Options: Homework 6.1 (select 2 questions from below list, due next week):
Page 529: #1, #6
Page 536: #7
Page 546: #1
Page 566: #1
Homework 6.2 (select 2 questions from below list, due this week):
Page 576: #2
Page 585: #2
Page 591: #5
Page 599: #5
Page 529: #1, #6
In Problems 1-4, verify that the given function pair is a solution to the first-order system.
1.\(x = -e^{t}\) \(y = e^{t}\)
\(\frac{dy}{dt}=x\) \(\frac{dx}{dt}=y\)
Prove:
Differentiation of x gives us:
\(x' = -e^{t} = y\)
Differentiation of y gives us:
\(y' = e^{t} = x\)
So \(x = -e^{t}\) and \(y = e^{t}\) is the solution to the first-order system \(\frac{dy}{dt}=x\) \(\frac{dx}{dt}=y\)
In Problems 5-8, find and classify the rest points of the given autonomous system.
Solution:
If simultaneously \(\frac{dx}{dt}=-(y-1) = 0\) and \(\frac{dy}{dt}= x-2 = 0\). Thus (x= 2, y=1), is the rest point of the system.
Use the Substitution Method to find the solution for the first-order syste.
\(x'=1-y, x''=-y'= -x+2\) \(x''+x=2\)
\(y'=x-2, y''=-x'= 1-y\) \(y''+y=1\)
\(x'=a_{11}x+a_{12}y+b_1, a_{11}= 0, a_{12}=-1,b_1=1\)
\(x'=a_{21}x+a_{22}y+b_2, a_{11}= 1, a_{12}=0,b_2=-2\)
Solve the homogeneous part:
\(x^h(t)=c_1e^{r_1t}+c_2e^{r_2t}\)
\(y^h(t)=-r_1c_1e^{r_1t}-r_2c_2e^{r_2t}\)
Solve the steady-state part:
\(\overline{x}=\frac{a_{12}b_2 - a_{22}b_1}{a_{11}a_{22} - a_{12}a_{21}}\)
\(\overline{x}= \frac{(-1)\times(-2)-0}{-(-1)\times 1} = 2\)
\(\overline{y}=\frac{a_{21}b_2 - a_{11}b_1}{a_{11}a_{22} - a_{12}a_{21}}\)
\(\overline{y}=\frac{1\times 1-0}{0-(-1)\times 1} = 1\)
If \(x''+a_1x-2=0\), r_1,r_2 could be expressed as:
\(r_1,r_2 = \frac{- a_1 \pm \sqrt{a_1^2-4a_2}}{2}\)
\(a_1^2 =0, 4a_2=4\)
So
\(a_1^2 < 4a_2\)
So \(r_1 and r_2\) have Complex Roots & \(r_1 \neq r_2\)
\(r_1,r_2 = \pm i\)
In the case when$ r_1 and r_2$ have complex Roots &\(r_1 \neq r_2\), if the real part of roots h < 0, the steady state is stable. But now h = 0, so the steady state is unstable. In the other word, The rest point is unstable.
Page 536: #7
\(\frac{dy}{dx}=\frac{(m-nx)y}{(a-by)x}\)
Prove:
system (12.6)
\(\frac{dx}{dt}=(a-by)x\)
\(\frac{dy}{dt}=(m-nx)y\)
According to the chain rule, we have
\(\frac{dy}{dx}\frac{dx}{dt}=\frac{dy}{dt}\)
\(\frac{dx}{dy}(a-by)x=(m-nx)y\)
\(\frac{dx}{dy}=\frac{(m-nx)y}{(a-by)x}\)
\(y^ae^{-by}=Kx^me^{-nx}\)
where K is a constant of integration.
Solution:
Separate the variables and write
\(\frac{(a-by)}{y}dy=\frac{(m-nx)}{x}dx\)
$aln|y|-by + C_1 = mln|x|-nx + C_2 $
\(a\times ln|y|-by = m\times ln|x|-nx + C_2 - C_1\)
Applying the natural logarithm to each side:
\(e^{a\times ln|y|-by}=e^{m\times ln|x|-nx} \times e^{C_2 - C_1}\)
\(e^{ln|y|^a-by}=e^{ln|x|^m-nx} \times e^{C_2 - C_1}\)
Let
\(K= e^{C_2 - C_1}\)
Then
\(y^ae^{-by}=Kx^me^{-nx}\)
Prove:
When
\(f'(y)=ay^{a-1}e^{-by}+y^ae^{-by}(-b)= 0\)
y reach M_y.
From the above equation, we have
\(y=a/b\)
\(M_y=y^a/e^{by}=(a/b)^a/e^{b\frac{a}{b}}\)
\(M_y = (a/eb)^a\)
Similarly, when
\(g'(x)=my^{m-1}e^{-ny}+y^me^{-ny}(-n)= 0\)
x reach M_x.
From the above equation, we have
\(x=m/n\)
\(M_y=y^a/e^{by}=(a/b)^a/e^{b\frac{a}{b}}\)
\(M_x = (x/en)^m\)
Page 576: #2
Solution:
When the demand higer than offer, the company may use back ordering policy to retain the customers.Such a policy can reduce storage costs. But costs should also be assigned to stock-outs because the sales or revenue within a certain period of time will decrease when the commodity is out of stock. In addition, the stock-out may make the custormers irate or disappointed thus causes the cost of lost of future profit from customers or loss of goodwill.
The following notations are assigned to the cost of stock-outs:
s = storage costs per item per day
w = loss of goodwill cost per day
Some other notations are used for constructing the model:
h = handling fee in dollars per sale
r = demand rate of the item per day
Q = quantity of the orders
T = time in days
Assuming Q = q, is sold at time T = 0, and the item is sold out after T = t days. l = w dollars per unit per day. The same cycle is then repeated, as illustrated in Figure 13.7.
Average daily inventory is q/2
cost per cycle: \(h+s\frac{q}{2} t+w\frac{(-q)}{2}t\)
which, upon division by t , yields the average daily cost:
\(c = \frac{h}{t}+ s\frac{q}{2}+w\frac{(-q)}{2}\)
For a single cyclic period, the amount sold equals the amount demanded. So q = rt. Subtitution yield
\(c = \frac{h}{t}+ s\frac{rt}{2}+w\frac{(-rt)}{2}\)
to find the crytical point
\(c'= = \frac{h}{t^2}+ \frac{sr}{2}-\frac{wr}{2}= 0\)
\(T* = \sqrt{\frac{2h}{(s-w)r}}\)
\(c''= 2\frac{h}{t^3}\) is always positive. So at the crytical point, we can have the minimum cost.
The optimal order quantity \(Q*=rT*=r\sqrt{\frac{2h}{(s-w)r}}=\sqrt{\frac{2rh}{s-w}}\)
When the order quantity is \(\sqrt{\frac{2rh}{s-w}}\), we can have control the cost at lowest level and maximize the profit.
Page 585: #2
\(f(x, y) = 3x^2 + 6xy+ 7y^2 -2x+ 4y\)
Solution:
\(\frac{\partial f(x,y)}{\partial x} = 6x + 6y -2 =0\)
\(\frac{\partial f(x,y)}{\partial y} = 6x14y +4=0\)
x=13/12, y=-3/4
Reference: 1. Simultaneous Systems of Differential Equations (http://web.uvic.ca/~kumara/econ351/)