$e^{z\sqrt{1-t}}=\sum \limits_{k=0}^\infty \frac{u_k(z)t^k}{k!}$
$\frac{\partial}{\partial t }(e^{z\sqrt{1-t}})=\frac{\partial}{\partial t }(\sum \limits_{k=0}^\infty \frac{u_k(z)t^k}{k!})$
$\frac{-z}{2\sqrt{1-t} }e^{z\sqrt{1-t}}=\sum \limits_{k=1}^\infty \frac{u_k(z)t^{k-1}}{{(k-1)}!}=\sum \limits_{k=0}^\infty \frac{u_{k+1}(z)t^{k}}{{k}!}$
$\frac{-1}{2\sqrt{1-t} }e^{z\sqrt{1-t}}=\sum \limits_{k=0}^\infty \frac{u_{k+1}(z)}{z}\frac{t^{k}}{{k}!}$
$\frac{\partial}{\partial z }(\frac{-1}{2\sqrt{1-t} }e^{z\sqrt{1-t}})=\frac{\partial}{\partial z }(\sum \limits_{k=0}^\infty \frac{u_{k+1}(z)}{z}\frac{t^{k}}{{k}!})$
$\frac{-e^{z\sqrt{1-t}}}{2}=\sum \limits_{k=0}^\infty \frac{\partial}{\partial z }(\frac{u_{k+1}(z)}{z})\frac{t^{k}}{{k}!}$
$\sum \limits_{k=0}^\infty \frac{u_k(z)t^k}{k!}=-2\sum \limits_{k=0}^\infty \frac{\partial}{\partial z }(\frac{u_{k+1}(z)}{z})\frac{t^{k}}{{k}!}$
$u_k(z)=-2\frac{\partial}{\partial z }(\frac{u_{k+1}(z)}{z})$
for $t=0$,$u_0(z)=e^z$
How to find the general formula of $u_k(z)$ ? I would like to learn the methods to solve such differential equations.
Thanks a lot for answers.
EDIT:
We can find $u_1(z)$ as shown below
$\frac{\partial}{\partial t }(e^{z\sqrt{1-t}})=\frac{-z}{2\sqrt{1-t} }e^{z\sqrt{1-t}}=\sum \limits_{k=1}^\infty \frac{u_k(z)t^{k-1}}{{(k-1)}!}$
for $t=0$, $u_1(z)=\cfrac{-ze^{z}}{2} $
If we continue to derivate in such way, we can find all $u_k(z)$ but it seems long method. I am looking for easier method.