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The characteristic equation is a key step in solving many differential equations, especially linear ones. It's derived from the homogeneous part of the equation, which means you consider the differential equation without the non-homogeneous term (often a function of x like 10x-27 in our example). For our differential equation, we start with the homogeneous counterpart: \( y^{\text{''}} - 7y^{\text{'}} + 10y = 0 \) By assuming a solution of the form \( y = e^{rt} \), where \( r \) is a constant, we substitute and get: \( r^2 e^{rt} - 7r e^{rt} + 10 e^{rt} = 0 \) Since \( e^{rt} eq 0 \), it simplifies to the characteristic equation: \[ r^2 - 7r + 10 = 0 \] This is a quadratic equation, and solving it will give us the roots \( r \), which are crucial for constructing the homogeneous solution.

The particular solution addresses the part of the differential equation that includes the non-homogeneous term. The goal is to find a specific solution that satisfies the entire non-homogeneous equation. Let's continue with our example. For the differential equation: \( y^{\text{''}} - 7y^{\text{'}} + 10y = 10x - 27 \) We assume a particular solution \( y_p \) that fits the form of the non-homogeneous part. In this case, because the non-homogeneous term is a linear polynomial, we can try a particular solution of the same form: \( y_p = Ax + B \) We then calculate the derivatives needed for substitution: \( y_p' = A \) \( y_p^{\text{''}} = 0 \) Substituting \( y_p \), \( y_p' \), and \( y_p^{\text{''}} \) back into the original differential equation gives: \( 0 - 7A + 10(Ax + B) = 10x - 27 \) Simplifying, we get an equation involving \( A \) and \( B \): \( 3Ax + (10B - 7A) = 10x - 27 \) By matching coefficients, we solve for \( A \) and \( B \): \( 3A = 10 \) so \( A = \frac{10}{3} \) \( 10B - 7A = -27 \), which leads to \( B = -\frac{19}{3} \) Thus, our particular solution is: \( y_p = \frac{10}{3}x - \frac{19}{3} \).

The general solution to a non-homogeneous differential equation combines both the homogeneous solution and the particular solution. This is because the solution to the non-homogeneous equation is made up of two parts: one that satisfies the homogeneous equation and another that takes into account the non-homogeneous term. From our example, we have already derived: The homogeneous solution \( y_h \): \( y_h = C_1 e^{2x} + C_2 e^{5x} \) And the particular solution \( y_p \): \( y_p = \frac{10}{3}x - \frac{19}{3} \) Hence, the general solution \( y \) is the sum of both: \( y = y_h + y_p \) So, substituting our solutions, the general solution to the original differential equation is: \( y = C_1 e^{2x} + C_2 e^{5x} + \frac{10}{3}x - \frac{19}{3} \) This general solution encompasses all possible solutions to the differential equation, thanks to the constants \( C_1 \) and \( C_2 \) which can be adjusted based on initial conditions or boundary values.