The differential operator on the left is $(D-3{)}^2$, where $D=d/dt$. Its repeated characteristic root suggests writing $y=e^{3t}v$. By the product rule,so the ordinary differential equation reduces to $v^{\prime \prime }=3(t+1)$. Two antiderivatives giveThe boundary conditions imply $v(0)=1$ and $v(1)=0$, hence $B=1$ and $A=-3$. ThereforeThe calculation of $v^{\prime \prime }$ verifies the ordinary differential equation, and the bracket takes the values one and zero at the two endpoints. These boundary conditions determine both constants uniquely.

Let $X_1$ and $X_2$ denote the first and second labels. Sampling with replacement restores the bag before the second draw, so conditioning on $X_1=x$ only specifies the threshold. For $1\le x\le g$, there are $x$ qualifying green balls and $x$ qualifying red balls. For $g<x\le r$, every green ball and the first $x$ red balls qualify. Hence the requested conditional probability isThe formulas agree at $x=g$ and give one at $x=r$. Labels outside the possible first-draw range do not define a conditioning event of positive probability.

Write $N=g+r$. Successive conditional probabilities account for the changing bag under sampling without replacement:On this event no equality of the two colour counts has occurred, so the game's stopping rule does not interrupt these three draws.