Let $$g(x,t)=P(N_t=x)$$ where $$N_t $$ refers to the number of times a particular case occurs from 0 to t. (Poisson process) $$\\$$ How to solve this differential equation below $$\frac{d}{dt}\{e^{\lambda t} g(x,t)/\lambda^x\} = e^{\lambda t}g(x-1,t)/\lambda^{x-1}$$
given $$g(0,0)=1 ,\ \ \ g(1,0) = g(2,0)=...=0$$
That's just the generator equation of the poisson process, so if you know the formula for a poisson random variable, then you have the explicit solution.
I'll assume that the point of the exercise is to assume that you don't know the explicit formula, and then derive it using that PDE as the starting point. Here's how to do that. Let $h(t,x) = e^{\lambda t} g(t,x)/\lambda^x$. The equation reads $$\partial_t h = Jh,$$ where $J$ is the linear operator given by $Jh(x):= h(x-1)$. This is just a linear equation in infinite variables so the solution at time $t$ will simply be given by $$h_t = e^{ t J} h_0.$$ Here $e^A:=\sum_{k \ge 0} \frac{A^k}{k!}$ denotes the matrix exponential. On our case $A = tJ$ so that $A^nh(x) = t^n h(x-n)$. Thus $$h_t(x) = \sum_{n \ge 0} \frac{ t^n}{n!} h_0(x-n).$$In your case $h_0(y) = 1_{\{y=0\}}$, so that we simply get $h_t(x) = t^x/x!$ for $x \ge 0$. Thus $g(t,x) = \lambda^x e^{-\lambda t}h(t,x) = \lambda^xt^xe^{-\lambda t}/x!$, which is the correct formula for a Poisson variable.