Substituting the expressions for Ct and It into
Yt
= C
t + It
gives
Yt
= (0.8Yt−1 + 100) + 200
= 0.8Y
t−1 + 300
The complementary function is given by
CF = A(0.8)t
and for a particular solution we try
Yt
= D
for some constant D. Substituting this into the difference equation gives
D = 0.8D + 300
which has solution D = 1500. The general solution is therefore
Yt
= A(0.8)t + 1500
The initial condition,
Y0
= 1700
gives
1700 = A(0.8)0 + 1500 = A + 1500
and so A is 200. The solution is
Yt
= 200(0.8t) + 1500
As t increases, (0.8)t converges to zero, and so Yt eventually settles down at the
equilibrium level of 1500. The system is therefore stable. Note also that because 0.8 lies
between 0 and 1, the time path displays uniform convergence.
Drag each expression to the correct location on the solution. not all expressions will be used. consider the polynomial 8x + 2x2 - 20x - 5. factor by grouping to write the polynomial in factored form.