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. \(y^{\prime \prime}-y=0, \quad x_{0}=0\)

Short answer

Question: Find the first four terms in each of the two linearly independent power series solutions for the following differential equation: $$y^{\prime\prime}(x) - y(x) = 0$$ Answer: The first four terms of the two linearly independent power series solutions are given by: $$ y_1(x) = 1 + 0\cdot x + \frac{1}{2}x^2 + 0\cdot x^3 + \frac{1}{24}x^4 + \cdots $$ $$ y_2(x) = 0 + x + 0\cdot x^2 + \frac{1}{6}x^3 + 0\cdot x^4 + \cdots $$

Step-by-step solution

Write down the assumed solution

Assume a solution of the form: $$ y(x) = \sum_{n=0}^{\infty}a_n x^n $$ where \(n\) is a non-negative integer and each \(a_n\) is a constant.

Calculate the first and second derivatives of y(x)

Compute the first derivative of the assumed solution with respect to x: $$ y^{\prime}(x) = \frac{d}{dx} \left(\sum_{n=0}^{\infty}a_n x^n\right) = \sum_{n=1}^{\infty}na_n x^{n-1} $$ Compute the second derivative, y''(x): $$ y^{\prime\prime}(x) = \frac{d^2}{dx^2} \left(\sum_{n=1}^{\infty}na_n x^{n-1}\right) = \sum_{n=2}^{\infty}n(n-1)a_n x^{n-2} $$

Substitute y(x), y'(x), and y''(x) into the given equation

Plug the expressions from steps 1 and 2 for y(x), y'(x), and y''(x) into the given differential equation: $$ \sum_{n=2}^{\infty}n(n-1)a_n x^{n-2} - \sum_{n=0}^{\infty}a_n x^n = 0 $$

Match the coefficients

Now, match the coefficients of the same powers of x on both sides of the equation.

Find the recurrence relation

Equating the coefficients of the same powers of x on both sides, we get: $$ n(n-1)a_n = a_{n-2} $$ This is the recurrence relation. Rearrange to find \(a_n\) as a function of \(a_{n-2}\): $$ a_n = \frac{a_{n-2}}{n(n-1)} $$

Find the first four terms in each of two linearly independent solutions

To find the first four terms of each linearly independent solution, we can set \(a_0\) and \(a_1\) as arbitrary constants and compute the other \(a_n\)s using the recurrence relation. For the first solution, let \(a_0 = 1\) and \(a_1 = 0\). Then we compute the terms \(a_2, a_3, a_4\) using the recurrence relation: $$ a_2 = \frac{a_0}{2(2-1)} = \frac{1}{2} $$ $$ a_3 = \frac{a_1}{3(3-1)} = 0 $$ $$ a_4 = \frac{a_2}{4(4-1)} = \frac{1}{2(4)(3)} = \frac{1}{24} $$ So the first solution has the form: $$ y_1(x) = 1 + 0\cdot x + \frac{1}{2}x^2 + 0\cdot x^3 + \frac{1}{24}x^4 + \cdots $$ For the second solution, let \(a_0 = 0\) and \(a_1 = 1\). Then we compute the terms \(a_2, a_3, a_4\) using the recurrence relation: $$ a_2 = \frac{a_0}{2(2-1)} = 0 $$ $$ a_3 = \frac{a_1}{3(3-1)} = \frac{1}{6} $$ $$ a_4 = \frac{a_2}{4(4-1)} = 0 $$ So the second solution has the form: $$ y_2(x) = 0 + x + 0\cdot x^2 + \frac{1}{6}x^3 + 0\cdot x^4 + \cdots $$