   Chapter 11.9, Problem 34E

Chapter
Section
Textbook Problem

# Show that the function f ( x ) = ∑ n = 0 ∞ ( − 1 ) n x 2 n ( 2 n ) ! is a solution of the differential equation f ″ ( x ) + f ( x ) = 0

To determine

To show: The function f(x)=n=0(1)nx2n(2n)! is a solution of the differential equation f(x)+f(x)=0 .

Explanation

Given:

The function is f(x)=n=0(1)nx2n(2n)!

Calculation:

Let f(x)=n=0(1)nx2n(2n)!

Differentiate the function as shown below,

f(x)=n=1(1)n(2n)x2n1(2n)!(The first term disappears)

Differentiate the function.

f(x)=n=1(1)n(2n)(2n1)x2n2(2n)!=n=1(1)n(2n)(2n1)x2n2(2n)(2n1)!=n=1(1)n(2n1)x2n2(2n1)(2n2)!f(x)=n=1(1)nx2n2(2n2)!

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