Question

Find a particular solution. $$ 2 y^{(4)}+y^{\prime \prime \prime}-2 y^{\prime}-y=3 e^{-x / 2}(1-6 x) $$

Short answer

Question: Determine the particular solution to the inhomogeneous ordinary differential equation (ODE) below using the method of undetermined coefficients: $$ 2y^{(4)} + y^{\prime\prime\prime} - 2y^{\prime} - y = 3e^{-x/2}(1-6x) $$ Answer: The particular solution to the given ODE is: $$ Y(x) = (-4 + 12x)e^{-x/2} $$

Step-by-step solution

Understand the given ODE

The given ODE is: $$ 2y^{(4)} + y^{\prime\prime\prime} - 2y^{\prime} - y = 3e^{-x/2}(1-6x) $$ It is an inhomogeneous linear ODE of order 4, with an exponential function multiplied by a linear function on the right-hand side (RHS).

Make a guess for the form of the particular solution

We are given that the RHS of the ODE has the form \(3e^{-x/2}(1-6x)\). Therefore, our guess for the form of the particular solution will look like: $$ Y(x) = (A + Bx)e^{-x/2} $$ Here, A and B are the undetermined coefficients, which we will determine by substituting the guess into the ODE and solving for A and B.

Calculate derivatives of the guess for the particular solution

We need to substitute the guess and its derivatives up to the fourth order into the ODE. We will calculate the first, second, third, and fourth derivatives of Y(x): First derivative: $$ Y^{\prime}(x) = (-\frac{1}{2}A - \frac{1}{2}Bx + B)e^{-x/2} $$ Second derivative: $$ Y^{\prime\prime}(x) = (\frac{1}{4}A + \frac{1}{4}Bx - B)e^{-x/2} $$ Third derivative: $$ Y^{\prime\prime\prime}(x) = (-\frac{1}{8}A - \frac{1}{8}Bx + \frac{3}{4}B)e^{-x/2} $$ Fourth derivative: $$ Y^{(4)}(x) = (\frac{1}{16}A + \frac{1}{16}Bx - \frac{3}{8}B)e^{-x/2} $$

Plug the guess and its derivatives into the ODE

We will now substitute our guess and its derivatives into the ODE: $$ 2(\frac{1}{16}A + \frac{1}{16}Bx - \frac{3}{8}B)e^{-x/2} + (-\frac{1}{8}A - \frac{1}{8}Bx + \frac{3}{4}B)e^{-x/2} - 2(\frac{1}{4}A + \frac{1}{4}Bx - B)e^{-x/2} - (A + Bx)e^{-x/2} = 3e^{-x/2}(1-6x) $$ This ODE must hold for all x, so the coefficients of like terms must be equal.

Equate coefficients and solve for A and B

Equate coefficients of \(e^{-x/2}\) and \(xe^{-x/2}\) and simplify: For terms with \(e^{-x/2}\): $$ -\frac{3}{16}A + \frac{1}{8}A - \frac{1}{2}A = 3 $$ $$ \Rightarrow A = -4 $$ For terms with \(xe^{-x/2}\): $$ -\frac{3}{8}B + \frac{3}{4}B - \frac{1}{2}B + B = -18 $$ $$ \Rightarrow B = 12 $$

Write the particular solution

Now that we have the values for A and B, our particular solution is: $$ Y(x) = (-4 + 12x)e^{-x/2} $$ This is the particular solution to the given ODE.

Key concepts

Method of Undetermined Coefficients

The method of undetermined coefficients is a classic means of finding a particular solution to a certain type of differential equation known as a linear nonhomogeneous ordinary differential equation (ODE). This method can only be applied when the inhomogeneous term (the term without the dependent variable) is a simple function, like a polynomial, exponential, sine, cosine, or a combination of these.
To use this method, you first propose a form for the particular solution where the coefficients are left undetermined, hence the name. These coefficients are then found by substituting the proposed solution back into the ODE and matching the coefficients on both sides.
In the given exercise, an educated guess is made based on the right-hand side being an exponential function multiplied by a linear function. By proposing a solution of similar form and finding its derivatives, we systematically determine the appropriate coefficients that satisfy the original equation. It's vital to remember that this strategy directly relates to the type of function you're working with on the inhomogeneous side of the equation.

Differential Equations

At the heart of many scientific disciplines lies differential equations. These equations are pivotal since they relate a function with its derivatives, representing change. Differential equations can describe the relationship between various rates of change, allowing us to model real-world phenomena across physics, engineering, biology, economics, and more.
Differential equations come in many forms. They can be ordinary, with one independent variable, or partial, with several independent variables. They're also classified based on linearity, order, and whether they're homogeneous or inhomogeneous. In our context, an inhomogeneous linear ordinary differential equation (ODE) means that the equation is non-uniform across the dependent variable and that it includes terms without the variable that we're solving for (hence, 'inhomogeneous'), often mandating a specific technique, like the method of undetermined coefficients, to find a solution.

Particular Solution

In the realm of differential equations, the particular solution is a specific solution to a nonhomogeneous equation that includes in its form the non-repeating solution of the inhomogeneous part.
When finding a particular solution, one aims to solve the ODE for a certain function that satisfies both the equation and initial or boundary conditions if they're present. Unlike the general solution, which includes the complementary solution (associated with the homogeneous part) and a constant that can take on infinite values, the particular solution is single and relates to the unique nature of the nonhomogeneous equation.
In practice, obtaining the particular solution involves guessing a function form close to the nonhomogeneous part, determining its necessary derivatives, and plugging these into the original equation to isolate the undetermined coefficients, culminating in a solution that fits the equation precisely. This process is artfully illustrated in our example problem, where a particular solution to an ODE with an exponential-right hand side is deduced.

Ordinary Differential Equations

Ordinary Differential Equations (ODEs) are equations involving derivatives of a function of a single independent variable. They are called 'ordinary' to distinguish them from 'partial' differential equations, which involve partial derivatives of a function of multiple independent variables.
An ODE encompasses an assortment of derivatives, possibly of different orders, equal to a function involving the independent variable(s). The highest derivative present determines the order of the ODE. ODEs are generally tougher to solve as the order increases and they can be linear or nonlinear. A linear ODE has solutions that can be added together or multiplied by constants to produce other solutions (principle of superposition), while nonlinear ODEs lack this property and are typically more challenging to solve.
In our exercise, we deal with a fourth-order linear ODE, which means the highest derivative is to the fourth power, and the solution can take advantage of linear ODE techniques. The steps outlined in the provided exercise systematically demonstrate the approach to tackling such an equation, emphasizing the methodical nature of handling ODEs whilst adhering to the innate characteristics of the equation at hand.