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In the realm of differential equations, a **homogeneous equation** is characterized by all terms involving the function and its derivatives being set to zero. These equations take the form:

• The homogeneous part of our example is: \[y'' - 2y' + y = 0\]
• Such equations are fundamental because they help us understand the underlying structure of the solution to more complex problems.

We tackle homogeneous equations by finding solutions that satisfy the equation when the right-hand side is zero. For linear differential equations with constant coefficients, like in our example, we derive the solution using the characteristic equation. Homogeneous equations are simpler to solve because they involve fewer terms, and understanding them provides a foundation for solving non-homogeneous equations.

A **particular solution** specifically solves the non-homogeneous differential equation. It accounts for the 'extra' parts of the equation that are not zero. For our equation: \[y'' - 2y' + y = 3 + e^x\]The right-hand side shows that it is non-homogeneous. The form of our particular solution is given as:\[y_p = A + Bx^2 e^x\]This is a speculative step where we select a form that mirrors the structure of the non-homogeneous part we are tackling. The coefficients, such as \(A\) and \(B\), are unknowns we must determine.

• Insert this particular solution into the original equation to find values for these coefficients.
• As shown, isolating and comparing coefficients on both sides helps find these values.

By finding this solution, we address the specific nature of the non-zero right-hand side of the equation.

The key to solving homogeneous linear differential equations like ours is the **characteristic equation**. Formed by replacing the derivatives with powers of an unknown (usually \(r\)), it converts a differential equation into an algebraic one. For \[y'' - 2y' + y = 0\]the characteristic equation would be:\[r^2 - 2r + 1 = 0\]This simplifies the process, as solving this quadratic equation gives us the roots, which inform the general solution for the homogeneous case.

• In our example, the equation factors to \((r - 1)^2 = 0\), leading to repeated roots.
• Repeated roots require a slight modification of solutions.

Instead of simple exponentials like \(e^{rx}\), we use repeated terms, leading to solutions like \(C_1 e^x + C_2 xe^x\). Understanding this connection is critical to mastering differential equations.

Combining solutions from both the homogeneous and non-homogeneous equations offers the **general solution**. This solution accounts for all possible behaviors of the system described by the differential equation. The general solution is the sum of the homogeneous solution \(y_h\) and the particular solution \(y_p\):\[y = y_h + y_p\]For our example, the homogeneous solution is:\[y_h = C_1 e^x + C_2 xe^x\]The particular solution, as found, is:\[y_p = 3\]Therefore, the complete general solution is:\[y = C_1 e^x + C_2 xe^x + 3\]This expression incorporates all constants of integration and ensures that the solution satisfies every initial condition or boundary condition that may apply.