Question

In Exercises \(1-14\) find a particular solution. $$ y^{\prime \prime}+y^{\prime}-12 y=e^{3 x}(-6+7 x) $$

Short answer

Based on the given second-order non-homogeneous differential equation and the step-by-step solution provided, answer the following question: Q: Find the particular solution of the differential equation \(y^{\prime\prime} + y^{\prime} - 12y = e^{3x}(-6 + 7x)\) A: The particular solution is \(y_p(x) = 3x e^{3x} + 4 e^{3x}\).

Step-by-step solution

Solve the Homogeneous Equation

First, let's consider the homogeneous equation, which is the one without the right side: $$ y^{\prime\prime} + y^{\prime} - 12y = 0 $$ To find the complementary function, we will assume the solution is in the form of \(y = e^{rx}\), where \(r\) is a constant. Plugging this into the homogeneous equation, we get the characteristic equation: $$ r^2 + r - 12 = 0 $$ Factor the characteristic equation and solve for r: $$ (r + 4)(r - 3) = 0 $$ We have two roots, \(r_1 = -4\) and \(r_2 = 3\). Therefore, the complementary function is given by: $$ y_c(x) = c_1e^{-4x} + c_2e^{3x} $$ where \(c_1\) and \(c_2\) are constants.

Assume a Form for the Particular Solution

Next, we need to find a particular solution for the given non-homogeneous equation. Let's assume the form of the particular solution as: $$ y_p(x) = A x e^{3x} + B e^{3x} $$ An extra x term is added in front of \(A\) to avoid duplicating terms from the complementary function.

Find the Derivatives and Substitute into the Equation

Find the first and second derivatives of \(y_p(x)\): First derivative: $$ y_p'(x) = A e^{3x} (3x + 1) + 3B e^{3x} $$ Second derivative: $$ y_p''(x) = 9A e^{3x} (x + 1) + 9B e^{3x} $$ Now, substitute \(y_p(x)\), \(y_p'(x)\), and \(y_p''(x)\) into the given non-homogeneous differential equation: $$ (9A e^{3x} (x + 1) + 9B e^{3x})+(A e^{3x} (3x + 1) + 3B e^{3x})-12(A x e^{3x} + B e^{3x})=e^{3x}(-6+7x) $$

Solve for the Coefficients A and B

Simplify and match the coefficients: $$ [(7A - 12)x + 12A - 9B + 3A] e^{3x} = (-6+7x) e^{3x} $$ So we have the following system of equations: $$ 7A -12 = 7\\ 12A - 9B + 3A = -6 $$ Solving for \(A\) and \(B\), we get \(A = 3\) and \(B = 4\).

Write the Particular Solution

Plug the values of \(A\) and \(B\) into the assumed form of the particular solution: $$ y_p(x) = 3x e^{3x} + 4 e^{3x} $$ This is the particular solution for the given non-homogeneous differential equation.

Key concepts

Particular Solution

In the world of differential equations, a "particular solution" refers to a solution that satisfies the entire non-homogeneous differential equation. This solution includes both the steady-state (or constant) and the varying components of the system described by the equation. To find a particular solution, we typically assume a form that resembles the non-homogeneous part of the equation.

In the given exercise, we assume that the particular solution is of the form \(y_p(x) = Ax e^{3x} + B e^{3x}\). This assumption is strategic to ensure that the solution balances the non-homogeneous part \(e^{3x}(-6+7x)\).

To find the coefficients \(A\) and \(B\), we differentiate our assumed solution, plug these derivatives back into the original differential equation, and then equate coefficients with those from the non-homogeneous part. This method simplifies the system of equations, allowing us to solve for \(A = 3\) and \(B = 4\). Thus, the particular solution becomes \(y_p(x) = 3x e^{3x} + 4 e^{3x}\).

Homogeneous Equation

The homogeneous equation is derived from a differential equation by setting the non-homogeneous part equal to zero. This simplifies the problems and focuses on the intrinsic behavior of the differential equation system. For the given exercise, the homogeneous equation is \(y'' + y' - 12y = 0\). Finding the solution to this equation gives us important insights into the natural response of the system.

Solving the homogeneous equation involves finding the complementary function, which represents the general solution to the homogeneous part. We assume this solution takes the form of \(y = e^{rx}\), where \(r\) is a constant. Plugging this assumed solution into the homogeneous equation leads us to develop the characteristic equation, a fundamental tool in differential equations.

Complementary Function

The complementary function is crucial in solving differential equations because it represents the general solution to the associated homogeneous equation. It models how the system behaves over time when there are no external forces (non-homogeneous part).

For the homogeneous equation \(y'' + y' - 12y = 0\), the complementary function is determined after finding the roots of the characteristic equation. Here, assuming a solution of the form \(y = e^{rx}\) gives us the characteristic equation \(r^2 + r - 12 = 0\). Solving this by factoring results in roots \(r_1 = -4\) and \(r_2 = 3\).

These roots thus give us the complementary function \(y_c(x) = c_1e^{-4x} + c_2e^{3x}\), where \(c_1\) and \(c_2\) are constants determined by initial conditions or specific boundary values related to the problem.

Characteristic Equation

The characteristic equation is a polynomial equation obtained from a homogeneous differential equation by assuming a solution of the form \(y = e^{rx}\). This is where the constant \(r\) becomes pivotal, as solving this equation provides the necessary components for determining the complementary function.

In the provided exercise, we transform the homogeneous differential equation \(y'' + y' - 12y = 0\) into a characteristic equation. By substituting \(y = e^{rx}\) into the homogeneous equation, we arrive at \(r^2 + r - 12 = 0\). Solving this quadratic equation by factoring, we derive the roots \(r = -4\) and \(r = 3\).

These roots form the foundation of our complementary function, with each root corresponding to an exponential term in the function \(e^{rx}\). This provides the structural basis for the solution to the homogeneous equation, integral in completing the solution to the entire differential equation.