Question

How do I show that... [check attached image]
when the 2nd derivative is y''= -800pi^2 (sin10pi t)

Answer 1

why don't you substitute it in?

\[y'' = -800\pi^2 \sin(10\pi t)\]\[y'' +100\pi^2 y = 0 \]\[-800\pi^2 \sin(10\pi t) + 100\pi^2 y = 0\]Now factor out the common factors:
\[-8\sin(10\pi t) + y = 0\]

Answer 2

I think if you solve that for \(y\) and differentiate it twice, you'll get what you have for y''