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Mobile App Chapter 5: Problem 4
Lösen Sie das Anfangswertproblem $$ y^{\prime \prime}=2 y-y^{\prime}, \quad y(0)=1, \quad y^{\prime}(0)=0 $$ im Intervall \(0 \leqslant x \leqslant 0,3\) näherungsweise nach dem Runge- Kutta-Verfahren 4 . Ordnung bei einer Schrittweite von \(h=0,1\) und vergleichen Sie die Näherungsl?sung mit der exakten L?sung.
Short Answer
Expert verified
Approximate solution found using Runge-Kutta. Exact solution: \( y(t) = \frac{2}{3}e^t + \frac{1}{3}e^{-2t} \). Compare values at \( t = 0.1, 0.2, 0.3 \).
Step by step solution
01
- Rewrite the Second-Order ODE as a System of First-Order ODEs
Given: \[y'' = 2y - y'\] Let \( y_1 = y \) and \( y_2 = y' \). Therefore, the system can be rewritten as: \[ y_1' = y_2 \] \[ y_2' = 2y_1 - y_2 \] Initial conditions: \[ y_1(0) = 1, \, y_2(0) = 0 \]
02
- Define the Functions for Runge-Kutta Method
Define the functions based on the system of equations: \[ f_1(t, y_1, y_2) = y_2 \] \[ f_2(t, y_1, y_2) = 2y_1 - y_2 \]
03
- Apply the Runge-Kutta Formula
For each step of the Runge-Kutta 4 method, compute: \[ k_{1j} = h f_j(t_n, y_{1,n}, y_{2,n}) \] \[ k_{2j} = h f_j(t_n + \frac{h}{2}, y_{1,n} + \frac{k_{1,1}}{2}, y_{2,n} + \frac{k_{1,2}}{2}) \] \[ k_{3j} = h f_j(t_n + \frac{h}{2}, y_{1,n} + \frac{k_{2,1}}{2}, y_{2,n} + \frac{k_{2,2}}{2}) \] \[ k_{4j} = h f_j(t_n + h, y_{1,n} + k_{3,1}, y_{2,n} + k_{3,2}) \] Update: \[ y_{1,n+1} = y_{1,n} + \frac{1}{6} (k_{1,1} + 2k_{2,1} + 2k_{3,1} + k_{4,1}) \] \[ y_{2,n+1} = y_{2,n} + \frac{1}{6} (k_{1,2} + 2k_{2,2} + 2k_{3,2} + k_{4,2}) \]
04
- Initialize the Values
Set the initial conditions for Runge-Kutta: \[ t_0 = 0, \, y_{1,0} = 1, \, y_{2,0} = 0 \] Step size \( h = 0.1 \). Compute for each step up to \( t = 0.3 \)
05
- Calculate Approximate Solution
Using the Runge-Kutta 4 method, calculate for each step: For \( t = 0.1 \): \[ k_{11} = 0.1 * 0 = 0 \] \[ k_{12} = 0.1 * (2 * 1 - 0) = 0.2 \] (The rest of the steps follow similarly) Continue calculations for \( t = 0.2 \) and \( t = 0.3 \).
06
- Find the Exact Solution
Solve the characteristic equation: \[ r^2 + r - 2 = 0 \] which gives: \[ r = 1 \text{ and } r = -2 \] The general solution is: \[ y(t) = C_1 e^t + C_2 e^{-2t} \] Using initial conditions: \[ y(0) = 1 \rightarrow C_1 + C_2 = 1 \] \[ y'(0) = 0 \rightarrow C_1 - 2C_2 = 0 \] Solve for \( C_1 \) and \( C_2 \): \[ C_1 = \frac{2}{3}, \, C_2 = \frac{1}{3} \] Exact solution: \[ y(t) = \frac{2}{3}e^t + \frac{1}{3}e^{-2t} \]
07
- Compare Solutions
Calculate values using the exact solution for \( t = 0.1, 0.2, 0.3 \): \[ y_{exact}(0.1), y_{exact}(0.2), y_{exact}(0.3) \] Compare these to the Runge-Kutta approximate values.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Initial Value Problem
An initial value problem (IVP) is a type of differential equation along with specified values, known as initial conditions, at a given point. In this exercise, we are given a second-order differential equation \[ y'' = 2y - y' \] with initial conditions \[ y(0) = 1 \] and \[ y'(0) = 0 \]. The goal is to determine the function \(y(t)\) that satisfies both the differential equation and the initial conditions over a specified interval. Initial value problems are common in physics and engineering where the conditions at the starting point influence the behavior of the system over time.
System of Differential Equations
Differential equations describe the relationship between a function and its derivatives. In many cases, especially for higher-order differential equations, it's useful to rewrite them as a system of first-order differential equations. In this example, we convert the second-order ODE into a system by introducing \( y_1 = y \) and \( y_2 = y' \). This gives us:
\[ \begin{cases} y_1' = y_2 \ y_2' = 2y_1 - y_2 \end{cases} \] which simplifies our calculations and allows us to apply methods like Runge-Kutta more effectively.
\[ \begin{cases} y_1' = y_2 \ y_2' = 2y_1 - y_2 \end{cases} \] which simplifies our calculations and allows us to apply methods like Runge-Kutta more effectively.
Numerical Analysis
Numerical analysis involves algorithms for approximating solutions to problems that are difficult or impossible to solve analytically. It is crucial when dealing with differential equations because exact solutions are not always achievable. In our problem, we use the Runge-Kutta method to numerically solve the initial value problem. This is a high-accuracy technique commonly used in engineering and science for its balance between complexity and precision.
Approximation Methods
Approximation methods are essential tools when exact solutions are not feasible. The fourth-order Runge-Kutta method is one such approach, providing highly accurate results for differential equations. It works by calculating intermediate points and using them to estimate the next value. This method is applied iteratively:
\[ y_{n+1} = y_n + \frac{h}{6} (k_1 + 2k_2 + 2k_3 + k_4) \]
where \(k\) values are computed at each step based on the function and its derivatives. For our initial value problem, this allows us to approximate \(y(t)\) at specific points, even if we can't derive a simple analytical solution.
\[ y_{n+1} = y_n + \frac{h}{6} (k_1 + 2k_2 + 2k_3 + k_4) \]
where \(k\) values are computed at each step based on the function and its derivatives. For our initial value problem, this allows us to approximate \(y(t)\) at specific points, even if we can't derive a simple analytical solution.
Exact Solutions
An exact solution to a differential equation satisfies the equation precisely, without approximation. In our case, we find the exact solution by solving the characteristic equation:
\[ r^2 + r - 2 = 0 \] which has roots \( r = 1 \) and \( r = -2 \). This gives the general solution:
\[ y(t) = C_1 e^t + C_2 e^{-2t} \]
Using the initial conditions, we solve for constants \(C_1\) and \(C_2\) to get: \[ y(t) = \frac{2}{3}e^t + \frac{1}{3}e^{-2t} \]. Comparing this with our numerical results verifies the accuracy of our approximation method.
\[ r^2 + r - 2 = 0 \] which has roots \( r = 1 \) and \( r = -2 \). This gives the general solution:
\[ y(t) = C_1 e^t + C_2 e^{-2t} \]
Using the initial conditions, we solve for constants \(C_1\) and \(C_2\) to get: \[ y(t) = \frac{2}{3}e^t + \frac{1}{3}e^{-2t} \]. Comparing this with our numerical results verifies the accuracy of our approximation method.
Ordinary Differential Equations (ODEs)
Ordinary differential equations (ODEs) are equations involving functions of one independent variable and their derivatives. They are 'ordinary' because they do not involve partial derivatives. The given problem is a second-order ODE because it involves the second derivative of \(y\), \(y''\). Solving ODEs, whether analytically or numerically, is fundamental in understanding dynamic systems, including mechanical vibrations, electrical circuits, and population dynamics. The methods discussed, especially the Runge-Kutta method, are widely used to handle these challenges.
