Solve ODE by Laplace transform y''+y=sin3t

The initial condition y′(0) is missing in the question (scan problem). The evident version is y″ + y = sin 3t with y(0) = 0 and y′(0) = 0, and it is solved here. (If y′(0) = k, add the term k sin t to the answer.)

Step 1: Take the Laplace transform. Let Y(s) = L[y].

L[y″] = s²Y − s·y(0) − y′(0) = s²Y
s²Y + Y = 3/(s² + 9)

Step 2: Solve for Y.

Y = 3 / [(s² + 1)(s² + 9)]

Partial fractions: 3/[(s²+1)(s²+9)] = (3/8)[1/(s²+1) − 1/(s²+9)].

Step 3: Inverse transform.

y(t) = (3/8) sin t − (3/8)(1/3) sin 3t
     = (3/8) sin t − (1/8) sin 3t

Check: y(0) = 0 and y′(0) = 3/8 − 3/8 = 0. Also y″ + y = −(3/8) sin t + (9/8) sin 3t + (3/8) sin t − (1/8) sin 3t = sin 3t. ✓

Answer: y(t) = (3 sin t − sin 3t)/8