Question

In Exercises \(1-14\) find a particular solution. $$ y^{\prime \prime}-6 y^{\prime}+5 y=e^{-3 x}(35-8 x) $$

Short answer

Question: Determine the general solution of the following non-homogeneous second-order linear differential equation. $$ y^{\prime \prime}-6 y^{\prime}+5 y=e^{-3 x}(35-8 x) $$ Answer: The general solution to the given differential equation is: $$ y(x) = C_1e^x + C_2e^{5x} + e^{-3x}(-2x^2 + 2x - 5) $$ where \(C_1\) and \(C_2\) are arbitrary constants.

Step-by-step solution

Finding the complementary function

First, we will find the complementary function for the homogeneous equation: $$ y^{\prime \prime}-6 y^{\prime}+5 y=0 $$ The characteristic equation for this equation is: $$ r^2 - 6r + 5 = 0 $$ Upon solving, we get \(r_1 = 1\) and \(r_2 = 5\). Therefore, the complementary function is: $$ y_c(x) = C_1e^x + C_2e^{5x} $$

Choosing the form of the particular integral

We will use the undetermined coefficients method to find the particular integral. The non-homogeneous term is of the form: $$ e^{-3x}(35-8x) $$ As per the non-homogeneous term, we can assume the particular integral to be of the following form: $$ y_p(x) = e^{-3x}(Ax^2 + Bx + C) $$ Now, we need to find the first and second derivatives of \(y_p(x)\) with respect to x.

Computing the derivatives

The first and second derivatives of the assumed particular integral are as follows: $$ y_p^{\prime}(x) = e^{-3x}(-3Ax^2 - (6A + 3B)x + (2B - 3C)) $$ and $$ y_p^{\prime \prime}(x) = e^{-3x}(9Ax^2 + (18A + 9B)x - (6B - 9C)) $$

Substituting the derivatives into the given equation

Substitute \(y_p(x)\), \(y_p^{\prime}(x)\) and \(y_p^{\prime \prime}(x)\) into the given differential equation: $$ y^{\prime \prime}-6 y^{\prime}+5 y=e^{-3 x}(35-8 x) $$ So we get: $$ e^{-3x}(9Ax^2 + (18A + 9B)x - (6B - 9C)) - 6e^{-3x}(-3Ax^2 - (6A + 3B)x + (2B - 3C)) + 5e^{-3x}(Ax^2 + Bx + C) = e^{-3x}(35-8x) $$

Comparing the coefficients and solving for A, B, and C

Comparing the coefficients on the left and right-hand side, we get the following linear system of equations: $$ \begin{cases} 9A-18A+5A=0 \\ 18A+9B+6A+3B=0 \\ -6B-9C+2B-3C=-35 \\ \end{cases} $$ Solving this linear system, we find \(A=-2\), \(B=2\), and \(C=-5\). Thus, the particular integral is $$ y_p(x) = e^{-3x}(-2x^2 + 2x - 5) $$

Combining the complementary function and particular integral

The general solution to the given differential equation is the sum of the complementary function and the particular integral: $$ y(x) = y_c(x) + y_p(x) = C_1e^x + C_2e^{5x} + e^{-3x}(-2x^2 + 2x - 5) $$ This is the general solution to the given non-homogeneous second-order linear differential equation.

Key concepts

Complementary Function

When solving a second-order linear differential equation, one of the first steps is to find the complementary function, which is the solution to the associated homogeneous equation (where the right-hand side is zero). The complementary function, often denoted as y_c(x), is critical because it represents the general solution to the homogeneous part of the differential equation.

The complementary function is constructed using the solutions to the characteristic equation, which typically results in a combination of exponentials, sines, and cosines. For example, in our exercise, the characteristic equation r^2 - 6r + 5 = 0 yielded roots r_1 = 1 and r_2 = 5. This provided us with the complementary function y_c(x) = C_1e^x + C_2e^{5x} where C_1 and C_2 are constants determined by initial conditions or boundary values.

Particular Integral

The particular integral is a specific solution to the non-homogeneous differential equation. Its purpose is to account for the non-zero right-hand side of the equation. To determine the particular integral, denoted as y_p(x), you must first identify the form of the non-homogeneous term and then assume a function that reflects this form with undetermined coefficients.

For instance, the non-homogeneous term in our exercise was e^{-3x}(35-8x), resulting in an assumed particular integral of y_p(x) = e^{-3x}(Ax^2 + Bx + C). The correct form to assume for the particular integral is crucial, and it often mirrors the non-homogeneous term but with added flexibility (in the form of undetermined coefficients) to satisfy the equation.

Undetermined Coefficients Method

The undetermined coefficients method is a technique used to find the particular integral of a non-homogeneous linear differential equation. Once a suitable form for the particular integral is assumed – based on the non-homogeneous term – the method involves calculating its derivatives and substituting them back into the differential equation. This process helps to find values for the undetermined coefficients.

After substituting the derivatives of the assumed particular integral into the original differential equation, as shown in our exercise, we equate the coefficients of like terms on both sides of the equation. This equating generates a system of algebraic equations that, once solved, provides the values for the coefficients A, B, and C. By meticulously following this process, one can effectively derive a particular integral that fully satisfies the given non-homogeneous equation.

Characteristic Equation

The characteristic equation is a cornerstone concept while solving second-order homogeneous linear differential equations. It is a polynomial equation obtained by substituting a trial solution of the form e^{rx} into the homogeneous equation, where r is the eigenvalue or root of the equation.

For our exercise, the characteristic equation derived from the homogeneous counterpart of the provided differential equation is r^2 - 6r + 5 = 0. Solving the characteristic equation yields the roots, which are then used to construct the complementary function. It's important to solve the characteristic equation accurately, as these roots are pivotal in forming the basis of the complementary function and ultimately in finding the full solution to the differential equation.