Question

Solve the initial value problem and graph the solution, given that \(y_{1}\) satisfies the complementary equation. C/G \(\left(x^{2}-4\right) y^{\prime \prime}+4 x y^{\prime}+2 y=x+2, \quad y(0)=-\frac{1}{3}, \quad y^{\prime}(0)=-1 ; \quad y_{1}=\frac{1}{x-2}\)

Short answer

#Question# Based on the given solution, briefly describe the steps involved in solving an initial value problem for a second-order non-homogeneous linear differential equation. #Answer# 1. Identify the mixing method as the solving approach. 2. Apply the mixing method to find another linearly independent solution, \(y_2\). 3. Determine the particular solution, \(y_p\), using the method of undetermined coefficients. 4. Determine the general solution as the combination of \(y_1\), \(y_2\), and \(y_p\). 5. Apply the initial conditions to find the constants in the general solution. 6. Graph the solution in a suitable domain.

Step-by-step solution

Identify the mixing method as solving approach

We are given the complementary equation and a solution, \(y_1\). Now our task is to find \(y_2\) and a particular solution, \(y_p\). Then the general solution will be the sum of the general solution of the complementary equation and the particular solution. \(y(x) = c_1y_1 + c_2y_2 + y_p\).

Apply mixing method

We'll use the mixing method to find another linearly independent solution, \(y_2\). Using the given solution \(y_1=\frac{1}{x-2}\), we will determine \(y_2\) by solving the following equation: \(\frac{y_2'}{y_2} = -\frac{y_1'}{x^2-4}\). Integrate this equation to find the second solution, \(y_2\).

Determine the particular solution

The particular solution \((y_p)\) can be found using the method of undetermined coefficients. Here, we will assume that \(y_p = Ax+B\) and we will plug this into the given differential equation to find the values of A and B.

Determine the general solution

Once we have \(y_2\) and the particular solution, \(y_p\), we can find the general solution \(y(x)=c_1y_1+c_2y_2+y_p\). Combine these solutions to get the general solution, where \(c_1\) and \(c_2\) are constants.

Apply initial conditions

Apply the initial conditions \(y(0)=-\frac{1}{3}\) and \(y^{\prime}(0)=-1\) to the general solution to solve for \(c_1\) and \(c_2\).

Graph the solution

Finally, after finding the solution to the initial value problem, graph the function \(y(x)\) in a suitable domain.

Key concepts

Complementary Equation

When solving certain types of differential equations known as linear differential equations, one part of the solution is associated with the 'homogeneous equation' or 'complementary equation'. The complementary equation is the part where the original differential equation is set to zero. In other words, if you have an equation of the form \(ay'' + by' + cy = f(x)\), the complementary equation would be \(ay'' + by' + cy = 0\).

The solutions to this complementary equation are crucial as they form the basis for the 'general solution' to the full differential equation. These solutions are typically found by identifying characteristic polynomials and their roots, or by looking for specific types of functions that satisfy the equation. In our given exercise, the solution \(y_1 = \frac{1}{x-2}\) satisfies the complementary equation, which means it's one constituent of the general solution to the differential equation.

Method of Undetermined Coefficients

To find a specific type of solution called the 'particular solution' to a non-homogeneous differential equation, the 'method of undetermined coefficients' is often used. This method works well when the non-homogeneous term, like \(f(x)\) in \(ay'' + by' + cy = f(x)\), is a polynomial, exponential, sine, or cosine function.

The general strategy involves guessing a form for the particular solution where the 'coefficients' are literally undetermined. You then substitute this guess into the differential equation and solve for the coefficients, making the educated guess into a bona fide solution. For instance, if \(f(x)\) is a polynomial, the guess for \(y_p\) would also be a polynomial of the same degree. In the case of the exercise, the assumption is \(y_p = Ax + B\). After substitution, A and B are determined, yielding a particular solution.

Particular Solution

Referring to the solution found through the method of undetermined coefficients, the 'particular solution' is the part of the general solution to a non-homogeneous differential equation that accounts for the presence of the non-homogeneous term. It is designed to satisfy the original differential equation exactly, whereas the complementary solution only satisfies the homogeneous version of the equation.

In layman's terms, you can think of the particular solution as the puzzle piece that precisely completes the picture formed by the complementary solution when dealing with the entire puzzle of a differential equation. It addresses the specific structure of the equation and provides the necessary component to reach the equation's true solution.

General Solution

The 'general solution' to a differential equation represents the entire family of possible solutions, encompassing every specific scenario that fulfills the equation. It typically involves arbitrary constants that can be later defined by initial or boundary conditions. The general solution combines the complementary solution, representing the solution to the homogeneous equation, with the particular solution that deals directly with the specifics of the non-homogeneous equation.

In the initial value problem provided, the general solution is expressed as \(y(x) = c_1y_1 + c_2y_2 + y_p\), with \(c_1\) and \(c_2\) being constants to determine and \(y_p\) being the particular solution. By applying the initial conditions given, \(c_1\) and \(c_2\) can be uniquely identified, turning the general solution into the specific solution of the problem.

Differential Equations

Differential equations are mathematical equations that involve derivatives of a function and the function itself. They are fundamental in describing various phenomena across physical sciences, engineering, and even in finance and economics. These equations can be classified according to their properties, such as order, degree, and linearity.

An 'initial value problem' in differential equations is when you have a differential equation along with the value of the function (and possibly its derivatives) at a particular point, referred to as the initial conditions. The solution to such a problem is a function that satisfies both the differential equation and the initial conditions, providing a specific trajectory among the family of possible solutions described by the general solution.