Problem 17 - Advanced Engineering Mathematics

Question

In Problems 17-20, the indicated function \(y_{1}(x)\) is a solution of the associated homogeneous equation. Use the method of reduction of order to find a second solution \(y_{2}(x)\) of the homogeneous equation and a particular solution of the given nonhomogeneous equation. $$ y^{\prime \prime}-4 y=2 ; \quad y_{1}=e^{-2 x} $$

Step-by-Step Solution

Answer

The second solution is \( y_2(x) = xe^{-2x} \) and a particular solution is \( y_p = -\frac{1}{2} \).

Step 1: Establish the Problem

We have a differential equation: \( y'' - 4y = 2 \) with one given solution to the homogeneous equation: \( y_1 = e^{-2x} \). We need to find another solution to the homogeneous equation and a particular solution to the full equation.

Step 2: Use Reduction of Order

Assume a second solution of the form \( y_2 = v(x) e^{-2x} \), where \( v(x) \) is a function to be determined. Then, plug \( y_2 = v(x) e^{-2x} \) into the homogeneous equation \( y'' - 4y = 0 \). Calculate \( y_2' = v'e^{-2x} - 2ve^{-2x} \) and \( y_2'' = v''e^{-2x} - 4v'e^{-2x} + 4ve^{-2x} \).

Step 3: Substitute into the Homogeneous Equation

Substitute \( y_2'', y_2', \) and \( y_2 \) into the homogeneous equation. The terms involving \( v' \) and \( v'' \) will guide us to a differential equation that doesn't repeat the existing \( v \). Solve this for \( v \). You will find that a solution is \( v'' = 0 \), suggesting \( v' = c_1 \) and \( v = c_1x + c_2 \).

Step 4: Solve for a Second Solution

Since we need a second, linearly independent solution, choose \( v(x) = x \) which gives \( y_2(x) = x e^{-2x} \). This confirms the second linearly independent solution to the homogeneous equation is \( y_2(x) = x e^{-2x} \).

Step 5: Find a Particular Solution

To find a particular solution \( y_p \) for \( y'' - 4y = 2 \), use trial solutions or the method of undetermined coefficients. Assume \( y_p = A \). Substitute into the equation and solve for \( A \). You get \( y_p = -\frac{1}{2} \).

Step 6: General Solution for Nonhomogeneous Equation

Combine the solutions to write the general solution as \( y(x) = c_1 e^{-2x} + c_2 x e^{-2x} - \frac{1}{2} \).

Key Concepts

Differential Equations

Differential equations are mathematical equations that involve derivatives of a function. They describe the relationship between a function and its derivatives. In engineering, physics, economics, and many other fields, differential equations are used to model how a system changes over time. These models help us understand complex real-world phenomena.

In this particular exercise, we are dealing with a second-order differential equation, which involves the second derivative of the function: \( y'' - 4y = 2 \). This tells us how the rate of change of the rate of change of the function \( y \) relates to the function itself. By solving this differential equation, we find special functions \( y_1 \) and \( y_2 \) which make the equation true for particular values of \( x \). These solutions can be used to understand the behavior of the system described by the equation.

Homogeneous Equation

A homogeneous equation is a type of differential equation where all terms involve the unknown function or its derivatives, and there is no constant or external forcing term. For a linear homogeneous differential equation, like \( y'' - 4y = 0 \), the right side of the equation is zero.

Solving a homogeneous equation involves finding solutions for \( y \) that satisfy the equation when it is set to zero. The given solution \( y_1 = e^{-2x} \) is an example of a solution to the homogeneous equation. Solving homogeneous equations is a crucial step because it provides the groundwork for building the general solution of non-homogeneous equations, which also include an extra term, such as the constant 2 in our original differential equation.

Particular Solution

A particular solution is a solution to a non-homogeneous differential equation that includes terms not present in the corresponding homogeneous equation. In our exercise, we needed to find a particular solution for the equation \( y'' - 4y = 2 \).

A particular solution can sometimes be guessed based on the form of the non-homogeneity (in this case the constant 2). One common method for finding such a solution is the method of undetermined coefficients, where we assume a form for the solution and then solve for the coefficients that satisfy the equation. Here, a trial solution of the form \( y_p = A \) was used. When substituted back into the differential equation, it helped us find that \( y_p = -\frac{1}{2} \), effectively counteracting the non-zero term in the original equation to balance it out.

Linearly Independent Solutions

Linearly independent solutions of a differential equation are solutions that cannot be expressed as a linear combination of one another. This concept is important because it allows us to construct the general solution of the equation.

For the homogeneous equation \( y'' - 4y = 0 \), we have one solution \( y_1 = e^{-2x} \). Using the method of reduction of order, we find a second solution \( y_2 = x e^{-2x} \). These two solutions are linearly independent and form the fundamental set of solutions for the homogeneous equation.

• Linearly independent solutions provide the basis needed to build any general solution for the equation.
• The general solution is a linear combination of all linearly independent solutions, often involving arbitrary constants.

In this case, the general solution of the equation captures the behavior of the fullest range of solutions possible, combining both homogeneous and particular solutions to describe the system comprehensively.