Question

2. Non-linear Shooting Method for a Two-point Boundary Value Problem Consider the differential equation y'' = -(y')^2 - y + ln(x), 1 <= x <= 2 with the boundary conditions y(1) = 0, y(2) = ln 2. Show that the exact solutions is y(x) = ln x. Implement the shooting method for this problem in Matlab. Use Matlab solver ode45. Note that this is a non-linear problem, so you need to use a secant iteration. Since the secant iteration converges quickly if the initial guess is good, it is crucial to get a good initial guess. Try the values z1 = 1, z2 = 0.5. You may choose the tolerance to be 10^-9, and maximum number of iterations for the secant method to be 5. Plot the approximate solutions together with the exact solution. Plot also the error.

          2. Non-linear Shooting Method for a Two-point Boundary Value Problem

Consider the differential equation
y'' = -(y')^2 - y + ln(x), 1 <= x <= 2

with the boundary conditions
y(1) = 0, y(2) = ln 2.

Show that the exact solutions is
y(x) = ln x.

Implement the shooting method for this problem in Matlab. Use Matlab solver ode45. Note that this is a non-linear problem, so you need to use a secant iteration. Since the secant iteration converges quickly if the initial guess is good, it is crucial to get a good initial guess. Try the values z1 = 1, z2 = 0.5.

You may choose the tolerance to be 10^-9, and maximum number of iterations for the secant method to be 5. Plot the approximate solutions together with the exact solution. Plot also the error.

Show more…

2. Non-linear Shooting Method for a Two-point Boundary Value Problem

Consider the differential equation
y” = -(y')^2 - y + ln(x), 1 <= x <= 2

with the boundary conditions
y(1) = 0, y(2) = ln 2.

Show that the exact solutions is
y(x) = ln x.

Implement the shooting method for this problem in Matlab. Use Matlab solver ode45. Note that this is a non-linear problem, so you need to use a secant iteration. Since the secant iteration converges quickly if the initial guess is good, it is crucial to get a good initial guess. Try the values z1 = 1, z2 = 0.5.

You may choose the tolerance to be 10^-9, and maximum number of iterations for the secant method to be 5. Plot the approximate solutions together with the exact solution. Plot also the error.

Added by Robert C.

Calculus: Early Transcendentals

James Stewart 8th Edition

Thanks for your feedback!

Non-linear Shooting Method for a Two-point Boundary Value Problem Consider the differential equation y'' = -(y')^2 - y + ln(x), 1 <= x <= 2 with the boundary conditions y(1) = 0, y(2) = ln 2. Show that the exact solutions is y(x) = ln x. Implement the shooting method for this problem in Matlab. Use Matlab solver ode45. Note that this is a non-linear problem, so you need to use a secant iteration. Since the secant iteration converges quickly if the initial guess is good, it is crucial to get a good initial guess. Try the values z1 = 1, z2 = 0.5. You may choose the tolerance to be 10^-9, and maximum number of iterations for the secant method to be 5. Plot the approximate solutions together with the exact solution. Plot also the error.

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00:01 In this giving question first of all we have to understand what kind of diagram we will have.

00:05 So here let's take this as a cue.

00:12 So here is the q like this and now our what axis will be this will be the y -axis.

00:27 Here we have c -axis and this is our third x -axis.

00:34 So now the plank will be like this way.

00:38 From here to here let me shade this as this way and another diagram will be again we will create a q so this is a q and with a similar way this is our pi axis here we have our z -axis and this way we will get our x -axis and now the position of the plank is from the next side here to here.

01:18 This will be the position of the plank, this way.

01:26 Right? and now thirdly, this is our first diagram, second, or we can say 011, then 0, 1, and thirdly we will create here another diagram for the q representation.

01:52 So in this way, here we have our y -axis, this is our x -exes, y is z -exes, and x -axis.

02:04 So now the position of the plank is between these two diagonals.

02:12 Here is the position of the plank.

02:17 So we are done with the diagram of 1 -1 -0 also.

02:22 So now at the very first point, answer for 1, we need to determine the crystal system.

02:33 Crystal system so this is a unit cell this unit cell belongs to the orthormarabic because we have a does not equal to b does not equal to c and here failure of a is 0 .35 nanometers b is 0 .4 nanometers c is 0 .3 nanometers.

03:22 So alpha will be equal to beta is equal to karma and all are equal to 90.

03:28 So this is a solution for the first.

03:32 I'll answer to first part...

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