Given the boundary-value problem $y^{″}-2y^{\prime }+2y=2x-2$, with $y\left(0\right)=0$ and $y\left(\pi \right)=\pi$, which of the following is the correct solution for $y\left(x\right)$?

• A. $y\left(x\right)=x+1+e^x(C\cos x+D\sin x)$
• B. $y\left(x\right)=x+1+e^x(\cos x+\sin x)$
• C. $y\left(x\right)=x+1+e^x(\cos x-\sin x)$
• D. $y\left(x\right)=x+e^x(A\cos x+B\sin x)$

Answer: (B)

Verified step by step guidance

1. Step 1: Start by analyzing the given differential equation y'' - 2y' + 2y = 2x - 2. This is a second-order linear differential equation with constant coefficients. The solution will consist of two parts: the complementary solution (y_c) and the particular solution (y_p).
2. Step 2: Solve for the complementary solution y_c by setting the right-hand side of the equation to zero: y'' - 2y' + 2y = 0. Assume a solution of the form y = e^{rx}, substitute it into the equation, and solve the characteristic equation r^2 - 2r + 2 = 0.
3. Step 3: Solve the characteristic equation r^2 - 2r + 2 = 0 using the quadratic formula r = (-b ± √(b^2 - 4ac)) / 2a. Here, a = 1, b = -2, and c = 2. Compute the discriminant (b^2 - 4ac) to determine the nature of the roots. Since the discriminant is negative, the roots will be complex: r = 1 ± i.
4. Step 4: Write the complementary solution y_c based on the complex roots r = 1 ± i. The general form for complex roots is y_c = e^{αx} (C \, \cos(βx) + D \, \sin(βx)), where α = 1 and β = 1. Thus, y_c = e^{x} (C \, \cos x + D \, \sin x).
5. Step 5: Find the particular solution y_p by guessing a solution based on the form of the non-homogeneous term (2x - 2). Since the right-hand side is a polynomial, guess y_p = Ax + B, substitute it into the original equation, and solve for A and B. Combine y_c and y_p to form the general solution y(x). Finally, apply the boundary conditions y(0) = 0 and y(π) = π to solve for the constants C and D.