Solve the differential equation using variation of parameters: $y^{″}-16y=16x{e}^{4x}$. Which of the following is the general solution?

• A. $y=C_1{e}^{4x}+C_2{e}^{-4x}+(x-\frac{1}{4})e^{4x}$
• B. $y=C_1{e}^{4x}+C_2{e}^{-4x}+x^2{e}^{4x}$
• C. $y=C_1{e}^{4x}+C_2{e}^{-4x}+(x-\frac{1}{4})x{e}^{4x}$
• D. $y=C_1{e}^{4x}+C_2{e}^{-4x}+x{e}^{4x}$

Answer: (C)

Verified step by step guidance

1. Step 1: Identify the type of differential equation. The given equation is a second-order linear non-homogeneous differential equation: y'' - 16y = 16x e^{4x}. The solution will consist of the complementary solution (solution to the homogeneous equation) and a particular solution.
2. Step 2: Solve the homogeneous equation y'' - 16y = 0. Assume a solution of the form y = e^{rx}, substitute into the equation, and solve the characteristic equation r^2 - 16 = 0. The roots are r = 4 and r = -4, so the complementary solution is y_c = C_1 e^{4x} + C_2 e^{-4x}.
3. Step 3: Use the method of variation of parameters to find the particular solution. For this method, assume the particular solution has the form y_p = u_1(x)e^{4x} + u_2(x)e^{-4x}, where u_1(x) and u_2(x) are functions to be determined.
4. Step 4: Compute u_1(x) and u_2(x) using the formulas derived from variation of parameters. These formulas involve integrating expressions that include the Wronskian of the solutions e^{4x} and e^{-4x}, as well as the non-homogeneous term 16x e^{4x}. Perform the necessary integrations to find u_1(x) and u_2(x).
5. Step 5: Combine the complementary solution y_c and the particular solution y_p to form the general solution. After simplifying, the general solution is y = C_1 e^{4x} + C_2 e^{-4x} + (x - 1/4) x e^{4x}.