Question

verify that ( y = sin 2t + 3cos 2t ) is a solution to the initial value problem ( 3y + 12y = 0 ), ( y(0) = 3 ), ( y(0) = 2 ). find the maximum of ( |y(t)| ) for ( -infty < t < infty ). find the derivatives of ( y ). ( y = square )

Explanation:

Step1: Differentiate \( \sin 2t \)

Using the chain rule, the derivative of \( \sin u \) with respect to \( t \) is \( \cos u \cdot u' \). For \( u = 2t \), \( u' = 2 \), so the derivative of \( \sin 2t \) is \( \cos 2t \cdot 2 = 2\cos 2t \).

Step2: Differentiate \( 3\cos 2t \)

Using the chain rule, the derivative of \( \cos u \) with respect to \( t \) is \( -\sin u \cdot u' \). For \( u = 2t \), \( u' = 2 \), so the derivative of \( \cos 2t \) is \( -\sin 2t \cdot 2 = -2\sin 2t \). Then multiply by 3: \( 3 \cdot (-2\sin 2t) = -6\sin 2t \).

Step3: Combine the derivatives

The derivative of \( y = \sin 2t + 3\cos 2t \) is the sum of the derivatives of its terms: \( y' = 2\cos 2t - 6\sin 2t \).

Answer:

\( 2\cos 2t - 6\sin 2t \)