= Solution
Eliminate the juvenile variable from the <stage-structured population model>. Since $y_{t+1}=2z_t$,
$$
z_{t+2}=\tfrac34 y_{t+1}+\tfrac12z_{t+1}
=\tfrac32z_t+\tfrac12z_{t+1}.
$$
The <characteristic polynomial> of this <linear recurrence> is
$$
q^2-\tfrac12q-\tfrac32=(q-\tfrac32)(q+1),
$$
so $z_t=A(3/2)^t+B(-1)^t$. The <initial conditions> give $z_0=0$ and $z_1=(3/4)10=15/2$, hence $A+B=0$ and $(3/2)A-B=15/2$. Therefore
$$
\boxed{z_t=3\left[\left(\frac32\right)^t-(-1)^t\right],\qquad t=0,1,2,\ldots}.
$$
The alternating term records the initial imbalance between life stages; the growing $(3/2)^t$ term dominates eventually.
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