Problem 20 - Advanced Engineering Mathematics

Question

Solve each differential equation by variation of parameters subject to the initial conditions \(y(0)=1, y^{\prime}(0)=0\). $$ 2 y^{\prime \prime}+y^{\prime}-y=x+1 $$

Step-by-Step Solution

Answer

The solution is \( y(x) = e^{\frac{1}{2}x} - \frac{2}{3}x e^{\frac{1}{2}x} + \frac{2}{3}e^{-x} \).

Step 1: Identify the Differential Equation

The given differential equation is \( 2y'' + y' - y = x + 1 \). Our task is to solve it using the method of variation of parameters.

Step 2: Solve the Homogeneous Equation

First, find the complementary solution \( y_c \) by solving the homogeneous equation \( 2y'' + y' - y = 0 \). The characteristic equation is \( 2r^2 + r - 1 = 0 \), which factors to \((2r - 1)(r + 1) = 0\). The roots are \( r_1 = \frac{1}{2}, r_2 = -1 \). Thus, the complementary solution is \( y_c = C_1 e^{\frac{1}{2}x} + C_2 e^{-x} \).

Step 3: Set up the Variation of Parameters Method

We need a particular solution \( y_p \) of the form \( y_p = u_1(x)e^{\frac{1}{2}x} + u_2(x)e^{-x} \) where \( u_1' \) and \( u_2' \) are functions to be determined.

Step 4: Calculate the Wronskian

The Wronskian \( W \) of solutions \( e^{\frac{1}{2}x} \) and \( e^{-x} \) is computed. \[ W = \begin{vmatrix} e^{\frac{1}{2}x} & e^{-x} \ \frac{1}{2}e^{\frac{1}{2}x} & -e^{-x} \end{vmatrix} = e^{\frac{1}{2}x}(-e^{-x}) - e^{-x}\left(\frac{1}{2}e^{\frac{1}{2}x}\right) = -\frac{3}{2}e^{-\frac{1}{2}x}. \]

Step 5: System of Equations Using Variation Parameters

Using the variation parameters method, establish the following system: 1. \( u_1'(x) \cdot e^{\frac{1}{2}x} + u_2'(x) \cdot e^{-x} = 0 \)2. \( u_1'(x) \cdot \( \frac{1}{2} \) e^{\frac{1}{2}x} + u_2'(x) \cdot (-e^{-x}) = x + 1\).

Step 6: Solve for \( u_1'(x) \) and \( u_2'(x) \)

Utilize the Wronskian to find \( u_1' \) and \( u_2' \): - \( u_1' = \frac{-e^{-x}(x+1)}{W} = \frac{-2e^{\frac{1}{2}x}(x+1)}{3} \).- \( u_2' = \frac{e^{\frac{1}{2}x}(x+1)}{W} = \frac{-2e^{-x}(x+1)}{3} \).Integrate these expressions to find \( u_1 \) and \( u_2 \).

Step 7: Integrate to Find \( u_1(x) \) and \( u_2(x) \)

Integrate \( u_1' \) and \( u_2' \):- \( u_1 = \int \frac{-2e^{\frac{1}{2}x}(x+1)}{3} \,dx = -\frac{2}{3}e^{\frac{1}{2}x}(x+1) + \text{constant} \).- \( u_2 = \int \frac{-2e^{-x}(x+1)}{3} \,dx = \frac{2}{3}e^{-x}(x+1) + \text{constant} \).

Step 8: Write the Particular Solution

The particular solution \( y_p \) is given by substituting these integrals back:\[ y_p = -\frac{2}{3}(e^x(x+1)e^{\frac{1}{2}x}) + \frac{2}{3}(e^{-x}(x+1)e^{-x}) = -\frac{2}{3}x e^{\frac{1}{2}x} + \frac{2}{3} \( x+1 \) e^{-x}. \]

Step 9: General Solution and Initial Conditions

Combine \( y_c \) and \( y_p \) to form the general solution:\( y = C_1 e^{\frac{1}{2}x} + C_2 e^{-x} - \frac{2}{3}x e^{\frac{1}{2}x} + \frac{2}{3}e^{-x} \).Use initial conditions \( y(0) = 1 \) and \( y'(0) = 0 \) to find \( C_1 \) and \( C_2 \).

Step 10: Apply Initial Conditions

Solve for \( C_1 \) and \( C_2 \) using initial conditions:- From \( y(0) = 1 \): \( C_1 + C_2 + \frac{2}{3} = 1 \) implies \( C_1 + C_2 = \frac{1}{3} \).- From \( y'(0) = 0 \): \( \frac{1}{2}C_1 - C_2 - \frac{1}{3} = 0 \) implies \( C_1 = 2C_2 + \frac{2}{3} \).Solve the system to find \( C_1 = 1 \) and \( C_2 = 0 \).

Step 11: Final Solution

Substitute \( C_1 \) and \( C_2 \) into the general solution gives:\( y(x) = e^{\frac{1}{2}x} - \frac{2}{3}x e^{\frac{1}{2}x} + \frac{2}{3}e^{-x} \). Thus, this is the particular solution satisfying the given initial conditions.

Key Concepts

Differential Equations

Differential equations are mathematical equations that involve rates of change represented by derivatives. They are used to describe a variety of phenomena in fields like physics, engineering, and economics. A differential equation relates a function with its derivatives.
In our problem, we deal with a second-order linear differential equation: \( 2y'' + y' - y = x + 1 \). The equation includes terms of the function \( y \), its first derivative \( y' \), and its second derivative \( y'' \). The goal is to find the function \( y \) that satisfies this equation.
By using methods like variation of parameters, we aim to find both the complementary and particular solutions to the differential equation, looking for functions that satisfy both the homogeneous part \( 2y'' + y' - y = 0 \) and the non-homogeneous part \( x + 1 \).

Initial Conditions

Initial conditions in differential equations are specific values that solutions must satisfy. They play a crucial role in determining the unique solution of a differential equation.
For the equation \( 2y'' + y' - y = x + 1 \), we are given the initial conditions \( y(0) = 1 \) and \( y'(0) = 0 \). These conditions specify the initial state of the system represented by our differential equation.
To apply the initial conditions, we substitute them into our general solution. This allows us to solve for any constants within the solution, ensuring that the function not only satisfies the equation but also starts from the specified initial state. Without these conditions, an infinite number of solutions might fit the differential equation.

Complementary Solution

The complementary solution is a key concept when solving non-homogeneous differential equations. It solves the associated homogeneous differential equation \( 2y'' + y' - y = 0 \).
To find the complementary solution, we first form the characteristic equation from the homogeneous part, which is \( 2r^2 + r - 1 = 0 \). By solving this quadratic equation, we find the roots \( r_1 = \frac{1}{2} \) and \( r_2 = -1 \). These roots guide us to the complementary solution \( y_c = C_1 e^{\frac{1}{2}x} + C_2 e^{-x} \).
The constants \( C_1 \) and \( C_2 \) are determined once we apply the initial conditions to the general solution. This complementary solution captures the behavior of the system without external forces represented by the non-homogeneous part of the original equation.

Wronskian

The Wronskian is a determinant used to check whether a set of solutions is linearly independent, which is important when using the variation of parameters method.
For our differential equation, we compute the Wronskian of the functions \( e^{\frac{1}{2}x} \) and \( e^{-x} \). The Wronskian \( W \) is given by the determinant:\[ W = \begin{vmatrix} e^{\frac{1}{2}x} & e^{-x} \ \frac{1}{2}e^{\frac{1}{2}x} & -e^{-x} \end{vmatrix} = e^{\frac{1}{2}x}(-e^{-x}) - e^{-x}\left(\frac{1}{2}e^{\frac{1}{2}x}\right) = -\frac{3}{2}e^{-\frac{1}{2}x}. \]The nonzero Wronskian indicates that our chosen solutions \( e^{\frac{1}{2}x} \) and \( e^{-x} \) are indeed linearly independent.
These linearly independent solutions are then used to construct the particular solution using variation of parameters, allowing us to find the complete solution to the differential equation.