Consider the initial value problem: $y^{\mathrm{\text{'}\text{'}}}+16y=\cos \left(4t\right)$, with $y\left(0\right)=0$ and $y^{\mathrm{\text{'}}}\left(0\right)=0$. What is the particular solution $y\left(t\right)$?

• A. $y\left(t\right)=\frac{1}{16}\cos \left(4t\right)$
• B. $y\left(t\right)=\frac{1}{16}\sin \left(4t\right)$
• C. $y\left(t\right)=\frac{1}{32}t\sin \left(4t\right)$
• D. $y\left(t\right)=\frac{1}{32}\sin \left(4t\right)$

Answer: (C)

Verified step by step guidance

1. Step 1: Recognize that this is a second-order linear differential equation with constant coefficients. The equation is y'' + 16y = cos(4t), and we are tasked with finding the particular solution y(t). The initial conditions are y(0) = 0 and y'(0) = 0.
2. Step 2: Solve the homogeneous equation y'' + 16y = 0. The characteristic equation is r^2 + 16 = 0, which has roots r = ±4i. This implies the general solution to the homogeneous equation is y_h(t) = C_1 cos(4t) + C_2 sin(4t).
3. Step 3: To find the particular solution y_p(t), use the method of undetermined coefficients. Since the right-hand side is cos(4t), propose a solution of the form y_p(t) = t(A cos(4t) + B sin(4t)). The factor of t is included because cos(4t) is a solution to the homogeneous equation.
4. Step 4: Substitute y_p(t) into the original differential equation y'' + 16y = cos(4t). Compute y_p'(t) and y_p''(t), then substitute these into the equation. Collect terms involving cos(4t) and sin(4t) to solve for the coefficients A and B.
5. Step 5: After determining A and B, combine the particular solution y_p(t) with the homogeneous solution y_h(t) to form the general solution. Apply the initial conditions y(0) = 0 and y'(0) = 0 to solve for the constants C_1 and C_2. The final particular solution is y(t) = (1/32)t sin(4t).