/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 4 Convert the following system of ... [FREE SOLUTION] | 91Ó°ÊÓ

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Convert the following system of second-order equations to a larger system of firstorder equations. It arises from studying the gravitational attraction of one mass by another. $$ x^{\prime \prime}(t)=\frac{-c x(t)}{r(t)^{3}}, \quad y^{\prime \prime}(t)=\frac{-c y(t)}{r(t)^{3}}, \quad z^{\prime \prime}(t)=\frac{-c z(t)}{r(t)^{3}} $$ with \(c\) positive constant and \(r(t)=\left[x(t)^{2}+y(t)^{2}+z(t)^{2}\right]^{1 / 2}, t=\) time.

Short Answer

Expert verified
The second-order system can be converted to a first-order system by introducing new variables for the first derivatives, resulting in six first-order equations.

Step by step solution

01

Define New Variables

Introduce new variables to represent the first derivatives of the original variables. Let \(x_1(t) = x(t)\), \(x_2(t) = x'(t)\), \(y_1(t) = y(t)\), \(y_2(t) = y'(t)\), \(z_1(t) = z(t)\), and \(z_2(t) = z'(t)\).
02

Express First Derivatives

Write down the equations representing the first derivatives of the new variables: \(x_1'(t) = x_2(t)\), \(y_1'(t) = y_2(t)\), and \(z_1'(t) = z_2(t)\).
03

Formulate Second Derivatives

Rewrite the second-order equations in terms of the newly defined variables: \(x_2'(t) = \frac{-c x_1(t)}{r(t)^3}\), \(y_2'(t) = \frac{-c y_1(t)}{r(t)^3}\), and \(z_2'(t) = \frac{-c z_1(t)}{r(t)^3}\). Here, \(r(t) = \sqrt{x_1(t)^2 + y_1(t)^2 + z_1(t)^2}\).
04

Combine into Larger System

Combine all the first-order equations into a single system. We have: 1. \(x_1'(t) = x_2(t)\)2. \(x_2'(t) = \frac{-c x_1(t)}{r(t)^3}\)3. \(y_1'(t) = y_2(t)\)4. \(y_2'(t) = \frac{-c y_1(t)}{r(t)^3}\)5. \(z_1'(t) = z_2(t)\)6. \(z_2'(t) = \frac{-c z_1(t)}{r(t)^3}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Second-Order Differential Equations
Second-order differential equations involve derivatives of a function up to the second order. They can describe a variety of phenomena in physics and engineering, such as motion, heat transfer, and vibrations. In our exercise, the second-order equations represent how the positions coordinates \( x(t) \), \( y(t) \), and \( z(t) \) change over time due to gravitational attraction.
An example of a second-order equation is:
\[ x''(t) = \frac{-c x(t)}{r(t)^3} \]
Here, \( x''(t) \) signifies the second derivative of position with respect to time, representing acceleration.
Gravitational Attraction
Gravitational attraction describes how objects with mass attract each other. Sir Isaac Newton formulated the law of universal gravitation, stating that the force between two masses m and M is:
\[ F = G \frac{mM}{r^2} \]where G is the gravitational constant, and r is the distance between the centers of the two masses.
In our exercise, the gravitational attraction of one mass by another is given by the second-order system:
\[ x''(t) = \frac{-c x(t)}{r(t)^3}, \ y''(t) = \frac{-c y(t)}{r(t)^3}, \ z''(t) = \frac{-c z(t)}{r(t)^3} \]
Here, c is a positive constant related to the gravitational force, and r(t) calculates the distance between the objects over time as:


\[ r(t) = \ \sqrt{x(t)^2 + y(t)^2 + z(t)^2} \]
Variable Substitution
Variable substitution is a common technique in solving differential equations. It simplifies complex equations into more manageable ones by introducing new variables. This makes it easier to apply standard methods for solving first-order systems.
In the given problem, we introduce new variables to assist with the conversion from a second-order to a first-order system. Let:
\[ x_1(t) = x(t), \ x_2(t) = x'(t), \ y_1(t) = y(t), \ y_2(t) = y'(t), \ z_1(t) = z(t), \ z_2(t) = z'(t) \]
This substitution facilitates rewriting the second-order equations as:
\[ x_1'(t) = x_2(t), \ y_1'(t) = y_2(t), \ z_1'(t) = z_2(t), \ x_2'(t) = \frac{-c x_1(t)}{r(t)^3}, \ y_2'(t) = \frac{-c y_1(t)}{r(t)^3}, \ z_2'(t) = \frac{-c z_1(t)}{r(t)^3} \]
Numerical Methods
Numerical methods are employed to approximate solutions to differential equations that cannot easily be solved by analytical methods. These techniques are particularly useful for complex systems involving multiple variables and non-linear relationships.
Common numerical methods include Euler's Method, Runge-Kutta Methods, and the Finite Difference Method. These approaches work by discretizing time or space and solving the equations iteratively using step-by-step calculations.
For a first-order system, such as:
\[ x_1'(t) = x_2(t), x_2'(t) = \frac{-c x_1(t)}{r(t)^3}, \ y_1'(t) = y_2(t), y_2'(t) = \frac{-c y_1(t)}{r(t)^3}, \ z_1'(t) = z_2(t), z_2'(t) = \frac{-c z_1(t)}{r(t)^3} \], numerical methods can approximate solutions at discrete time intervals. This allows us to predict the future states of a system based on initial conditions.

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Most popular questions from this chapter

Consider the motion of a particle of mass in falling vertically under the earth's gravitational field, and suppose the downward motion is opposed by a frictional force \(p(v)\) dependent on the velocity \(v(t)\) of the particle. Then the velocity satisfies the equation $$ m v^{\prime}(t)=-m g+p(v), \quad t \geq 0, \quad v(0) \text { given } $$ Let \(m=1 \mathrm{~kg}, g=9.8 \mathrm{~m} / \mathrm{sec}^{2}\), and \(v(0)=0\). Solve the differential equation for \(0 \leq t \leq 20\) and for the following choices of \(p(v)\) : (a) \(p(v)=-0.1 v\), which is positive for a falling body. (b) \(\quad p(v)=0.1 v^{2}\). Find answers to at least three digits of accuracy. Graph the functions \(v(t) .\) Compare the solutions.

In this exercise, we consider a method with third-order truncation errors that is not convergent or stable. (a) If \(Y(x)\) is three times continuously differentiable, show that $$ \begin{aligned} Y\left(x_{n+1}\right)=& 3 Y\left(x_{n}\right)-2 Y\left(x_{n-1}\right)+\frac{h}{2}\left[Y^{\prime}\left(x_{n}\right)-3 Y^{\prime}\left(x_{n-1}\right)\right] \\ &+\frac{7}{12} h^{3} Y^{\prime \prime \prime}\left(x_{n}\right)+O\left(h^{4}\right) \end{aligned} $$ Thus, a numerical method for solving the differential equation $$ Y^{\prime}(x)=f(x, Y(x)) $$ is $$ y_{n+1}=3 y_{n}-2 y_{n-1}+\frac{h}{2}\left[f\left(x_{n}, y_{n}\right)-3 f\left(x_{n-1}, y_{n-1}\right)\right], \quad n \geq 1 $$ This is a numerical method whose truncation error is \(O\left(h^{3}\right) .\) It is an example of a multistep method (cf. Section 8.6). To use the method, we need a value for \(y_{1}\), called an artificial initial value, in addition to the initial value \(y_{0}=Y_{0}\). Hint: To prove (8.65), use Taylor expansions about the point \(x_{n}\). (b) Now apply the method to solve the very simple initial value problem $$ Y^{\prime}(x) \equiv 0, \quad Y(0)=1 $$ whose solution is \(Y(x) \equiv 1\). Show that if the initial values are chosen to be \(y_{0}=1, y_{1}=1+h\), then the numerical solution is \(y_{n}=1-h+h 2^{n}\). Note that \(\left|y_{1}-Y(h)\right|=h \rightarrow 0\) as \(h \rightarrow 0\). Let \(x_{n}=1\). Show that \(\left|Y(1)-y_{n}\right| \rightarrow\) \(\infty\) as \(h \rightarrow 0\). Thus, the method is not convergent. (c) A slight variant of the arguments of (b) can be used to show the instability of the method. Show that with the initial values \(y_{0}=y_{1}=1\), the numerical solution is \(y_{n}=1\) for all \(n\), whereas if the initial values are perturbed to \(y_{\epsilon, 0}=\) 1, \(y_{\epsilon, 1}=1+\epsilon\), then the numerical solution becomes \(y_{\epsilon, n}=1-\epsilon+\epsilon 2^{n}\). Show that at any fixed node point \(x_{n}=\bar{x}>0,\left|y_{\epsilon, n}-y_{n}\right| \rightarrow \infty\) as \(h \rightarrow 0\). Hence, the method is unstable.

Consider the solutions \(Y(x)\) of $$ Y^{\prime}(x)+a Y(x)=d e^{-b x} $$ with \(a, b, d\) constants and \(a, b>0 .\) Calculate $$ \lim _{x \rightarrow \infty} Y(x) $$ Hint: \(\quad\) Consider separately the cases \(a \neq b\) and \(a=b\).

Verify that any function of the form \(Y(x)=c_{1} x+c_{2} x^{2}\) satisfies the equation $$ x^{2} Y^{\prime \prime}(x)-2 x Y^{\prime}(x)+2 Y(x)=0 $$ Determine the solution of the equation with the boundary conditions $$ Y(1)=0, \quad Y(2)=1 $$ Use the MATLAB program ODEBVP to solve the boundary value problem for \(h=\) \(0.1,0.05,0.025\), print the errors of the numerical solutions at \(x=1.2,1.4,1.6\) and 1.8. Comment on how errors decrease when \(h\) is halved. Do the same for the extrapolated solutions.

Let $$ \begin{gathered} A=\left[\begin{array}{ll} 1 & -2 \\ 2 & -1 \end{array}\right], \quad \boldsymbol{Y}=\left[\begin{array}{l} Y_{1} \\ Y_{2} \end{array}\right] \\ \boldsymbol{G}(x)=\left[\begin{array}{l} -2 e^{-x}+2 \\ -2 e^{-x}+1 \end{array}\right], \quad \boldsymbol{Y}_{0}=\left[\begin{array}{l} 1 \\ 1 \end{array}\right] \end{gathered} $$ Write out the two equations that make up the system $$ \boldsymbol{Y}^{\prime}=A \boldsymbol{Y}+\boldsymbol{G}(x), \quad \boldsymbol{Y}\left(x_{0}\right)=\boldsymbol{Y}_{0} $$ The true solution is \(\boldsymbol{Y}=\left[e^{-x}, 1\right]^{T}\).

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