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Chapter 2

Question 1

In[1]:= Clear@x, yD (i) (a)

dydx

= ky and (b) y =c kx. Differentiating (b)

In[2]:= D@c Ek x , xDOut[2]= ck x k

Substitutingyinto dy/dx, then

dydx

=k c

k

xtherefore k c

k

x=

k c

k

x. Hence true.

(ii) (a)dydx

= - xy

, (b) y=x2 +y2 =c. Fromythen

In[3]:= Dt@x ^ 2 + y^2, xDOut[3]= 2 x + 2 y Dt@y, xD

which is equal to zero, since cis a constant. Solving we have

In[4]:= Solve@2 x + 2 y Dt@y, xD == 0, Dt@y, xDDOut[4]=

99Dt

@y, x

D

x

y==Therefore - x

y = -

xy

. Hence, true.

(iii) (a)dydx

= -2yx

, (b) y= ax2

. From (b) differentiate with respect tox, then

In[5]:= D@a x^2, xDOut[5]=

2 ax3

Buty=2 ax2

so thatdydx

= -2yx

. Therefore -2yx

= -2yx

. Hence true.

Question 2

Solving and checking limiting values

This first series of equations solve forp(t). It is then checked for the limit t->0. The upper limit cannot be

checked because no assumptions can be made about the values of aand k.

In[6]:= Clear@p, slope, t, Gompequ, Gompequ2, a, k, p0DIn[7]:= eq1 :=p '@tD ==k p@tD Ha Log@p@tDDL

Chapter 02.nb 1

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In[8]:= DSolve@eq1, p@tD, tDOut[8]= 99p@tD k t Hak t +k C@1DL==

In[9]:= DSolve@ 8eq1, p@0D ==p0 DOut[12]= LimitAa+k t+Log@a+Log@p0DD , t E

Properties of growth equation

The stationary values are obtained. The programme gives the upper stationary value and warns about the

existence of the lower one.

In[13]:= Solve@k pHa Log@pDL ==0, pDSolve::verif : Potential solution 8p 0< Hpossibly

discarded by verifierL should be checked by hand. May require use of limits.

Out[13]=

88p a

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In[21]:= Gompequ2

Out[21]= 5+ 0.8 t+Log@5Log@20DD

In[22]:= Plot@Gompequ2,8t, 0, 10 DOut[23]= 5

In[24]:= N@%DOut[24]= 148.413

In[25]:= N@E ^ 4DOut[25]= 54.5982

In[26]:= Plot@Gompequ,8p, 0.1, 150

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In[27]:= 11 y2

y

Out[27]= 12

Log@1 + yD + 12

Log@1 + yD

In[28]:= xx

Out[28]=x2

2

Hence, 12

Log@1 + yD + 12

Log@1 + yD= x22

+cwhere cis the constant of integration. Solving for y we

obtain

In[29]:= SolveA 12

Log@1 + yD + 12

Log@1 + yD == x22

+ c, yESolve::verif : Potential solution 8y < Hpossibly

discarded by verifierL should be checked by hand. May require use of limits.

Solve::ifun : Inverse functions are being used by Solve, so some solutions may not be found.

Out[29]= 99y 1 + 2 c+x2

1 + 2 c+x2

==

(ii)dydx

=y2 - 2y + 1. But y2 - 2y + 1=H1 -yL2. Hence dyH1-yL2 = dx. Integrating both sides

In[30]:= 1H1 yL2 y

Out[30]= 1

1 + y

In[31]:= 1x

Out[31]= x

Solving for y

In[32]:= SolveA 11 + y

+ c ==x, yE

Out[32]= 99y 1 + c xc x

==

(iii)

dy

dx =

y2

x2 hence

dy

y2 =

dx

x2 . Integrating both sides

In[33]:= 1y2

y

Out[33]= 1y

In[34]:= 1x2

x

Out[34]= 1x

Introducing the constant of integration and solving, we have

Chapter 02.nb 4

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In[35]:= SolveA 1y

== 1x+ c, yE

Out[35]= 99y x1 + c x

==

Question 4

(i)dydx

=x2 - 2x + 1 and y=1 when x=0. Hence dy=Hx2 - 2x + 1L dx. Integrating both sides

In[36]:= 1y

Out[36]= y

In[37]:= Hx2 2x + 1Lx

Out[37]= x x2 + x3

3

Adding the constant of integration and solving forywe obtain

In[38]:= SolveAy ==x x2 + x33

+ c, yE

Out[38]= 99y c + x x2 + x3

3==

Including initial values we can solve for c

In[39]:= SolveAy == 13 H3 c + 3 x 3 x2 + x3L, cE .8y >1, x >001.2Out[86]= H0.930233 + 0.216299 0.3225tL1.33333

But k=s aka- (n+d)k hence

In[87]:= kdot =0.1 4k^0.25 H0.03 + 0.4LkOut[87]= 0.4 k0.25 0.43 k

In[88]:= sol =NSolve@kdot ==0, kDOut[88]= 88k 0. All, AxesOrigin >80, 0.58"t", "k"

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In[92]:= Plot@kdot,8k, 0, 1keOut[94]= 0.3225

Letf(k) =s aka- (n+d)k then the linear approximation is given by:

k=f(k) +f'(k)(k-ke) =f'(k)(k-ke)

In[95]:= fprime =D@kdot, kD . k >keOut[95]= 0.3225

Since |fprime| < 1 then keis a stable equilibrium.

Question 15

(i)

YHtL

agains Yis simply a linear line through the origin with slope sv

> 0. The phase line is horizontal with

fixed point at the origin, and arrows indicating an ever increasing value of Yfor some YH0L> 0.

(ii)

Given the differential equation Y '(t) - (s/v)Y(t) = 0, we can solve as follows:

In[96]:= Clear@s, YDIn[97]:= DSolve@ 8Y '@tD Hs vLY@tD ==0, Y@0D ==Y0

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Question 16

Given Y[t] = Y0er t. SolvingD '[t] = kY(t) = kY0e

r t. SinceDis protected, we usexin its place.

In[98]:= Clear@x, r, YDIn[99]:= sol =DSolve@ 8x '@tD ==k Y0 E^Hr tL, x@0D ==x0

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Question 18

(a)

In[105]:= Clear@fDIn[106]:= f@g_D:=100 Log@2D gIn[107]:= Map@f,82.7, 5.0, 2.5, 2.0, 1.4, 2.4, 2.0, 0.2, 0.2

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In[111]:= 11620000000.0173H20001992L

Out[111]= 1.33448 109

Question 20

In[112]:= 50000.05 40 + H2000 0.05L HH0.0540L 1LOut[112]= 292508.

Chapter 02.nb 15