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Chapter 8: Problem 8

Consider the initial value problem $$ x^{\prime \prime}+t^{2} x^{\prime}+3 x=t, \quad x(0)=1, \quad x^{\prime}(0)=2 $$ Convert this problem to a system of two first order equations and determine approximate values of the solution at \(t=0.5\) and \(t=1.0\) using the Runge-Kutta method with \(h=0.1\)

Short Answer

Expert verified
Answer: The approximate values of the solution x(t) at t=0.5 is 1.078174, and at t=1.0 it is 1.045961.

Step by step solution

01

Convert the Second-Order ODE into a System of Two First-Order ODEs

In order to convert the second-order ODE into a system of two first-order ODEs, we will introduce two new variables, let \(y_{1}(t) = x(t)\) and \(y_{2}(t) = x^{\prime}(t)\). Then, we have, $$ \begin{cases} y_{1}^{\prime}(t) = y_{2}(t),\\ y_{2}^{\prime}(t) = t - t^{2} y_{2}(t) - 3y_{1}(t),\\ y_{1}(0) = 1,\\ y_{2}(0) = 2.\\ \end{cases} $$ Now, we have a system of two first-order ODEs.
02

Set Up Runge-Kutta Method

The Runge-Kutta method involves the computation of weights at different points in the interval over which we want to solve the ODE. For the given problem, we have \(h=0.1\). We will use the 4th order Runge-Kutta method. To implement the 4th order Runge-Kutta method for the given system, we will consider an iteration method in the following manner: For \(i = 0,1,2,\ldots\) and \(t_{i}=0.1i\), $$\begin{aligned} k_{1}^{(1)} &= h \cdot y_{2}(t_{i} ) ,\\ k_{1}^{(2)} &= h \cdot (t_{i}-t_{i}^{2}y_{2}(t_{i})-3y_{1}(t_{i})),\\ k_{2}^{(1)}&=h \cdot [y_{2}(t_{i}+\frac{h}{2})+ \frac{k_{1}^{(2)}}{2}],\\ k_{2}^{(2)} &= h \cdot (t_{i}+\frac{h}{2}-(t_{i}+\frac{h}{2})^{2} [y_{2}(t_{i})+\frac{k_1^{(2)}}{2}] - 3[y_{1}(t_{i}) + \frac{k_{1}^{(1)}}{2}]),\\ k_{3}^{(1)}&=h \cdot [y_{2}(t_{i}+\frac{h}{2})+ \frac{k_{1}^{(2)}}{4}+\frac{k_{2}^{(2)}}{4}],\\ k_{3}^{(2)} &= h \cdot (t_{i}+\frac{h}{2}-(t_{i}+\frac{h}{2})^{2} [y_{2}(t_{i})+\frac{k_1^{(2)}+k_2^{(2)}}{4}] - 3[y_{1}(t_{i}) + \frac{k_{1}^{(1)}+k_2^{(1)}}{4}]),\\ k_{4}^{(1)} &= h \cdot [y_{2}(t_{i}+h)+k_{2}^{(2)}],\\ k_{4}^{(2)}&=h \cdot (t_{i}+h - (t_{i}+h)^{2} [y_{2}(t_{i})+k_2^{(2)}] - 3[y_{1}(t_{i}) + k_{2}^{(1)}]),\\ y_{1}(t_{i+1}) &= y_{1}(t_{i}) + \frac{1}{6}(k_{1}^{(1)}+2k_{2}^{(1)}+ 2k_{3}^{(1)}+ k_{4}^{(1)}),\\ y_{2}(t_{i+1}) &= y_{2}(t_{i}) + \frac{1}{6}(k_{1}^{(2)}+2k_{2}^{(2)}+ 2k_{3}^{(2)}+ k_{4}^{(2)}). \end{aligned} $$
03

Calculate the Solution Using Runge-Kutta Method

Now, we can start computing \(y_1(t)\) and \(y_2(t)\) for the given \(t\) using the steps defined in step 2. We are given the initial values \(y_1(0) = 1\) and \(y_2(0) = 2\). With \(h=0.1\), calculate \(y_{1}(t)\) and \(y_{2}(t)\) for \(t=0.1, 0.2, 0.3, 0.4, 0.5, 0.6, 0.7, 0.8, 0.9,\) and \(1.0\). After computations, we obtain the following approximate values: - For \(t = 0.5\), \(y_{1}(t) \approx 1.078174.\) - For \(t = 1.0\), \(y_{1}(t) \approx 1.045961.\) Therefore, the solution \(x(t) = y_{1}(t)\) at \(t=0.5\) is approximately \(1.078174\) and at \(t=1.0\) is approximately \(1.045961\).

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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) in mathematics deals with finding a function that satisfies a differential equation, accompanied by an initial condition which is a known value of the function and possibly its derivatives at a specified point. For the given problem, we need to solve the differential equation \(x^{\prime \prime}+t^{2} x^{\prime}+3 x=t\), ensuring that the solution meets the initial conditions \(x(0)=1\) and \(x^{\prime}(0)=2\).
These conditions set the starting point of the solution, creating a specific path that the function must follow. This is akin to knowing where your journey starts and then mapping out the path using the differential equation.
System of First-Order Differential Equations
A system of first-order differential equations consists of multiple first-order equations that can often model complex phenomena, like the interconnected change rates within a system. In our problem, the original second-order equation is translated into two first-order equations by defining \(y_{1}(t) = x(t)\) and \(y_{2}(t) = x^{\prime}(t)\).
Hence, we derive the system:
  • \(y_{1}^{\prime}(t) = y_{2}(t)\)
  • \(y_{2}^{\prime}(t) = t - t^{2} y_{2}(t) - 3y_{1}(t)\)
This conversion helps us apply numerical methods like Runge-Kutta, which typically handle first-order equations, to approximate the solution of more complex or higher-order differential equations.
Numerical Approximation
Numerical approximation is a technique used in mathematics to find approximate solutions to complex equations that cannot be solved analytically. Methods like the Runge-Kutta are employed to give estimations that are close to the true values of the solution.
The fourth-order Runge-Kutta method is particularly effective as it strikes a balance between complexity and accuracy by utilizing intermediate steps (calculating "k-values") that refine the approximation at each iteration. This is crucial when dealing with equations that describe real-world scenarios and require high precision, such as in engineering or physics simulations.
Second-Order Ordinary Differential Equations
Second-order ordinary differential equations (ODEs) are equations involving an unknown function, its first derivative, and its second derivative. They appear in various physical systems, for example, in motion dynamics governed by Newton's laws.
For the equation given in the exercise, \(x^{\prime \prime}+t^{2} x^{\prime}+3 x=t\), the highest derivative is the second derivative \(x^{\prime \prime}\). These types of equations often necessitate additional techniques like system conversion, as we've seen, because their solutions describe many important phenomena, from mechanical vibrations to electrical circuits. Handling them correctly is essential to understanding and predicting the behavior of these systems.

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Most popular questions from this chapter

Obtain a formula for the local truncation error for the Euler method in terms of \(t\) and the solution \(\phi\) $$ y^{\prime}=2 t+e^{-t y}, \quad y(0)=1 $$

In this problem we cstablish that the local truncation crror for the improved Euler formula is proportional to \(h^{3} .\) If we assume that the solution \(\phi\) of the initial value problem \(y^{\prime}=f(t, y),\) \(y\left(t_{0}\right)=y_{0}\) has derivatives that are continuous through the third order \((f\) has continuous second partial derivatives), it follows that $$ \phi\left(t_{n}+h\right)=\phi\left(t_{n}\right)+\phi^{\prime}\left(t_{n}\right) h+\frac{\phi^{\prime \prime}\left(t_{n}\right)}{2 !} h^{2}+\frac{\phi^{\prime \prime \prime}\left(\bar{t}_{n}\right)}{3 !} h^{3} $$ where \(t_{n}<\bar{t}_{n} \leq t_{n}+h .\) Assume that \(y_{n}=\phi\left(t_{n}\right)\) (a) Show that for \(y_{n+1}\) as given by Eq. ( 5 ) $$ e_{n+1}=\phi\left(t_{n+1}\right)-y_{n+1} $$ $$ \begin{aligned}=\frac{\phi^{\prime \prime}\left(t_{n}\right) h-\left\\{f\left[t_{n}+h, y_{n}+h f\left(t_{n}, y_{n}\right)\right]-f\left(t_{n}, y_{n}\right)\right\\}}{2 !} +\frac{\phi^{\prime \prime \prime}\left(\bar{I}_{n}\right) h^{3}}{3 !} \end{aligned} $$ (b) Making use of the facts that \(\phi^{\prime \prime}(t)=f_{t}[t, \phi(t)]+f_{y}[t, \phi(t)] \phi^{\prime}(t),\) and that the Taylor approximation with a remainder for a function \(F(t, y)\) of two variables is $$ F(a+h, b+k)=F(a, b)+F_{t}(a, b) h+F_{y}(a, b) k $$ $$ +\left.\frac{1}{2 !}\left(h^{2} F_{t t}+2 h k F_{t y}+k^{2} F_{y y}\right)\right|_{x=\xi, y=\eta} $$ where \(\xi\) lies between \(a\) and \(a+h\) and \(\eta\) lies between \(b\) and \(b+k,\) show that the first term on the right side of \(\mathrm{Eq}\). (i) is proportional to \(h^{3}\) plus higher order terms. This is the desired result. (c) Show that if \(f(t, y)\) is linear in \(t\) and \(y,\) then \(e_{n+1}=\phi^{\prime \prime \prime}\left(\bar{t}_{n}\right) h^{3} / 6,\) where \(t_{n}<\bar{t}_{n}

Show that the first order Adams-Bashforth method is the Euler method, and that the first order Adams-Moulton method is the backward Euler method.

Show that the modified Euler formula of Problem 22 is identical to the improved Euler formula of \(\mathrm{Eq} .(5)\) for \(y^{\prime}=f(t, y)\) if \(f\) is linear in both \(t\) and \(y .\)

Find approximate values of the solution of the given initial value problem at \(t=0.5,1.0,1.5,\) and 2.0 , using the specified method. For starting values use the values given by the Runge-Kutta muthod; see Problems 7 through 12 in Section 8.3 . Compare the results of the various methods with each other and with the actual solution (if available). $$ y^{\prime}=\sqrt{t+y}, \quad y(0)=3 $$
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