Chapter 11

Needs["Graphics`PlotField`"]

Clear[y, β]

Question 1

(i)

sol = 
      Simplify[DSolve[{y'[
                t] == β f[u] - β (1 - ξ) y[t], 
            y[0] == y0}, y[t], t]]

Simplify[
      Solve[0 == β f[u] - β (1 - ξ) ystar, 
        ystar]]

x[t_] := sol[[1, 1, 2]]

x[t]

x2[t_] := 
      Collect[x[t], {(E^(t β (-1 + ξ))), y0}]

x2[t]

Using the algebraic manipulations palette, namely the simplify command, we readily obtain the following expression:

x3[t_] := 
      f[u]/(1 - ξ) + 
        (E^(t β (-1 + ξ))) (y0 + 
              f[u]/(-1 + ξ))

x3[t]

(ii)

Note that this expression is,
	\(TraditionalForm
      \`y(t) = \(y\^*\) + 
          \(E\^\(\(-\(β(1 - ξ)\)\) t\)\)(y0 - \(y\^*\))\)
where
	\(TraditionalForm\`\(y\^*\) = \(f(u)\)\/\(1 - ξ\)\)
and is asymptotically stable so long as β(1-ξ) < 1.

Question 2

For the system,
	(Subscript[y, t - 1]) = 9 + 0.2 (5 - (Subscript[p, t - 1]))(Subscript[π, t]) = α((((Subscript[y, t - 1]) - 6)/6))
we can solve for (Subscript[p, t]).

Simplify[
      Solve[((p - p1)/p1) == α ((((9 + (1/5) (5 - p1)) - 6)/6)), p]]

We can solve for p as follows,

Solve[p == p + ((2  p α)/3) - ((α (p^2))/30), 
      p]

which is independent of α.
	The difference equation in terms of p and α is given by,
	\(TraditionalForm
      \`p\_t = \(f(p\_\(t - 1\)) = 
          \((1 + \(2  α\)\/3)\) p\_\(t - 1\) - 
            \((α\/30)\) p\_\(t - 1\)\%2\)\)
whose stability we can investigate by considering,

slope = 
      D[p1 + ((2 p1 α)/3) - (((p1^2) α)/30), p1]

eqslope = slope  /. p1 -> 20

Plot[1 - ((2  α)/3), {α, 0, 6}];

Solve[1 - ((2  α)/3) == 0, α]

Solve[1 - ((2  α)/3) == -1, α]

Solve[1 - ((2  α)/3) == 1, α]

Thus, the system is asymptotically stable if 0 < α < 3. The system converges steadily on \(TraditionalForm\`\(p\^*\) = 20\), if 0 < α < 3/2 and oscillates to equilibrium if 3/2 < α < 3. Beyond α = 3 the system gives rise to an explosive oscillation. To see these alternatives, let α = {1.2, 2, 3.5}.

Clear[f]

f[p_] := p + ((2  p α)/3) - ((α (p^2))/30)

pathp1 = NestList[f, 10, 10]  /. α -> 1.2;

ListPlot[pathp1, PlotJoined -> True,  
        AxesLabel -> {"t", "p(t)"},  
        PlotLabel -> "α=1.2"];

pathp2 = N[NestList[f, 10, 10]]  /. α -> 2;

ListPlot[pathp2, PlotJoined -> True,  
        AxesLabel -> {"t", "p(t)"},  
        PlotLabel -> "α=2"];

pathp3 = N[NestList[f, 10, 10]]  /. α -> 3.5;

ListPlot[pathp3, PlotJoined -> True,  
        AxesLabel -> {"t", "p(t)"},  
        PlotLabel -> "α=3.5"];

Question 3

Let

matrixA=(({{-1.85, -10}, {0.3, 0}}))

sol = Eigenvalues[matrixA]

Re[ sol[[1]] ]

Im[ sol[[1]] ]

Re[ sol[[2]] ]

Im[ sol[[2]] ]

Clear that the real part of the complex conjugate roots is α = -0.925.

Question 4

Clear[rhs, p, pe]

pe1 = 
      Solve[pe[t + 1] == pe[t] + (1 - λ) (p[t] - pe[t]), 
        pe[t + 1]]

newpe1 = Collect[pe1[[1, 1, 2]], p[t]]

rhs[
        i_] := (1 - λ) p[t - i] + λ pe[t - i]

rhs[i]

Table[rhs[i], {i, 0, 3}]  // TableForm

Collect[
      Simplify[(1 - λ) 
            p[t] + λ ((1 - λ) p[-1 + 
                      t] + λ pe[-1 + t])], {p[t], 
        p[t - 1]}]

Collect[
      Simplify[(1 - λ) λ p[-1 + 
                t] + (1 - λ) p[
              t] + (λ^2) ((1 - λ) p[-2 + 
                      t] + λ pe[-2 + t])], {p[t], p[t - 1], 
        p[t - 2]}]

Collect[ 
      Simplify[(1 - λ) (λ^2) p[-2 + 
                t] + (1 - λ) λ p[-1 + 
                t] + (1 - λ) p[
              t] + (λ^3) ((1 - λ) p[-3 + 
                      t] + λ pe[-3 + t])], {p[t], p[t - 1], 
        p[t - 2], p[t - 3]}]

This is the same as the expression,

(1 - λ) (∑_(k = 0)^3) (λ^k) 
            p[t - k] + (λ^4) pe[t - 3]

Taking the series to the limit, we have
	\(TraditionalForm
      \`\(E\_t\) P\_\(t + 1\) = 
        \((1 - λ)\) 
          \(∑\+\(k = 0\)\%∞\( λ\^k\) P[t - k]\)\)

Solve[b ((ep/p) - 1) == a - (M/p), p]

((b ep + M)/(a + b))  /.  
      ep -> (1 - λ) (∑_(k = 0)^∞) (λ^k) p[t - k]

Therefore,
	(Subscript[P, t]) = ((Subscript[M, t])/(a + b)) + ((b/(a + b)))\(TraditionalForm
      \`\((1 - λ)\) 
        \(∑\+\(k = 0\)\%∞\( λ\^k\) P\_\(t - k\)\)\)

Question 5

(i)

Setting (Subscript[y, t]^d) = (Subscript[y, t]^s) = (Subscript[y, t]) and solving for (Subscript[y, t]) and (Subscript[p, t]) under the assumption of fixed expectations. In what follows we let ep denote the expression (Subscript[E, t - 1]) (Subscript[p, t]).

solpy = 
      Simplify[Solve[{y == a0 + a1 (m - p), 
            y == yn + b1 (p - ep)}, {y, p}]]

Next we impose the money supply following a systematic component m = μ0.

solpy  /. {m -> μ0}

Since,
	\(TraditionalForm
      \`p\_t = \(a0 - yn\)\/\(a1 + b1\) + \(a1\ μ0\)\/\(a1 + b1\) + 
          \(b1\ \(E\_\(t - 1\)\) p\_t\)\/\(a1 + b1\)\)
then
	(Subscript[E, t - 1]) (Subscript[p, t]) = \(TraditionalForm
      \`\(a0 - yn\)\/\(a1 + b1\) + \(a1\ μ0\)\/\(a1 + b1\) + 
        \(b1\ \(E\_\(t - 1\)\) p\_t\)\/\(a1 + b1\)\)

Simplify[
      Solve[ep == ((a0 + b1 ep - yn + a1 μ0)/(a1 + b1)), 
        ep]]

or
	\(TraditionalForm
      \`\(E\_\(t - 1\)\) p\_t = \((\(a0 - yn\)\/a1)\) + μ0\)
Substituting this into the solution for y, we obtain

Simplify[
      Solve[{y == ((a0 b1 + a1 (yn + b1 (-ep + μ0)))/(a1 + b1)), ep == ((a0 - yn + a1 μ0)/a1)}, {y, 
          ep}]]

(ii)

solpy  /. {m -> μ0 + z}

(Subscript[E, t - 1]) (Subscript[p, t]) is as before since (Subscript[E, t - 1]) (Subscript[z, t]) = 0.

Simplify[ 
      Solve[{y == ((a0 b1 + a1 (yn + b1 (-ep + z + μ0)))/(a1 + b1)), ep == ((a0 - yn + a1 μ0)/a1)}, {y, 
          ep}] ]

Or,
	(Subscript[y, t]) = yn + ((a1 b1 (Subscript[z, t]))/(a1 + b1)).
Income deviates from its natural level only in so far as changes in the money supply are unexpected.

Question 6

(i)

Since,
	\(TraditionalForm
      \`r\ q = \ 
        \(\(max\ \)\+λ\) {
            x(e) - a\ λ - w - s\ q + \(\(q\ }\)\& . \)\)
and
	w = b + (r + s + h(q, e)) (b/λ)
then
	\(TraditionalForm
      \`r\ q = x(e) - b - s\ q + \(q\& . \) - 
          \(min\+λ\) {
              a\ λ + \((r + s + h(q, e))\) \((b/λ)\)\)}
Consequently the value of λ which maximises r q is that which minimises the expression in curly braces. Let this be denoted z. Then,

z = a λ + (r + s + h[q, e]) (b/λ)

D[z, λ]

D[z, {λ, 2}]

This last expression is positive. So the result of setting the first derivative to zero gives a minimum.

Simplify[Solve[D[z, λ] == 0, λ]]

Ignoring the negative value of λ, then
	\(TraditionalForm
      \`λ = \(\((b\/a)\)\^\(1\/2\)\)[r + s + h(q, e)]\^\(1\/2\)\).

(ii)

Simplify[
      Solve[ {w == 
            b + (r + s + h[q, e]) 
                (b/λ), λ == (Sqrt[(b/a)]) (Sqrt[r + s + h[q, e]])}, {w, λ}]]

Or,
	\(TraditionalForm
      \`w = b + \(\((a\ b)\)\^\(1\/2\)\)[r + s + h(q, e)]\^\(1\/2\)\)

To solve for profit π(q, e) we note that,
	\(TraditionalForm
      \`π(q, e) = \(MRP\_L - w = x(e) - a\ λ - b - \)\)\(TraditionalForm
      \`\(\((a\ b)\)\^\(1\/2\)\)[r + s + h(q, e)]\^\(1\/2\)\)

Clear[x]

Simplify[
      x[e] - a ((Sqrt[(b/a)]) (Sqrt[r + s + h[q, e]])) - 
        b - (Sqrt[a b]) (Sqrt[r + s + h[q, e]]) ]

Hence,
	\(TraditionalForm
      \`π(q, e) = 
        x(e) - b - 2 \(\((a\ b)\)\^\(1\/2\)\)[r + s + h(q, e)]\^\(1\/2\)\).

Question 7

(i)

Clear[m, u, v, e, q, h, c]

Solve[{m == ((u v)^(1/4)), ((m q)/v) == c }, {m, v}]

Taking only the postive value for v, we have
	v = ((((q/c)))^((4/3))) (u^((1/3))).

(ii)

Since,
	\(TraditionalForm
      \`h(q, e) = 
        \(\(m(u, v)\)\/u = 
          \(\((u\^\(1\/4\)\ v\^\(1\/4\))\)/u = 
            \(u\^\(\(-3\)\/4\)\) v\^\(1\/4\)\)\)\),	v = ((((q/c)))^((4/3))) (u^((1/3))),	u = 1 - e
then h(q, e) is equal to

h[q, e] = 
      Simplify[(((1 - e))^(-3/4)) (((q/c))^(1/
                3)) (((1 - e))^(1/12))]

(iii)

D[h[q, e], q]

D[h[q, e], e]

which are both positive.

(iv)

Since,
	\(TraditionalForm\`\(e\& . \) = \((1 - e)\) \(h(q, e)\) - s\ e\)
then

edot = (1 - e) h[q, e] - s e

Simplify[Solve[edot == 0, q]]

Question 8

Clear[m, u, v, e, c, q]

(i)

Given,
	m = (Sqrt[u v])	and	((m/u)) q = c
and noting that u = 1 - e, then

Simplify[
      Solve[{((Sqrt[u v])/u) q == c, u == 1 - e}, {v, u}]]

Hence,
	v = (1 - e) ((((c/q)))^2)

(ii)

Simplify[
      Solve[{h == (Sqrt[(v/((1 - e)))]), v == (1 - e) ((((c/q)))^2)}, {h, 
          v}]]

Thus, h(q, e) = (c/q) and is independent of e.

Question 9

Clear[a, s, β, λ, δ, n, x]

kdot = s a (k^β) - (n + δ) k

kdot0 = 
      kdot  /. {a -> 2, 
          s -> 0.2, β -> 0.25, λ -> 0.05, δ -> 0.03, 
          n -> 0.02}

NSolve[kdot0 == 0, k]

Since,
	x = y (r^(((-1)/4))),	y = f(k) = a (k^β)	\(TraditionalForm
      \`\(f\^′\)(k) = β\ a\ k\^\(β - 1\)\)
then

Solve[x == a (k^β) (r^(((-1)/4))), r]

xdot = (β a (k^(β - 1)) + λ - δ - 
            n - (((a^4) (k^(4 β)))/(x^4))) x

xdot0 = 
      xdot  /. {a -> 2, 
          s -> 0.2, β -> 0.25, λ -> 0.05, δ -> 0.03, 
          n -> 0.02}

NSolve[(0.5/(k^0.75)) - ((16 (k^1.) )/(x^4)) == 0, x]

The only positive real solution for x is \(TraditionalForm
      \`x = \(2.37841  k\^\(7/16\) = 2.37841  k\^0.4375\)\).

NSolve[{x == 2.37841  (k^0.4375), k == 16}, {k, x}]

Equilibrium k is shown by the intersection of s f (k) and (n + δ) k.

initdiag = 
        Plot[{0.05  k,  0.4  (k^0.25)}, {k, 0, 30},  
          AxesLabel -> {"k", "y"}];

The dynamics are shown by the direction-field diagram. In plotting this we first derive the isoclines and plot these along with the equilibrium point. These are then superimposed on the direction-field diagram. It is important when displaying the direction field to include the option ScaleFunction->(1&). What this does is make all vectors of equal length.

xdot0 = Plot[2.37841  (k^0.4375), {k, 0, 40}];

linept0 = 
        Graphics[{Line[{{16, 0}, {16, 12}}], PointSize[0.02], 
            Point[{16, 8}]}];

isoclines0 = 
        Show[linept0, xdot0,  Axes -> True, 
          AxesLabel -> {"k", "x"}];

field0 = 
        PlotVectorField[{0.4  (k^0.25) - 
              0.05  k, ((0.5/(k^0.75)) - ((16  k)/(x^4))) x}, {k, 0.1, 
            40}, {x, 0.1, 12}, Frame -> True, ScaleFunction -> (1 &), 
          AspectRatio ->  .8];

Show[isoclines0, field0];

(i) Rise in s from 0.2 to 0.3

kdot1 = 
      kdot  /. {a -> 2, 
          s -> 0.3, β -> 0.25, λ -> 0.05, δ -> 0.03, 
          n -> 0.02}

NSolve[kdot1 == 0, k]

xdot1 = 
      xdot  /. {a -> 2, 
          s -> 0.3, β -> 0.25, λ -> 0.05, δ -> 0.03, 
          n -> 0.02}

NSolve[(0.5/(k^0.75)) - ((16 (k^1.) )/(x^4)) == 0, x]

Which is the same isocline for \(TraditionalForm\`\(x\& . \) = 0\).

NSolve[{x == 2.37841  (k^0.4375), k == 27.4731}, {k, x}]

diag1 = 
        Plot[{0.05  k, 0.4  (k^0.25), 0.6  (k^0.25)}, {k, 0, 40},  
          AxesLabel -> {"k", "y"}];

linept1 = 
        Graphics[{Line[{{16, 0}, {16, 12}}], 
            Line[{{27.4731, 0}, {27.4731, 12}}], PointSize[0.02], 
            Point[{16, 8}],  Point[{27.4731, 10.1347}]}];

isoclines1 = 
        Show[linept0, linept1, xdot0,  Axes -> True, 
          AxesLabel -> {"k", "x"}, 
          PlotLabel -> "Rise in s"];

field1 = 
        PlotVectorField[{0.6  (k^0.25) - 
              0.05  k, ((0.5/(k^0.75)) - ((16  k)/(x^4))) x}, {k, 0.1, 
            40}, {x, 0.1, 12}, Frame -> True, ScaleFunction -> (1 &), 
          AspectRatio ->  .8];

Show[isoclines1, field1];

Note that a rise in s shifts only s f(k) in (k,y)-space and only the \(TraditionalForm\`\(k\& . \) = 0\) isocline in (k,x)-space.

(ii) A rise in n from 0.02 to 0.03

kdot2 = 
      kdot  /. {a -> 2, 
          s -> 0.2, β -> 0.25, λ -> 0.05, δ -> 0.03, 
          n -> 0.03}

NSolve[kdot2 == 0, k]

xdot2 = 
      xdot  /. {a -> 2, 
          s -> 0.2, β -> 0.25, λ -> 0.05, δ -> 0.03, 
          n -> 0.03}

NSolve[(-0.01 + (0.5/(k^0.75)) - ((16 (k^1.) )/(x^4))) x == 0, 
      x]

NSolve[{x == (((4.47214 + 4.47214  I) (k^(7/16)))/(((-50 + (k^(3/4))))^(1/4))), k == 12.5471}, {k, x}]

diag2 = 
        Plot[{0.05  k, 0.06  k, 0.4  (k^0.25)}, {k, 0, 40}, 
          AxesLabel -> {"k", "y"}];

xdot2 = 
        Plot[(((4.47214 + 4.47214  I) (k^0.4375))/(((-50 + (k^0.75)))^0.25)), {k, 0, 40}];

\(\(linept2 = 
        Graphics[{
			Line[{{16, 0}, {16, 12}}], 
            Line[{{12.5471, 0}, {12.5471, 12}}], PointSize[0.02], 
            Point[{16, 8}], Point[{12.5471, 7.45483}]
		}];\)\)

isoclines2 = 
        Show[linept0, linept2, xdot0, xdot2, Axes -> True,  
          AxesLabel -> {"k", "x"},  
          PlotLabel -> "Rise in n"];

field2 = 
        PlotVectorField[{0.4  (k^0.25) - 
              0.06  k, (-0.01 + (0.5/(k^0.75)) - ((16  k)/(x^4))) 
              x}, {k, 0.1, 40}, {x, 0.1, 12}, Frame -> True, 
          ScaleFunction -> (1 &), AspectRatio -> 0.8];

Show[isoclines2, field2];

Note that a rise in n shifts only (n + δ) k in (k,y)-space but shifts both isoclines in (k,x)-space.

(iii) A rise in technology (a rise in a from 2 to 5)

kdot3 = 
      kdot  /. {a -> 5, 
          s -> 0.2, β -> 0.25, λ -> 0.05, δ -> 0.03, 
          n -> 0.02}

NSolve[kdot3 == 0, k]

xdot3 = 
      xdot  /. {a -> 5, 
          s -> 0.2, β -> 0.25, λ -> 0.05, δ -> 0.03, 
          n -> 0.02}

NSolve[(1.25/(k^0.75)) - ((625 k)/(x^4)) == 0, x]

NSolve[{x == 4.72871 (k^(7/16)), k == 54.2884}, {k, x}]

diag3 = 
        Plot[{0.05  k,  0.4  (k^0.25), (k^0.25)}, {k, 0, 60}, 
          AxesLabel -> {"k", "y"}];

xdot3 = Plot[4.72871  (k^0.4375), {k, 0, 60}];

\(\(linept3 = 
        Graphics[{
			Line[{{16, 0}, {16, 30}}], \ 
            Line[{{54.2884, 0}, {54.2884, 30}}], PointSize[0.02], 
            Point[{16, 8}], Point[{54.2884, \ 27.1442}]
		}];\)\)

isoclines3 = 
        Show[linept0, linept3, xdot0, xdot3, Axes -> True, 
          AxesLabel -> {"k", "x"}, 
          PlotLabel -> "Rise in technology"];

field3 = 
        PlotVectorField[{(k^0.25) - 
              0.05  k, ((1.25/(k^0.75)) - ((625  k)/(x^4))) x}, {k, 0.1, 
            60}, {x, 0.1, 30},  Frame -> True,  ScaleFunction -> (1 &), 
          AspectRatio -> 0.8];

Show[isoclines3, field3];

Note that a rise in technology shifts only s f(k) in (k,y)-space but shifts both isoclines in (k,x)-space.

Question 10

This is a continuation of the previous question and we shall identify all plots with the number 4.

kdot4 = 
      kdot  /. {a -> 2, 
          s -> 0.2, β -> 0.25, λ -> 0.06, δ -> 0.03, 
          n -> 0.02}

NSolve[kdot4 == 0, k]

xdot4 = 
      xdot  /. {a -> 2, 
          s -> 0.2, β -> 0.25, λ -> 0.06, δ -> 0.03, 
          n -> 0.02}

NSolve[0.01 + (0.5/(k^0.75)) - ((16  k)/(x^4)) == 0, x]

NSolve[{x == ((6.32456  (k^0.4375))/(((50 + (k^0.75)))^0.25)), 
        k == 16}, {k, x}]

Since a change in λ affects neither (n + δ) k nor s f(k), then there is no change to equilibrium k, and the diagram in (k,y)-space remains unaffected. However, we do note that the \(TraditionalForm\`\(x\& . \) = 0\) isocline shifts.

xdot4 = 
        Plot[((6.32456  (k^0.4375))/(((50 + (k^0.75)))^0.25)), {k, 0, 
            30}];

\(\(linept4 = 
        Graphics[{
			Line[{{16, 0}, {16, 12}}], PointSize[0.02], 
            Point[{16, 8}], Point[{16, 7.70861}]
		}];\)\)

isoclines4 = 
        Show[linept0, linept4, xdot0, xdot4, Axes -> True, 
          AxesLabel -> {"k", "x"}, 
          PlotLabel -> "Rise in λ"];

field4 = 
        PlotVectorField[{0.4  (k^0.25) - 
              0.05  k, (0.01 + (0.5/(k^0.75)) - ((16  k)/(x^4))) x}, {k, 
            0.1, 40}, {x, 0.1, 12},  Frame -> True, 
          ScaleFunction -> (1 &), AspectRatio -> 0.8];

Show[isoclines4, field4];

The rise in λ shifts the \(TraditionalForm\`\(x\& . \) = 0\) isocline down, equilibrium k remains at 16 while equilibrium x falls from 8 to 7.7. The economy's trajectory, therefore, is vertically down from one equilibrium to the other.

Question 11

Clear[AA, BB, CC, DD, EE, FF, G, H, J]

(i) and (ii)

AA = ((-α (a + i0 + g))/(1 - b (1 - t) + (k h/u))) + α yn

BB = ((-α (h/u))/(1 - b (1 - t) + (k h/u)))

CC = ((-α h)/(1 - b (1 - t) + (k h/u))) - 
        1

parameters = {a -> 100, b -> 0.8, i0 -> 600, 
        yn -> 3000, g -> 525, t -> 0.25,  k -> 0.25,  
        h ->  2.5,  
        u -> 5,  α -> 0.2,  β -> 
          0.05}

AA  /.  parameters

BB  /.  parameters

CC  /.  parameters

Simplify[
      Solve[0 == 133.333 - 0.190476  ms - 1.95238 pie, 
        pie]]

DD = ((α β (a + i0 + g))/(1 - b (1 - t) + (k h/u))) - α β yn

EE = ((α β (h/u))/(1 - b (1 - t) + (k h/u)))

FF = ((α β h)/(1 - b (1 - t) + (k h/u)))

DD  /.  parameters

EE  /.  parameters

FF  /.  parameters

Simplify[
      Solve[0 == -6.66667 + 0.00952381  ms + 0.047619 pie, 
        pie]]

Solve[{0 == 133.333 - 0.190476  ms - 1.95238  pie, 
        0 == -6.666670 + 0.00952381  ms + 0.047619  pie}, {ms, 
        pie}]

i.e., equilibrium values are \(TraditionalForm\`\((\(ms\^*\), π\^\(\(e\)\(*\)\))\) = \(\((700, 
            0)\)\(.\)\)\)

mA = {{-0.190476, -1.95238}, {0.00952381, 0.047619}}

Tr[mA]

Det[mA]

Eigenvalues[mA]

Eigenvectors[mA]

(iii)

G = (((k/u) (a + i0 + g))/(1 - b (1 - t) + (k h/u)))

H = (((h/u) (k/u))/(1 - b (1 - t) + (k h/u))) - 
        (1/u)

J = (((k h/u))/(1 - b (1 - t) + (k h/u)))

G  /.  parameters

H /.  parameters

J  /. parameters

Simplify[
      Solve[0 == 116.667 - 0.152381  ms + 0.238095  pie, 
        pie]]

dfield = 
        PlotVectorField[{133.333 - 0.190476  ms - 
              1.95238  pie, -6.66667 + 0.00952381  ms + 
              0.047619  pie}, {ms, 600, 800}, {pie, -5, 5}, 
          Frame -> True, PlotPoints -> 20, AspectRatio -> 1, 
          DisplayFunction -> Identity,  
          ScaleFunction -> (1 &)];

sol = 
      NDSolve[{ms'[t] == 
            133.333 - 0.190476  ms[t] - 1.95238  pie[t], pie'[
              t] == -6.66667 + 0.00952381  ms[t] + 0.047619  pie[t], 
          ms[0] == 625, pie[0] == 0}, {ms, pie}, {t, 0, 
          100}]

traj = 
        ParametricPlot[
          Evaluate[{ms[t], pie[t] } /. sol, {t, 0, 100}, 
            PlotPoints -> 500], DisplayFunction -> Identity, 
          PlotStyle -> {RGBColor[1, 0, 0], 
              Thickness[0.01]}];

lines = 
        Plot[{68.2925 - 0.0975609  ms, 140 - 0.2  ms}, {ms, 600, 800}, 
          PlotRange -> {-5, 5}, DisplayFunction -> Identity, 
          PlotStyle -> {{RGBColor[0, 0, 1], 
                Thickness[0.007]}, {RGBColor[0, 0, 1], 
                Thickness[0.007]}}];

\(\(Show[dfield, traj, lines, \ 
        AxesLabel -> {"ms", \*""\!\(π\^e\)""}, 
        PlotRange -> {\(-5\), 5}, 
        DisplayFunction -> $DisplayFunction];\)\)

Question 12

(i)

Since
\(TraditionalForm\`S = \(ΔM\/P = \(\(ΔM\/M\) M\/P = λ  M\/P\)\)\) then
Ln(
          S) = Ln(λ) + m - p
     =  
          Ln(λ) - α π     
since (π^e) = π
Hence
Ln(S) =  Ln(λ) - α λ

(ii)

Solve[
      D[Log[λ] - α λ, λ] == 
        0, λ]