Question

In Exercises \(1-14\) find a particular solution. $$ y^{\prime \prime}-2 y^{\prime}+y=e^{x}(1-6 x) $$

Short answer

Question: Find the particular solution of the given second-order linear ordinary differential equation: \(y^{\prime \prime}-2 y^{\prime}+y=e^{x}(1-6 x)\). Solution: Using the method of undetermined coefficients, the particular solution of the given ordinary differential equation is \(y_p(x) = xe^{x}\).

Step-by-step solution

Identify the coefficients of the ODE and the non-homogeneous term

Given the ODE: $$ y^{\prime \prime}-2 y^{\prime}+y=e^{x}(1-6 x) $$ We identify the coefficients as \(a = 1\), \(b = -2\), and \(c = 1\). The non-homogeneous term \(g(x) = e^{x}(1-6x)\).

Guess a particular solution with undetermined coefficients

We guess a particular solution of the form: $$ y_p(x)=Axe^{x}+Be^{x} $$ Where \(A\) and \(B\) are constants to be determined.

Find the first and second derivatives of the guessed particular solution

Now we need to find the first and second derivatives of our guessed particular solution, \(y_p(x)\). First derivative: $$ y_p^{\prime}(x) = A(xe^{x}+e^{x})+Be^{x} $$ Second derivative: $$ y_p^{\prime \prime}(x) = A(xe^{x}+2e^{x})+Be^{x} $$

Substitute the derivatives into the ODE and solve for the undetermined coefficients

Now we substitute \(y_p(x)\), \(y_p^{\prime}(x)\), and \(y_p^{\prime \prime}(x)\) into the ODE: $$ (A(xe^{x}+2e^{x})+Be^{x}) - 2(A(xe^{x}+e^{x})+Be^{x})+(Axe^{x}+Be^{x})=e^{x}(1-6x) $$ Which simplifies to: $$ -6Axe^{x}+e^{x} = e^{x}(1-6x) $$ Comparing the coefficients of \(e^{x}\) and \(xe^{x}\) terms, we get the following system of equations: \begin{align} -6A &= -6 \\ 1-2A+2B &= 1 \end{align} Solving this system of equations, we get \(A=1\) and \(B=0\).

Write the particular solution

Substituting the coefficients back into our guessed function, we get the particular solution: $$ y_p(x) = xe^{x} $$

Key concepts

Non-homogeneous Differential Equation

In the realm of differential equations, we often find ourselves working with both homogeneous and non-homogeneous equations. But what separates them? A non-homogeneous differential equation includes a term that does not have the function or its derivatives in it, a term often referred to as the 'non-homogeneous term'. This term is crucial because it adds a bit more complexity and requires us to find a *particular solution* in addition to the general solution for the homogeneous part.

To make sense of this, take the differential equation from the exercise: \( y'' - 2y' + y = e^x(1-6x) \). Here, \( e^x(1-6x) \) is the non-homogeneous component. Without it, the equation would be homogeneous, meaning it wouldn't be equal to anything but zero on the right side. Recognizing this allows us to approach solving the equation with methods specific to non-homogeneous equations.

Particular Solution

When we deal with non-homogeneous differential equations, we need to find a solution that specifically satisfies the non-homogeneous part - this is known as the *particular solution*. It's a function that you can plug into the equation that makes both sides equal, but it typically won't capture all possible behaviors of the system; that's where the general solution of the homogeneous part comes in.

In our exercise, the particular solution we found is \( y_p(x) = xe^x \). Once identified, it works together with solutions of the homogeneous part to form the complete solution of the differential equation. This demonstrates the necessity to experiment with forms of functions, such as polynomials, exponentials, or combinations thereof, to find a viable particular solution.

Undetermined Coefficients

The method of undetermined coefficients is a systematic approach for finding a particular solution when dealing with linear differential equations with constant coefficients. The technique involves "guessing" the form of a particular solution based on the non-homogeneous term and then determining the specific coefficients that make it work through substitution into the original equation.

In our exercise, the guess was \( y_p(x)=Axe^x+Be^x \). Here, \( A \) and \( B \) are the undetermined coefficients we solve for. By computing derivatives of our guessed function and substituting them back into the original differential equation, we match coefficients to solve for these unknowns. This process may involve simple algebraic manipulation and matching similar terms on both sides of the equation.

Second Order Differential Equation

Second order differential equations are those in which the highest derivative is the second derivative. These types of equations are prevalent across various domains due to their ability to describe phenomena involving acceleration, curvature, and other dimensions of dynamical systems.

For instance, the given differential equation in our exercise: \( y'' - 2y' + y = e^x(1-6x) \), is a second-order equation. Here, \( y'' \) represents the second derivative, which involves changes in the rate of the rate of change of the original function. This aspect makes them intriguing yet challenging, requiring robust techniques such as undetermined coefficients to solve fully. Understanding such equations is a cornerstone in the study of differential equations, providing insights into modeling and solving real-world problems.