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