$\text { Find the differential equation having the solution } \sum _ { n = 0 } ^ { \infty } \frac { ( - 1 ) ^ { n } ( 4 x ) ^ { 2 n + 1 } } { ( 2 n + 1 ) ! } \text {. }$
A) $y ^ {tt } - 16 y = 0$
B) $y ^ {tt } + 16 y ^ { \prime } = 0$
C) $y ^ { t } - 16 y = 0$
D) $y ^ { tt } + 16 y = 0$
E) $y ^ { t } + 16 y = 0$

Question

$\text { Find the differential equation having the solution } \sum _ { n = 0 } ^ { \infty } \frac { ( - 1 ) ^ { n } ( 4 x ) ^ { 2 n + 1 } } { ( 2 n + 1 ) ! } \text {. }$ 
A) $y ^ {tt } - 16 y = 0$
B) $y ^ {tt } + 16 y ^ { \prime } = 0$
C) $y ^ { t } - 16 y = 0$
D) $y ^ { tt } + 16 y = 0$
E) $y ^ { t } + 16 y = 0$

Quizplus · Accepted Answer

The given solution is in the form of a power series, and we can recognize this as the power series expansion for sin(2x). Taking the second derivative of sin(2x) gives us -4sin(2x), so we obtain the differential equation y'' + 16y = 0, which matches with Option D.

A) ytt−16y=0y ^ {tt } - 16 y = 0ytt−16y=0

A) $y ^ {tt } - 16 y = 0$