Question

In Exercises 49-54, show that the function represented by the power series is a solution of the differential equation. $$ y=\sum_{n=0}^{\infty} \frac{x^{2 n}}{(2 n) !}, \quad y^{\prime \prime}-y=0 $$

Short answer

The function \(y = \sum_{n=0}^{\infty} \frac{x^{2n}}{(2n)!}\) is indeed a solution of the differential equation \(y'' - y = 0\). This is proven by differentiating the power series twice to get \(y''\), then subtracting \(y\) from \(y''\) to form the given differential equation and showing that it evaluates to 0.

Step-by-step solution

Differentiation

Differentiate the power series twice to get \(y''\). Using the power rule, the first derivative \(y'\) is given by\[y' = \sum_{n=0}^{\infty} \frac{d}{dx} \left( \frac{x^{2n}}{(2n)!} \right) = \sum_{n=1}^{\infty} \frac{2n x^{2n - 1}}{(2n)!}\]. Differentiating once more gives\[y'' = \sum_{n=1}^{\infty} \frac{d}{dx} \left( \frac{2n x^{2n - 1}}{(2n)!} \right) = \sum_{n=1}^{\infty} \frac{2n(2n - 1) x^{2n - 2}}{(2n)!}\]

Subtraction

Subtract \(y\) from \(y''\) to form the equation \(y'' - y = 0\). This gives\[y'' - y = \sum_{n=1}^{\infty} \frac{2n(2n - 1) x^{2n - 2}}{(2n)!} - \sum_{n=0}^{\infty} \frac{x^{2n}}{(2n)!}\]. Here the task is to prove that the right hand side evaluates to 0.

Simplification

To simplify, we return the two series to their original indices of summation. The first sums from \(n = 1\) and the second from \(n = 0\), so replace \(n\) by \(n+1\) in the first series and regroup. We then get the same series, which can be subtracted term by term, leading to 0. Hence, \(y\) is a solution of the differential equation \(y'' - y = 0\) as claimed.