Solution Found!
?2E many of the following problems, it will be essential
Chapter 3, Problem 2E(choose chapter or problem)
In many of the following problems, it will be essential to have a calculator or computer available. You may use a software package† or write a program for solving initial value problems using the improved Euler’s method algorithms on pages 127 and 128. (Remember, all trigonometric calculations are done in radians.)
IMPROVED EULER’S METHOD SUBROUTINE |
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Purpose To approximate the solution \(\phi(x)\) to the initial value problem \(y^{\prime}=f(x, y)\) , \(y\left(x_{0}\right)=y_{0}\) , for \(x_{0} \leq x \leq c\). INPUT \(x_0,\ y_0,\ c\ ,N\) (number of steps), PRNTR (=1 to print a table) Step 1 Set step size \(h=\left(c-x_{0}\right) / N\), \(x=x_{0}\), \(x=x_{0}\) Step 2 For \(i=1\) to N, do steps 3-5 Step 3 Set \(F=f(x,\ y)\) \(G=f(x+h,\ y+hF)\) Step 4 Set \(x=x+h\) \(y=y+h(F+H)/2\) Step 5 If PRNTR = 1, print \(x,y\) |
IMPROVED EULER’S METHOD WITH TOLERANCE |
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Purpose To approximate the solution to the initial value problem \(y^{\prime}=f(x, y)\) , \(y\left(x_{0}\right)=y_{0}\) , at \(x=c\), with tolerance \(\varepsilon\) INPUT \(x_0,\ y_0,\ c\ ,\varepsilon\) , M (maximum number of halvings of step size) Step 1 Set \(z=y_0\), PRINTR = 0 Step 2 For \(m=0\) to M, do steps 3-7†† Step 3 Set \(N=2^m\) Step 4 Call IMPROVED EULER’S METHOD SUBROUTINE Step 5 Print \(h,y\) Step 6 If \(|y-z|<\varepsilon\), go to Step 10 Step 7 Set \(z=y\) Step 8 Print “\(\phi (c)\) is approximately”; y; “ but may not be within the tolerance”; \(\varepsilon\) Step 9 Go to Step 11 Step 10 Print “\(\phi (c)\) is approximately”; y; “with tolerance”; \(\varepsilon\) Step 11 STOP OUTPUT Approximations of the solution to the initial value problem at \(x=c\) using \(2^m) steps |
Show that when Euler’s method is used to approximate the solution of the initial value problem
\(y^{\prime}=-\frac{1}{2} y\), \(y(0)=3\),
at \(x=2\), then the approximation with step size 𝘩 is
\(3\left(1-\frac{h}{2}\right)^{2 / h}\).
Equation Transcription:
𝜙
𝜙
𝜙
Text Transcription:
phi(x)
y'=f(x,y) , y(x0=y0)
X_0 </= x </= c
x0,y0
h=(c-x_0)/N,x_0,y=y_0
i=1
F=f(x,y)
G=f(x+h,y+Hf)
x=x+h
y=y+h(F+G)/2
y'=(x,y), y(x_0)=y_0
x=c
epsilon
x_0,y_0
epsilon
z=y_0
m=0
N=2^m
|y-z|<epsilon
z=y
phi(c)
epsilon
phi(c)
epsilon
x=c
2^m
y’=-12y
y(0)=3
x=2
3(1- h over 2)^2/h
Questions & Answers
QUESTION:
In many of the following problems, it will be essential to have a calculator or computer available. You may use a software package† or write a program for solving initial value problems using the improved Euler’s method algorithms on pages 127 and 128. (Remember, all trigonometric calculations are done in radians.)
IMPROVED EULER’S METHOD SUBROUTINE |
|
Purpose To approximate the solution \(\phi(x)\) to the initial value problem \(y^{\prime}=f(x, y)\) , \(y\left(x_{0}\right)=y_{0}\) , for \(x_{0} \leq x \leq c\). INPUT \(x_0,\ y_0,\ c\ ,N\) (number of steps), PRNTR (=1 to print a table) Step 1 Set step size \(h=\left(c-x_{0}\right) / N\), \(x=x_{0}\), \(x=x_{0}\) Step 2 For \(i=1\) to N, do steps 3-5 Step 3 Set \(F=f(x,\ y)\) \(G=f(x+h,\ y+hF)\) Step 4 Set \(x=x+h\) \(y=y+h(F+H)/2\) Step 5 If PRNTR = 1, print \(x,y\) |
IMPROVED EULER’S METHOD WITH TOLERANCE |
|
Purpose To approximate the solution to the initial value problem \(y^{\prime}=f(x, y)\) , \(y\left(x_{0}\right)=y_{0}\) , at \(x=c\), with tolerance \(\varepsilon\) INPUT \(x_0,\ y_0,\ c\ ,\varepsilon\) , M (maximum number of halvings of step size) Step 1 Set \(z=y_0\), PRINTR = 0 Step 2 For \(m=0\) to M, do steps 3-7†† Step 3 Set \(N=2^m\) Step 4 Call IMPROVED EULER’S METHOD SUBROUTINE Step 5 Print \(h,y\) Step 6 If \(|y-z|<\varepsilon\), go to Step 10 Step 7 Set \(z=y\) Step 8 Print “\(\phi (c)\) is approximately”; y; “ but may not be within the tolerance”; \(\varepsilon\) Step 9 Go to Step 11 Step 10 Print “\(\phi (c)\) is approximately”; y; “with tolerance”; \(\varepsilon\) Step 11 STOP OUTPUT Approximations of the solution to the initial value problem at \(x=c\) using \(2^m) steps |
Show that when Euler’s method is used to approximate the solution of the initial value problem
\(y^{\prime}=-\frac{1}{2} y\), \(y(0)=3\),
at \(x=2\), then the approximation with step size 𝘩 is
\(3\left(1-\frac{h}{2}\right)^{2 / h}\).
Equation Transcription:
𝜙
𝜙
𝜙
Text Transcription:
phi(x)
y'=f(x,y) , y(x0=y0)
X_0 </= x </= c
x0,y0
h=(c-x_0)/N,x_0,y=y_0
i=1
F=f(x,y)
G=f(x+h,y+Hf)
x=x+h
y=y+h(F+G)/2
y'=(x,y), y(x_0)=y_0
x=c
epsilon
x_0,y_0
epsilon
z=y_0
m=0
N=2^m
|y-z|<epsilon
z=y
phi(c)
epsilon
phi(c)
epsilon
x=c
2^m
y’=-12y
y(0)=3
x=2
3(1- h over 2)^2/h
ANSWER:
Step 1:
In this problem, we have to show that the solution of Euler’s method is
at x = 2.