Solve the initial value problem y'' = 4sin 2x with y(0) = 0…

Solve the initial value problem $y'' = 4\sin 2x$ with $y(0) = 0$, $y'(0) = 0$.

Step 1: First integration

Integrate both sides with respect to $x$: \[y' = \int 4\sin 2x\,dx = -\frac{4}{2}\cos 2x + C_1 = -2\cos 2x + C_1.\]

Apply the initial condition $y'(0) = 0$: \[0 = -2\cos 0 + C_1 = -2 + C_1 \implies C_1 = 2.\]

Thus: \[y' = -2\cos 2x + 2. \tag{1}\]

Step 2: Second integration

Integrate (1): $y = \int\left(-2\cos 2x + 2\right)dx = -\sin 2x + 2x + C_2.$

Apply the initial condition $y(0) = 0$: \[0 = -\sin 0 + 0 + C_2 = C_2 \implies C_2 = 0.\]

Step 3: Particular solution

The particular solution satisfying the initial conditions is recorded in the result below.

Step 4: Verification

Differentiate twice: $y' = 2 - 2\cos 2x$, $y'' = 4\sin 2x$. $\checkmark$

Check initial conditions: \[y(0) = 0 - \sin 0 = 0,\qquad y'(0) = 2 - 2\cos 0 = 0.\ \checkmark\]