Problem 4 - Advanced Engineering Mathematics

Question

\(x=0\) is an ordinary point of a certain linear differential equation. After the assumed solution \(y=\sum_{n=0}^{\infty} c_{n} x^{n}\) is substituted into the \(\mathrm{DE}\), the following algebraic system is obtained by equating the coefficients of \(x^{0}, x^{1}, x^{2}\), and \(x^{3}\) to zero: $$ \begin{aligned} 2 c_{2}+2 c_{1}+c_{0} &=0 \\ 6 c_{3}+4 c_{2}+c_{1} &=0 \\ 12 c_{4}+6 c_{3}+c_{2}-\frac{1}{3} c_{1} &=0 \\ 20 c_{5}+8 c_{4}+c_{3}-\frac{2}{3} c_{2} &=0 \end{aligned} $$ Bearing in mind that \(c_{0}\) and \(c_{1}\) are arbitrary, write down the first five terms of two power series solutions of the differential equation.

Step-by-Step Solution

Answer

The power series solutions are: 1. \(y_1(x) = 1 - \frac{x^2}{2} + \frac{x^3}{3} + \frac{x^4}{36} + \frac{37x^5}{720}\). 2. \(y_2(x) = x - x^2 + \frac{x^3}{2} + \frac{5x^4}{12} + \frac{7x^5}{20}\).

Step 1: Analyze the Given System

The given system of equations represents the coefficients of the power series solution for the differential equation in terms of base functions. Notice that the coefficients like \(c_2\), \(c_3\), \(c_4\), and \(c_5\) depend on the arbitrary constants \(c_0\) and \(c_1\).

Step 2: Solve the First Equation

Solve for \(c_2\) from the first equation: \(2c_2 + 2c_1 + c_0 = 0\). Isolate \(c_2\) to get: \(c_2 = -c_1 - \frac{c_0}{2}\).

Step 3: Solve the Second Equation

Substitute \(c_2 = -c_1 - \frac{c_0}{2}\) into the second equation \(6c_3 + 4c_2 + c_1 = 0\). This gives \(6c_3 - 4c_1 - 2c_0 + c_1 = 0\), leading to \(6c_3 = 3c_1 + 2c_0\), and thus \(c_3 = \frac{c_1}{2} + \frac{c_0}{3}\).

Step 4: Solve the Third Equation

Substitute \(c_3 = \frac{c_1}{2} + \frac{c_0}{3}\) and \(c_2 = -c_1 - \frac{c_0}{2}\) into the third equation \(12c_4 + 6c_3 + c_2 - \frac{1}{3}c_1 = 0\). Solve to find \(c_4 = \frac{5c_1}{12} + \frac{c_0}{36}\).

Step 5: Solve the Fourth Equation

Replace \(c_4 = \frac{5c_1}{12} + \frac{c_0}{36}\), \(c_3 = \frac{c_1}{2} + \frac{c_0}{3}\), and \(c_2 = -c_1 - \frac{c_0}{2}\) into the fourth equation \(20c_5 + 8c_4 + c_3 - \frac{2}{3}c_2 = 0\). Solve to find \(c_5 = \frac{7c_1}{20} + \frac{37c_0}{720}\).

Step 6: Construct the Power Series Solutions

Use the coefficients found to construct the power series solutions. The first solution starts with \(c_0 = 1, c_1 = 0, etc.\) and the second with \(c_0 = 0, c_1 = 1, etc.\). This renders two solutions: 1. \(y_1(x) = c_0 + (-\frac{1}{2}c_0)x^2 + \frac{1}{3}c_0 x^3 + \frac{1}{36}c_0 x^4 + \frac{37}{720}c_0 x^5\). 2. \(y_2(x) = c_1 x - c_1 x^2 + \frac{1}{2}c_1 x^3 + \frac{5}{12}c_1 x^4 + \frac{7}{20}c_1 x^5\).

Key Concepts

ordinary point differential equation

An ordinary point in the context of differential equations is a value where the coefficients of the differential equation are analytic, meaning they can be expressed as a power series. This ensures that near this point, solutions can be expressed as power series as well. For the differential equation given in the exercise, the point \(x=0\) is identified as an ordinary point. This means around \(x=0\), our differential equation behaves "nicely" and we can use power series to express the solution.

The significance of finding an ordinary point lies in the fact that it guarantees the existence of convergent power series solutions in the neighborhood around that point. Since \(x=0\) is an ordinary point, we assume a solution of the form \( y = \sum_{n=0}^{\infty} c_n x^n \), where \(c_n\) are the coefficients of the power series. This approach simplifies complex differential equations by transforming them into algebraic systems, which are often easier to handle.

algebraic system of equations

By substituting the assumed power series solution into the differential equation, we derive an algebraic system of equations. These equations involve the coefficients of the power series. The equations seen in the original step-by-step solution arise from matching the coefficients of similar powers of \(x\) to zero. This ensures that the power series stands as a legitimate solution to the differential equation.

The system includes equations for each power, beginning with \(x^0, x^1, \) and so on, like the ones provided:

• \(2c_2 + 2c_1 + c_0 = 0\)
• \(6c_3 + 4c_2 + c_1 = 0\)
• \(12c_4 + 6c_3 + c_2 - \frac{1}{3}c_1 = 0\)
• \(20c_5 + 8c_4 + c_3 - \frac{2}{3}c_2 = 0\)

Solving this algebraic system gives us the relationships between different series coefficients, which ultimately help us construct the power series solutions.

coefficients of power series

The coefficients of the power series, denoted as \(c_n\), play a crucial role in determining the series solution. These coefficients are dependent on arbitrary constants, usually represented by initial terms such as \(c_0\) and \(c_1\). Solving the algebraic system of equations obtained helps figure out these coefficients based on assumptions in the series solution.

For instance, in the given equations:

• \(c_2\) is expressed as \(-c_1 - \frac{c_0}{2}\)
• \(c_3\) depends on \(\frac{c_1}{2} + \frac{c_0}{3}\)
• Subsequent coefficients are determined similarly by expressing them through previous coefficients and arbitrary constants

These coefficients allow us to effectively write down the power series solutions for the differential equation.

linear differential equation solutions

Linear differential equations are equations that can be expressed in the form \(a_n(x)y^{(n)} + a_{n-1}(x)y^{(n-1)} + ... + a_1(x)y' + a_0(x)y = f(x)\), where \(a_i(x)\) are functions of \(x\) and \(f(x)\) is a given function. These types of equations have systematic methods for finding solutions such as the power series method used here.

Solving the linear differential equation involves finding particular and general solutions which satisfy the equation. Since these equations are linear, solutions superimpose. That is, individual solutions can be added together to form a general solution. The provided power series solutions in the exercise are examples of how we can express the solutions:

• Solution 1 starts with \(c_0 = 1, c_1 = 0\)
• Solution 2 starts with \(c_0 = 0, c_1 = 1\)

Each represents a linear combination of functions, demonstrating how power series provide an effective mechanism to tackle differential equations at an ordinary point.