91Ó°ÊÓ

Differential equations can either be homogeneous or non-homogeneous. In the equation \( y^{\prime \prime}+y^{\prime}=3 e^{x} \), we focus on solving non-homogeneous differential equations. These equations have terms that are not necessarily zero on the right side of the equation.
For our specific equation, the term \( 3e^{x} \) makes it non-homogeneous. This part does not depend on the unknown function \( y \). Instead, it is an extra function added to the homogenous equation. Non-homogeneous equations are a bit more complex to solve because they require finding solutions that account for these additional terms or functions.

Finding a solution to a differential equation often begins with the characteristic equation. This equation helps solve the homogeneous part of a differential equation. For the equation derived from \( y^{\prime \prime}+y^{\prime}=3e^{x} \), the characteristic equation is created from the homogeneous form \( y^{\prime \prime} + y^{\prime} = 0 \).

• First, assume a solution of the form \( y = e^{rx} \).
• By substituting this form into the homogeneous equation, we arrive at the characteristic equation \( r^2 + r = 0 \).

Solving this equation involves determining the values of \( r \) that make the equation true. Here, factoring gives us roots \( r = 0 \) and \( r = -1 \), which are crucial in determining the complementary solution.

The complementary solution is the solution to the associated homogeneous differential equation. It makes use of the roots of the characteristic equation. For the equation \( y^{\prime \prime} + y^{\prime} = 0 \), with roots \( r = 0 \) and \( r = -1 \), the complementary solution takes the form:

• \( y_c = C_1 + C_2 e^{-x} \)

Each term in this solution corresponds to a root of the characteristic equation:

• \( C_1 \) is associated with the root \( r = 0 \).
• \( C_2 e^{-x} \) corresponds with the root \( r = -1 \).

These constants \( C_1 \) and \( C_2 \) are determined when initial conditions or additional information are provided. Until such conditions are known, they remain arbitrary.

To fully solve a non-homogeneous differential equation, identifying a particular solution is necessary. This solution complements the homogeneous part and depends on the non-homogeneous term, which in this case is \( 3e^x \).
Let’s assume a solution form inspired by this term: \( y_p = Ae^x \). Differentiating this form gives \( y_p' = Ae^x \) and \( y_p'' = Ae^x \).
Substituting \( y_p \), \( y_p' \), and \( y_p'' \) into the differential equation \( y^{\prime \prime}+y^{\prime}=3 e^{x} \), we find:

• \( 2Ae^x = 3e^x \)

This equation allows us to solve for \( A \). Here, \( A = \frac{3}{2} \). Thus, the particular solution is \( y_p = \frac{3}{2} e^x \). The particular solution accounts for the non-homogeneous part of the differential equation and is critical to writing the complete general solution.