Question

Find a particular solution. $$ 2 y^{(4)}-5 y^{\prime \prime \prime}+3 y^{\prime \prime}+y^{\prime}-y=e^{x}(11+12 x) $$

Short answer

Based on the given step by step solution, the process for finding the particular solution to a given ODE involves identifying the form of the particular solution, substituting this form into the ODE, solving the resulting equation for the coefficients, and finally writing the particular solution. In this example, the particular solution for the linear fourth-order inhomogeneous ODE $$2 y^{(4)}-5 y^{\prime \prime \prime}+3 y^{\prime \prime}+y^{\prime}-y=e^{x}(11+12 x)$$ is $$y_p(x) = (12x+23)e^x$$.

Step-by-step solution

Identify the form of the particular solution

The non-homogeneous term of the ODE is given by: \(e^{x}(11+12 x)\). In order to write the form of the particular solution to the given ODE, we will use a trial solution of the form: $$y_p(x) = (Ax + B)e^{x}$$ This is because the given ODE is linear and the non-homogeneous term is composed of the exponential function \(e^x\) multiplied by a first-degree polynomial.

Substitute the form of the particular solution into the ODE

Next, we will substitute the trial solution \(y_p(x)\) into the given ODE. First, we need to find its first four derivatives: $$y_p'(x) = (A + Ax + B)e^x$$ $$y_p^{\prime\prime}(x) = (2A + Ax + B)e^x$$ $$y_p^{\prime\prime\prime}(x) = (3A + Ax + B)e^x$$ $$y_p^{(4)}(x) = (4A + Ax + B)e^x$$ Now, we will substitute these derivatives into the given ODE: $$2(4A + Ax + B)e^x - 5(3A + Ax + B)e^x + 3(2A + Ax + B)e^x + (A + Ax + B)e^x - (Ax + B)e^x = e^x(11 + 12x)$$

Solve the resulting equation for the coefficients of the particular solution

Now we will simplify the left side of the equation: $$ [2(4A+B)-5(3A+B)+3(2A+B)+A]e^x + [2Ax-5Ax+3Ax+Ax-Ax]e^x= e^x(11+12x)$$ $$ [(-A+B)]e^x + [A]xe^x= e^x(11+12x)$$ Now, we can equate the coefficients of similar powers of \(x\) on both sides of the equation: For the \(e^x\) term: $$-A+B=11\implies B=A+11$$ For the \(xe^x\) term: $$A=12$$ Substituting the value of \(A\) into the equation for \(B\), we get: $$B=12+11\implies B=23$$

Write the final particular solution

Now that we have found the coefficients, we can write the final particular solution: $$y_p(x)=(12x+23)e^x$$ The particular solution to the given ODE is: $$y_p(x)=(12x+23)e^x$$

Key concepts

Linear Ordinary Differential Equation

A Linear Ordinary Differential Equation (ODE) is a type of differential equation where the function and its derivatives appear linearly. This means each term in the equation is either the unknown function itself or its derivative, multiplied by a constant or a low-degree polynomial. In the context of the problem provided, the ODE is linear because all terms of the form \( y, y', y'' \) and higher derivatives like \( y^{(4)} \) and \( y''' \) are linear, without any squared or nonlinear components.

Linear ODEs are fundamental in mathematical modeling across various fields, such as physics, engineering, and economics, because they often describe systems' behavior under change. Their solutions usually involve determining both a homogeneous solution (where the non-homogeneous part is \(0\)) and a particular solution to accommodate external forcing terms, in this case \( e^x(11+12x) \). Understanding the linearity helps in using systematic methods like the superposition principle and finding particular solutions using specific techniques.

In our problem, the linear nature permits the use of the Trial Solution Method, which is a systematic way to guess a particular solution based on the form of the non-homogeneous part.

Trial Solution Method

The Trial Solution Method, also known as the method of undetermined coefficients, is a technique to find a particular solution to linear ODEs with constant coefficients. When faced with a non-homogeneous differential equation, as in our exercise, one starts by proposing a 'trial solution' whose form resembles the non-homogeneous term in the equation.

The trial solution must be expressed with undetermined coefficients, which will be adjusted such that the equation holds true. In the given problem, the non-homogeneous term is \( e^x(11+12x) \), an exponential function multiplied by a first-degree polynomial. Thus, a sensible trial solution proposed is \( y_p(x) = (Ax + B)e^x \), where \( A \) and \( B \) are constants to be determined.

After choosing the trial form, the next step involves substituting the trial solution and its derivatives back into the original differential equation. The resulting equation is simplified by matching coefficients of like terms across both sides of the equation. In this exercise, doing so allows for solving the constants \( A \) and \( B \), thus determining the particular solution that satisfies the equation. This method is efficient specifically for linear ODEs with exponential or polynomial non-homogeneous parts.

Exponential Functions in ODEs

Exponential functions frequently appear in solutions of differential equations, especially when modeling growth processes, decay, or systems subject to constant change. In our exercise, the term \( e^x(11+12x) \) presents an exponential component, necessitating techniques that specifically address how exponential functions interact with linear ODEs.

The presence of \( e^x \) in the non-homogeneous term implies that any trial solutions should have \( e^x \) along with polynomial terms to match such a structure. Why exponential functions? They naturally arise when the function's rate of change is proportional to its current value, which is a common description in many physical systems. When you recognize \( e^{something} \), you immediately think of it as a distinctive part of both solutions and driving forces in ODEs.

Within the solution process for our problem, we calculated the derivatives of our trial solution involving \( e^x \), noticing how these derivatives kept contributing to the original form. This property makes exponentials convenient for simplifying complex equations and finding solutions where exponential growth or decay patterns are present.