Problem 18 - 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. $$ y^{\prime \prime}+y^{\prime}=1 ; \quad y_{1}=1 $$

Step-by-Step Solution

Answer

The second solution is \( y_2(x) = x \) and the particular solution is \( y_p(x) = x \).

Step 1: Setup the homogeneous equation

The given differential equation is \( y'' + y' = 1 \). The associated homogeneous equation is \( y'' + y' = 0 \). We know one solution is \( y_1(x) = 1 \). Our task is to find another solution \( y_2(x) \) using the method of reduction of order.

Step 2: Apply Reduction of Order

Assume the second solution takes the form \( y_2(x) = v(x) y_1(x) = v(x) \cdot 1 = v(x) \). Substitute \( y = v \) into the homogeneous equation to find the form of \( v(x) \).

Step 3: Find Derivatives of v(x)

Calculate the first and second derivatives: \( y_2' = v' \) and \( y_2'' = v'' \). Substitute these into the homogeneous equation \( y'' + y' = 0 \) to get \( v'' + v' = 0 \).

Step 4: Solve for v(x)

The equation \( v'' + v' = 0 \) can be rewritten as \( \frac{d}{dx}(v') + v' = 0 \). Integrate the differential equation \( v' = -C_1 \) to find \( v(x) = -C_1 x + C_2 \). For simplicity, choose constants such that \( v(x) = x \). Thus, the second solution is \( y_2(x) = x \).

Step 5: Solve the Nonhomogeneous Equation

To solve the nonhomogeneous equation \( y'' + y' = 1 \), use the particular solution method. Assume a particular solution of the form \( y_p = A \). Substitute back to get \( 0 + 0 = 1 \), which implies no contribution from a constant. Instead, try \( y_p = Ax \) to find \( 0 + A = 1 \), which gives \( A = 1 \). Thus, \( y_p(x) = 1 \cdot x = x \).

Step 6: General Solution

The general solution of the equation \( y'' + y' = 1 \) is the sum of the homogeneous solutions and particular solution: \( y(x) = C_1 \cdot 1 + C_2x + x = C_1 + C_2 x + x \).

Key Concepts

Homogeneous Equation

A homogeneous equation is a type of differential equation that has a zero on the right-hand side. In the context of second order differential equations, a homogeneous equation looks like this: \[ y'' + p(x) y' + q(x) y = 0 \] where the terms

• \( p(x) \) and \( q(x) \) are functions of \( x \),
• \( y'' \) is the second derivative of \( y \),
• and \( y' \) is the first derivative of \( y \).

In our exercise, the given differential equation is \[ y'' + y' = 1 \]. The homogeneous equation corresponding to this is \[ y'' + y' = 0 \]. The homogeneous equation is integral in the process of finding both the general and particular solutions for differential equations.

Nonhomogeneous Equation

A nonhomogeneous equation, unlike a homogeneous one, includes a non-zero term on the right-hand side. This allows the equation to describe a system under the influence of an external force or input. Formally, it looks like: \[ y'' + p(x) y' + q(x) y = g(x) \] where

• \( g(x) \) is a non-zero function representing external effects, and the rest are as in the homogeneous case.

In this problem, the given equation \[ y'' + y' = 1 \] is nonhomogeneous due to the constant on the right side. Solving nonhomogeneous equations typically involves finding a specific solution, called the particular solution, in addition to the solution of the associated homogeneous equation.

Second Order Differential Equations

Second order differential equations are equations involving the second derivative of a function. These are important for modeling systems where acceleration or curvature is involved, such as mechanical systems. They are typically expressed in the form: \[ y'' + p(x) y' + q(x) y = g(x) \] Here:

• \( y'' \) represents the acceleration or the rate of change of the rate of change.
• \( y' \) represents velocity or rate of change.
• \( y \) is the position or the function we aim to solve for.
• \( p(x) \), \( q(x) \), and \( g(x) \) are functions of \( x \).

Second order differential equations are solved using methods such as reduction of order, characteristic equation, or variation of parameters, depending on whether they are linear, homogeneous, or nonhomogeneous.

Particular Solution

In the context of nonhomogeneous differential equations, a particular solution is a specific solution that satisfies the entire equation. It accounts for the non-zero right-hand side value, representing external influence.To find a particular solution, assume a convenient form of \( y_p(x) \) that resembles the non-zero right side of the equation, often determined by the nature of \( g(x) \). In our exercise, the nonhomogeneous equation was \[ y'' + y' = 1 \]. Assuming \( y_p = Ax \), and substituting it back, we identified that \( A = 1 \), leading to the particular solution \( y_p(x) = x \).The particular solution is crucial because it helps us build the general solution of the differential equation when combined with the homogeneous solutions. This approach ensures that both the homogeneous behavior and the effects of the non-zero term \( g(x) \) are captured.