(g) y+1 = Yk + 6. %3D
Unitary Method
The word “unitary” comes from the word “unit”, which means a single and complete entity. In this method, we find the value of a unit product from the given number of products, and then we solve for the other number of products.
Speed, Time, and Distance
Imagine you and 3 of your friends are planning to go to the playground at 6 in the evening. Your house is one mile away from the playground and one of your friends named Jim must start at 5 pm to reach the playground by walk. The other two friends are 3 miles away.
Profit and Loss
The amount earned or lost on the sale of one or more items is referred to as the profit or loss on that item.
Units and Measurements
Measurements and comparisons are the foundation of science and engineering. We, therefore, need rules that tell us how things are measured and compared. For these measurements and comparisons, we perform certain experiments, and we will need the experiments to set up the devices.
Solve the question in the same way as a book
![Difference Equations
Example E
The graph of the nonlinear equation
Yk+1 = Yk
(2.203)
is shown in Figure 2.11. There are two fixed points, yk = 0 and yk = 1.
The slopes of the curve at these fixed points are, respectively, o and 2;
consequently, Yk = 0 is an unstable fixed point and yk = 1 is a stable fixed
point.
Note that the lower branch of the curve does not enter into consideration
since it does not correspond to real values of yk under continued iteration.
The transformation
Vk = log yk
(2.204)
will allow us to obtain the exact solution to equation (2.203). We find
2Uk+1 = Uk,
(2.205)
the solution of which is
Vk+1 =
= A(/½)*,
(2.206)
where A is an arbitrary constant. Therefore,
Yk = exp[A(/½)*],
(2.207)
which has the asymptotic expansion
Yk = 1+ A(/2)* + 1/24?(1/2)2k + 1/6A° (/2)3* + ....
(2.208)
Rewrite equation (2.203) as follows:
Yk+1 =
VYk,
(2.209)
and let
Yk = 1+uk;
(2.210)
where u is assumed to be small. Substitution of equation (2.210) into equation
(2.209) gives
Uk+1 =
1/2uk,
(2.211)
the solution of which is
Uk =
A(/2)*,
(2.212)
where A is an arbitrary constant. Therefore, the first approximation to yk is
Yk = 1+ A(2)*.
(2.213)
Now let
t = A(/2)*, z(t) = Yk.
(2.214)
%3D
FIRST-ORDER DIFFERENCE EQUATIONS
79
Consequently, equation (2.203) becomes
2(1/2t)2 = z(t).
(2.215)
Assuming a solution of the form
z(t) = 1+t+ A2t? + Azt +..
(2.216)
and substituting this result into equation (2.215) gives
z(/2t)? = 1+t+ (2A2 + 4)t2 + (1/4A3 +4A2)t³ +...
(2.217)
Comparison of equations (2.216) and (2.217) gives
1/2A2 + 4 = A2, 4A3 + 4A2 = A3
(2.218)
and
A2 = 1/2, A3 = 6.
(2.219)
Therefore,
z(t) = 1+t+ 1/2t + /6t° + ,
which, on using the definition of t in equation (2.214), gives the same asymp-
totic expansion as equation (2.208).](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fcf8ae0d8-6236-48e9-895b-2c58b4c18150%2F406d56f1-ecd6-438e-a619-385ecead4f5f%2Fylolr85_processed.jpeg&w=3840&q=75)
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