Chapter 14.2, Problem 64E

To determine

show that $\text{w}\left(\text{t}\right)\text{ = }\left(\text{sin }\left(\text{3t+4}\right)\text{, sin }\left(\text{3t-2}\right)\text{, cos }\left(\text{3t}\right)\right)$ satisfies the differential equation $\text{w''}\left(\text{t}\right)\text{ = -9w}\left(\text{t}\right)\text{.}$

To determine

To prove: $\text{w''}\left(\text{t}\right)\text{ = -9w}\left(\text{t}\right)\text{.}$

Proof:

$\text{w}\left(\text{t}\right)\text{ = }\left(\text{sin }\left(\text{3t+4}\right)\text{, sin }\left(\text{3t-2}\right)\text{, cos }\left(\text{3t}\right)\right)$

Differentiating the above equation we get,

$\text{w'}\left(\text{t}\right)\text{ =3cos }\left(\text{3t+4}\right)\text{ , 3cos}\left(\text{3t-2}\right)\text{ , }\left(\text{-3sin }\left(\text{3t}\right)\right)$

Differentiating the derivative vector to find the second derivative

$\text{w''}\left(\text{t}\right)\text{ = }\left(\text{-3×3 sin }\left(\text{3t+4}\right)\right)\text{ , }\left(\text{-3×3 sin }\left(\text{3t-2}\right)\right)\text{ , }\left(\text{-3×3 cos }\left(\text{3t}\right)\right)$

On rearranging we get,

$w″\left(t\right)= -9(\sin (3t+4),\sin (3t-2),\cos (3t))$

$\text{w''}\left(\text{t}\right)\text{ = -9w}\left(\text{t}\right)\text{.}$