To solve the given second-order linear differential equation with initial conditions, let's proceed systematically. The equation is:

$y'' - 2y' + y = 2x + 3e^x, \quad y(0) = 1, \quad y'(0) = 2$

Steps to Solve:

Step 1: Solve the homogeneous equation

The corresponding homogeneous equation is:

$y'' - 2y' + y = 0$

The characteristic equation is:

$r^2 - 2r + 1 = 0 \quad \Rightarrow \quad (r - 1)^2 = 0$

This gives a repeated root $r = 1$. Hence, the solution to the homogeneous equation is:

$y_h = (C_1 + C_2x)e^x$

Step 2: Find the particular solution

For the non-homogeneous part, $2x + 3e^x$, we use the method of undetermined coefficients. Let:

$y_p = Ax + B + Cxe^x$

where $Ax + B$ handles the polynomial term $2x$, and $Cxe^x$ handles $3e^x$ (since $e^x$ already appears in $y_h$).

Substitute $y_p$ into the original equation:

$y_p'' - 2y_p' + y_p = 2x + 3e^x$

First, compute the derivatives of $y_p$:

$y_p' = A + Ce^x + Cxe^x, \quad y_p'' = Ce^x + Ce^x + Cxe^x = 2Ce^x + Cxe^x$

Substitute into the equation:

$(2Ce^x + Cxe^x) - 2(A + Ce^x + Cxe^x) + (Ax + B + Cxe^x) = 2x + 3e^x$

Simplify:

$2Ce^x + Cxe^x - 2A - 2Ce^x - 2Cxe^x + Ax + B + Cxe^x = 2x + 3e^x$

Combine like terms:

1. For $e^x$: $2Ce^x - 2Ce^x = 0$
2. For $xe^x$: $Cxe^x - 2Cxe^x + Cxe^x = 0$
3. For $x$: $Ax = 2x \implies A = 2$
4. For constants: $B = 0$, $3e^x = 0 \implies C = 3$

Thus, the particular solution is:

$y_p = 2x + 3xe^x$

Step 3: General solution

The general solution is the sum of $y_h$ and $y_p$:

$y = y_h + y_p = (C_1 + C_2x)e^x + 2x + 3xe^x$

Simplify:

$y = C_1e^x + C_2xe^x + 2x + 3xe^x$

Step 4: Apply initial conditions

1. $y(0) = 1$: $y(0) = C_1e^0 + C_2 \cdot 0 \cdot e^0 + 2 \cdot 0 + 3 \cdot 0 \cdot e^0 = C_1 = 1$

2. $y'(0) = 2$: Differentiate $y$: $y' = C_1e^x + C_2e^x + C_2xe^x + 2 + 3e^x + 3xe^x$ At $x = 0$: $y'(0) = C_1e^0 + C_2e^0 + 0 + 2 + 3e^0 + 0 = C_1 + C_2 + 2 + 3 = 2$ Substitute $C_1 = 1$: $1 + C_2 + 2 + 3 = 2 \quad \Rightarrow \quad C_2 = -4$

Final Solution:

$y = e^x - 4xe^x + 2x + 3xe^x$

Simplify further:

$y = e^x(1 - xe^x) + 2x$

Let me know if you'd like a deeper dive or examples! Here's more to explore:

1. How is the method of undetermined coefficients applied to other equations?
2. Can this be solved using Laplace transforms?
3. Why is the complementary solution based on the characteristic equation?
4. What happens if initial conditions change?
5. What if the non-homogeneous part was a trigonometric function?

Tip: For repeated roots in characteristic equations, always include $C_2x$ in the solution to the homogeneous part.