 Solution

With $K=0$, a JK flip-flop obeys $Q^{+}=J\vee Q$ at its positive clock edge: $J=0$ holds its old state and $J=1$ sets it. Thus each original stage is a sticky record of a previously sampled high input until Start clears it. A D flip-flop instead obeys $Q^{+}=D$, so a later sample of zero can clear the output.

Connect each replacement D input where the corresponding J input was connected, preserving the clocks and asynchronous clear. The first valid sequence after Start still sets both outputs. Afterwards, let $X_0$ fall and apply another rising clock edge to $X_1$. The D version clears $Y_0$; the JK version retains $Y_0=1$. A subsequent rising clock edge of $X_2$ can therefore clear $Y_1$ only in the D version. Neither circuit continuously tracks its data input between clock edges.