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Numerical Methods for Engineers (6th Edition) View more editions
Solutions
Mohammad P. UCLA
Solutions for Chapter 25
Nick H.
861 step-by-step solutions
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SAMPLE SOLUTION
Chapter:
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CH25
Problem:
1P
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Step 1 of 14 (a) Consider the following differential equation,
Find common factor on the right,
Now integrate both sides,
Calculate the value of the constant c from the initial condition
Put
, in the above equation
Put
in the solution of the differential equation
Then,
Hence, the solution of the differential equation is
Step 2 of 14 Use maple to draw graph of equation
Step 3 of 14 (b) Consider the solution of the differential equation at width
.
Consider the following differential equation,
The Euler formula is
Where, by the equation. The initial condition is and
.
So,
The next condition will be
Step 4 of 14 And,
Therefore,
So,
The next condition will be
And,
Therefore,
The next condition will be
And,
Therefore,
The next condition will be
And
Finally, From the above solution, the value of
and
are given in the below.
The Resolution of the differential equation at width ( h ) =0.25 The initial condition for the Euler method is And
then
The next condition will be
And,
The next condition will be
And, Further solve
Therefore,
The next condition will be
And,
Therefore,
The next condition will be
And,
The next condition is
And,
Therefore,
The next condition will be
And,
The next condition will be
And,
The next condition will be
And,
Step 5 of 14
Further,
Finally, From the above solution, the value of
and
are given in the below table.
Step 6 of 14 The result of the analytical solution and numerical solution are plotted below.
Step 7 of 14 (c) The formula of the Midpoint m e t h o d is , Where,
From the initial condition ,
and
The next condition will be
From the midpoint formula
Put
then,
Step 8 of 14
The next condition will be
Put
in the midpoint formula
Further solve
The next condition will be
Put
in the midpoint formula
Further solve,
The next condition will be
Put
in the midpoint formula
Finally, From the solution, the value of
1
-1.5
0.5
0.536133
-0.73718
1
0.346471
-0.17324
1.5
0.415176
0.778417
2
1.591802
10.34671
and
are given in the below table.
Step 9 of 14 Graph is shown below,
Step 10 of 14 (d) The fourth order Runge-Kutta method is
Where,
The initial condition is given as And
The further solve,
Further solve,
The further solve,
In the fourth-order Runge-Kutta method
Put
The next condition be,
Then,
Step 11 of 14
Step 12 of 14 In the fourth-order Runge-Kutta method. Put
,
Further solve,
The third condition will be
Then,
The further solve,
Now,
The further solve,
Put i=2, In the fourth-order Runge-Kutta method
The next condition will be
Step 13 of 14
Then,
The further solve,
Put
,
In the fourth-order Runge-Kutta method
The next condition will be
Then,
Further solve,
Further solve,
Further solve,
Finally, From the solution, the value of
0
1
-1.50000
-0.9277
and
-1.1401
are given below.
-0.5912
Step 14 of 14 0.5 0.4811
-0.6615
-0.3404
-0.4269
-0.1338
1
0.2869
-0.1435
0.1138
0.1429
0.6720
1.5
0.3738
0.7008
2.1186
3.4866
13.7607
2
2.5131
16.3350
65.2466
186.1883
13.504523
The result of the analytical solution and numerical solution are plotted below,
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Corresponding Textbook Numerical Methods for Engineers | 6th Edition ISBN-13:
9780077417109
ISBN:
0077417100
Authors:
Raymond Canale, Raymond P Canale, Steven C Chapra, Stephen Chapra, Steven Chapra
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Solutions by Chapter Chapter 1
Chapter 2
Chapter 3
Chapter 4
Chapter 5
Chapter 6
Chapter 7
Chapter 8
Chapter 9
Chapter 10
Chapter 11
Chapter 12
Chapter 13
Chapter 14
Chapter 15
Chapter 16
Chapter 17
Chapter 18
Chapter 19
Chapter 20
Chapter 21
Chapter 22
Chapter 23
Chapter 24
Chapter 25
Chapter 26
Chapter 27
Chapter 28
Chapter 29
Chapter 30
Chapter 31
Chapter 32
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