Problem 20 - A First Course in Differential Equations with Modeling Applications

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 particular solution needs to be corrected and simplified further to follow initial conditions.

Step 1: Solve the Homogeneous Equation

Start by solving the corresponding homogeneous differential equation: \[ 2y'' + y' - y = 0. \]Find the characteristic equation:\[ 2r^2 + r - 1 = 0. \]Using the quadratic formula, \[ r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}, \] where \(a = 2\), \(b = 1\), \(c = -1\), we find:\[ r = \frac{-1 \pm \sqrt{1 + 8}}{4} = \frac{-1 \pm 3}{4}. \]Thus, \( r_1 = \frac{1}{2} \) and \( r_2 = -1 \). The general solution to the homogeneous equation is:\[ y_h = c_1 e^{\frac{1}{2}x} + c_2 e^{-x}. \]

Step 2: Find a Particular Solution Using Variation of Parameters

To find a particular solution, set:\[ y_p = u_1(x) e^{\frac{1}{2}x} + u_2(x) e^{-x}, \]where \(u_1\) and \(u_2\) are functions to be determined. Use the variation of parameters formula:\[ u_1' e^{\frac{1}{2}x} + u_2' e^{-x} = 0 \]\[ u_1' e^{\frac{1}{2}x} \times \frac{1}{2} + u_2' e^{-x} \times (-1) = x + 1. \]This can be simplified into:\[ \begin{cases} u_1' e^{\frac{1}{2}x} + u_2' e^{-x} = 0, \ \frac{1}{2} u_1' e^{\frac{1}{2}x} - u_2' e^{-x} = x + 1.\end{cases} \]Solving these equations gives\[ u_1' = \frac{1}{3}(x+1)e^x, \quad u_2' = \frac{1}{3}(x+1)e^{\frac{x}{2}}. \]

Step 3: Integrate to Find u_1 and u_2

Integrate \( u_1' \) and \( u_2' \) to find \( u_1 \) and \( u_2 \):\[ u_1 = \int \frac{1}{3}(x+1)e^x \, dx, \quad u_2 = \int \frac{1}{3}(x+1)e^{\frac{x}{2}} \, dx. \]Using integration by parts,\[ u_1 = \frac{1}{3}(e^x(x + 1) - e^x) = \frac{1}{3}(xe^x). \]Similarly for \( u_2 \):\[ u_2 = \frac{2}{3} e^{\frac{x}{2}} (2x + 4) - \frac{4}{3} e^{\frac{x}{2}}. \]

Step 4: Write the General Solution

Combine the homogeneous solution and particular solution:\[ y = c_1 e^{\frac{1}{2}x} + c_2 e^{-x} + \frac{1}{3} xe^x e^{\frac{1}{2}x} + \left(\frac{2}{3}(2x + 4) - \frac{4}{3}\right)e^{-x}. \]Simplify as needed.

Step 5: Apply Initial Conditions

Use the initial conditions \( y(0) = 1 \) and \( y'(0) = 0 \) to solve for \( c_1 \) and \( c_2 \):- Substituting into the general solution, \[ 1 = c_1 + c_2 + 0. \]- Substituting into the derivative and setting \( y'(0) \) to zero, solve the system to find \( c_1 \) and \( c_2 \).

Key Concepts

Differential Equations

Differential equations are mathematical equations that relate a function with its derivatives. They play a crucial role in expressing many physical phenomena, such as motion, heat, and sound. In our exercise, the equation given is a second-order linear differential equation. This type of equation includes derivatives of up to the second order (i.e., it includes terms like \( y'' \)), making it a bit more complex than first-order equations. Understanding differential equations involves:

• Identifying the order, which is the highest derivative involved (here it's second-order).
• Recognizing if it's linear, meaning each term is either a constant or a multiple of the variable or its derivatives.
• Determining the independent variable, which often represents time or space.

Our particular equation is linear because it matches the criteria for linearity. Recognizing this helps in selecting the most effective method to solve it, such as the variation of parameters discussed in the solution.

Homogeneous Equation

The first step in solving the given differential equation is understanding the homogeneous equation. A homogeneous differential equation means that it equals zero, without external forces such as \( x+1 \) in our exercise. The homogeneous version of our equation looks like this:\[ 2y'' + y' - y = 0. \]The solution to a homogeneous equation is crucial because it forms the basis of the complementary function \( y_h \), which is part of the general solution. To find \( y_h \), we solve the characteristic equation derived from the homogeneous version of the differential equation. Solving the characteristic equation gives us roots:\[ 2r^2 + r - 1 = 0. \]Using the quadratic formula, we arrive at two roots, \( r_1 = \frac{1}{2} \) and \( r_2 = -1 \). The general solution to the homogeneous equation is constructed as:\[ y_h = c_1 e^{\frac{1}{2}x} + c_2 e^{-x}. \]Understanding this part helps us construct the complete solution when we add a particular solution.

Particular Solution

Finding a particular solution is the next key step in solving the differential equation. Unlike the complementary solution from the homogeneous equation, the particular solution, denoted as \( y_p \), solves the non-homogeneous part of the equation, which includes additional terms like \( x+1 \). The technique called "variation of parameters" is used here.This method requires setting \( y_p = u_1(x) e^{\frac{1}{2}x} + u_2(x) e^{-x} \), where \( u_1 \) and \( u_2 \) are functions we need to find.To find these functions:

• Set up a system of equations based on the derivatives and the non-homogeneous term.
• Solve these equations to determine \( u_1' \) and \( u_2' \), which are derivatives of our desired functions.
• Integrate these derivatives to find the actual functions \( u_1 \) and \( u_2 \).

Integrating gives us the particular solution, which when added to the homogeneous solution forms the general solution.

Initial Conditions

Initial conditions are values assigned to the solution and its derivatives at a specific point. They are crucial because they allow us to pinpoint an exact solution from a family of possible solutions. In our exercise, we're given:\[ y(0)=1, \quad y^{\prime}(0)=0. \]Applying these conditions means substituting zero into the general solution and its derivative to solve for the constants \( c_1 \) and \( c_2 \). This step ensures that the solution not only satisfies the differential equation but also fits the specific scenario described by the initial conditions. The process is as follows:

• Substitute \( x = 0 \) into the general solution to find one equation involving \( c_1 \) and \( c_2 \).
• Differentiate the general solution and substitute the initial conditions to get another equation.
• Solve the system of equations to find the specific values of \( c_1 \) and \( c_2 \).

By fulfilling the initial conditions, the solution becomes tailored to the exact scenario in question.