta H(t)$. The effect of the original Hamiltonian $H^{(0)}$ has been removed from it.
Now using $(3)$,
$\boxed{i\hbar\frac{d}{dt}|\tilde\psi(t)\rangle=\delta H(t)|\tilde\psi(t)\rangle}\tag{4}$
We can see that $(2)$ and $(4)$ are not same. Why this is so? Derivation-wise both seems correct Also $(4)$ is much more intuitive than $(2)$ because as seen from the expression of $(2)$, the effect of $H^{(0)}$ has been removed.