Question

Solve the given differential equation by means of a power series about the given point \(x_{0} .\) Find the recurrence relation; also find the first four terms in each of two linearly independent solutions (unless the series terminates sooner). If possible, find the general term in each solution. \(\left(4-x^{2}\right) y^{\prime \prime}+2 y=0, \quad x_{0}=0\)

Short answer

Answer: The first four terms of each linearly independent solution are as follows: \(y_1(x) = 1 - x^2 + \dfrac{1}{6} x^4 - \dfrac{1}{90} x^6 + \cdots\) and \(y_2(x) = x - \dfrac{1}{3} x^3 + \dfrac{1}{10} x^5 - \dfrac{1}{210} x^7 + \cdots\).

Step-by-step solution

Assume a power series solution for y(x)

Assume the solution in the form of a power series, centered at \(x_0=0\): \(y(x) = \sum_{n=0}^{\infty} a_n x^n\) First, determine its first and second derivatives to substitute in the given differential equation: \(y'(x) = \sum_{n=1}^{\infty} n a_n x^{n-1}\) \(y''(x) = \sum_{n=2}^{\infty} n(n-1) a_n x^{n-2}\)

Substitute the series into the differential equation

Substitute the power series of \(y(x)\), \(y'(x)\) and \(y''(x)\) into the given equation: \(\left(4-x^2\right) \sum_{n=2}^{\infty} n(n-1)a_n x^{n-2} + 2 \sum_{n=0}^{\infty} a_n x^n = 0\)

Rewrite the equation with matching powers

Now rewrite the expression such that all terms have matching powers of \(x\): \(\sum_{n=2}^{\infty} n(n-1)a_n x^{n} (4-x^2) + \sum_{n=0}^{\infty} 2a_n x^n = 0\)

Find the recurrence relation

To find the recurrence relation, we need to equate coefficients of like terms, i.e., coefficients of \(x^n\) must be equal on both sides of the equation. To do this, we can rewrite the sums above as: \(\sum_{n=0}^{\infty} [2a_n + 4(n+2)(n+1)a_{n+2} - (n+2)(n+1) a_{n+2}] x^n = 0\) Equating the coefficients of like terms, we get: \(2a_n + 4(n+2)(n+1)a_{n+2} - (n+2)(n+1) a_{n+2} = 0\) Solving for \(a_{n+2}\), we get the following recurrence relation: \(a_{n+2} = \dfrac{-2a_n}{(n+1)(n+2)}\)

Finding the first four terms in each solution

Using the recurrence relation, calculate the first four terms for each of the linearly independent solutions: First solution (\(a_0 = 1, a_1 = 0\)): \(a_0 = 1\) \(a_2 = \dfrac{-2a_0}{1\cdot2} = -1\) \(a_4 = \dfrac{-2a_2}{3\cdot4} = \dfrac{1}{6}\) \(a_6 = \dfrac{-2a_4}{5\cdot6} = -\dfrac{1}{90}\) So, the first linearly independent solution is: \(y_1(x) = 1 - x^2 + \dfrac{1}{6} x^4 - \dfrac{1}{90} x^6 + \cdots\) Second solution (\(a_0 = 0, a_1 = 1\)): \(a_1 = 1\) \(a_3 = \dfrac{-2a_1}{2\cdot3} = -\dfrac{1}{3}\) \(a_5 = \dfrac{-2a_3}{4\cdot5} = \dfrac{1}{10}\) \(a_7 = \dfrac{-2a_5}{6\cdot7} = -\dfrac{1}{210}\) So, the second linearly independent solution is: \(y_2(x) = x - \dfrac{1}{3} x^3 + \dfrac{1}{10} x^5 - \dfrac{1}{210} x^7 + \cdots\)

Checking for a general term

Analyzing the solutions given above, no simple general formula for the terms in \(y_1(x)\) and \(y_2(x)\) can be deducted. Therefore, the answer will be in the form of the first four terms of each linearly independent solution: \(y_1(x) = 1 - x^2 + \dfrac{1}{6} x^4 - \dfrac{1}{90} x^6 + \cdots\) \(y_2(x) = x - \dfrac{1}{3} x^3 + \dfrac{1}{10} x^5 - \dfrac{1}{210} x^7 + \cdots\)

Key concepts

Differential Equations

A differential equation is a mathematical equation that involves the derivatives of a function. In simple terms, derivatives represent the rates at which things change. Differential equations are used to describe everything from the motion of particles to changes in population.When dealing with differential equations, one of the main goals is to find a function that satisfies the equation. This function is known as the solution. In the case presented, we are looking for a power series solution, which is a type of solution expressed as an infinite sum of terms:

• Each term involves powers of the variable, in this case, powers of \(x\).
• The coefficients of the terms need to be determined to ensure the series satisfies the differential equation.

Using a power series allows us to solve complex equations by breaking them down into simpler, infinite polynomial terms. This approach is especially useful when dealing with functions that do not have a closed-form solution.

Recurrence Relations

A recurrence relation is an equation that defines each term of a sequence using the preceding terms. In our problem, after substituting the power series into the differential equation and finding matching powers of \(x\), we derived a recurrence relation:\[ a_{n+2} = \dfrac{-2a_n}{(n+1)(n+2)} \]The recurrence relation acts like a recipe that tells us how to build the terms of the series step-by-step:

• Knowing the initial terms or conditions (such as \(a_0\) and \(a_1\)), we can compute subsequent terms.
• This process is really beneficial because it transforms the differential equation into a simpler, iterative problem.

In the context of the exercise, the recurrence relation helps us find terms like \(a_2, a_4, a_6\), and so forth. This approach highlights the elegance of using a power series to methodically handle differential equations.

Linearly Independent Solutions

When solving differential equations, we often come across multiple solutions. These solutions are termed 'linearly independent' if no solution can be expressed as a linear combination of others. To illustrate, consider our two solutions:

• First solution: \(y_1(x) = 1 - x^2 + \frac{1}{6} x^4 - \frac{1}{90} x^6 + \cdots\)
• Second solution: \(y_2(x) = x - \frac{1}{3} x^3 + \frac{1}{10} x^5 - \frac{1}{210} x^7 + \cdots\)

These are linearly independent because neither can be derived from the other using scalar multiplication and addition. Having such independent solutions is crucial because:

• They provide a complete set of solutions to the differential equation.
• Any other solution of the differential equation can be expressed as a combination of these basis solutions.

Linearly independent solutions form the foundation for solving higher-order differential equations effectively, providing a rich structure for understanding the solutions' behavior under different conditions.