Solve the differential equation $y^{\mathrm{\text{'}\text{'}}}+2y^{\mathrm{\text{'}}}=2x+3-e^{-2x}$ using the method of undetermined coefficients. Which of the following is a particular solution?

• A. $y_p\left(x\right)=2x+3+e^{-2x}$
• B. $y_p\left(x\right)=a{x}^2+bx+c+d{e}^{-2x}$, where $a=1$, $b=0$, $c=0$, $d=0$
• C. $y_p\left(x\right)=x^2+2x-e^{-2x}$
• D. $y_p\left(x\right)=x+\frac{3}{2}-\frac{1}{2}{e}^{-2x}$

Answer: (D)

Verified step by step guidance

1. Step 1: Recognize that the given differential equation is a second-order linear non-homogeneous differential equation of the form y'' + 2y' = 2x + 3 - e^{-2x}. The solution will be the sum of the complementary solution (y_c) and a particular solution (y_p).
2. Step 2: Solve for the complementary solution y_c by solving the associated homogeneous equation y'' + 2y' = 0. The characteristic equation is r^2 + 2r = 0, which factors as r(r + 2) = 0. This gives roots r = 0 and r = -2. Thus, the complementary solution is y_c = C_1 + C_2 e^{-2x}, where C_1 and C_2 are constants.
3. Step 3: Use the method of undetermined coefficients to find a particular solution y_p. The non-homogeneous term on the right-hand side is 2x + 3 - e^{-2x}, which is a combination of a polynomial (2x + 3) and an exponential term (-e^{-2x}). Assume a particular solution of the form y_p = Ax + B + De^{-2x}, where A, B, and D are constants to be determined.
4. Step 4: Substitute y_p = Ax + B + De^{-2x} into the original differential equation. Compute the first derivative y_p' = A - 2De^{-2x} and the second derivative y_p'' = 4De^{-2x}. Substitute these into y'' + 2y' = 2x + 3 - e^{-2x} and collect like terms to form an equation for A, B, and D.
5. Step 5: Solve the resulting system of equations for A, B, and D by equating coefficients of like terms (x, constant, and e^{-2x}) on both sides of the equation. This will yield the values A = 1, B = 3/2, and D = -1/2. Thus, the particular solution is y_p(x) = x + 3/2 - (1/2)e^{-2x}.