Solving The Differential Equation Y'' + Y' + 2y = 2e^x + 1 A Step-by-Step Guide

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This article delves into the process of solving a second-order linear non-homogeneous differential equation. Specifically, we will address the equation y'' + y' + 2y = 2e^x + 1. This type of equation frequently appears in various fields of science and engineering, including physics, electrical circuits, and control systems. Understanding the methods for solving these equations is crucial for modeling and analyzing these systems. We will explore a step-by-step approach, starting with finding the homogeneous solution and then proceeding to determine a particular solution. By combining these solutions, we obtain the general solution to the differential equation. This comprehensive guide aims to provide a clear and concise explanation of the solution process, making it accessible to students and professionals alike. Mastering the techniques presented here will empower you to tackle a wide range of differential equations and apply them to real-world problems.

1. Finding the Homogeneous Solution

The first step in solving the non-homogeneous differential equation y'' + y' + 2y = 2e^x + 1 is to find the solution to the corresponding homogeneous equation. The homogeneous equation is obtained by setting the right-hand side of the original equation to zero:

y'' + y' + 2y = 0

To solve this homogeneous equation, we assume a solution of the form y = e^(rx), where 'r' is a constant. This assumption is based on the property of exponential functions that their derivatives are proportional to themselves, making them suitable candidates for solutions to linear differential equations with constant coefficients. Substituting this assumed solution into the homogeneous equation, we obtain:

(e^(rx))'' + (e^(rx))' + 2(e^(rx)) = 0

Differentiating e^(rx) with respect to x once gives re^(rx), and differentiating it twice gives r2e(rx). Substituting these derivatives back into the equation, we have:

r2e(rx) + re^(rx) + 2e^(rx) = 0

We can factor out e^(rx) from each term:

e(rx)(r2 + r + 2) = 0

Since e^(rx) is never zero for any real value of x, we can divide both sides of the equation by e^(rx), leaving us with the characteristic equation:

r^2 + r + 2 = 0

This is a quadratic equation in 'r'. We can solve for 'r' using the quadratic formula:

r = (-b ± √(b^2 - 4ac)) / (2a)

where a = 1, b = 1, and c = 2 in our characteristic equation. Substituting these values into the quadratic formula, we get:

r = (-1 ± √(1^2 - 4 * 1 * 2)) / (2 * 1) r = (-1 ± √(1 - 8)) / 2 r = (-1 ± √(-7)) / 2 r = (-1 ± i√7) / 2

Thus, we have two complex conjugate roots:

r₁ = (-1 + i√7) / 2 r₂ = (-1 - i√7) / 2

Since the roots are complex conjugates of the form α ± iβ, where α = -1/2 and β = √7/2, the general solution to the homogeneous equation is given by:

y_h(x) = c₁e^(αx)cos(βx) + c₂e^(αx)sin(βx)

where c₁ and c₂ are arbitrary constants. Substituting the values of α and β, we obtain the homogeneous solution:

y_h(x) = c₁e^(-x/2)cos((√7/2)x) + c₂e^(-x/2)sin((√7/2)x)

This represents the general solution to the homogeneous part of the differential equation. The constants c₁ and c₂ will be determined by initial conditions if they are provided. The oscillatory nature of the solution is due to the presence of the complex roots in the characteristic equation. The exponential term e^(-x/2) indicates that the oscillations will decay as x increases, as the exponent is negative. This decaying oscillatory behavior is a common characteristic of systems with damping, and understanding this behavior is crucial in many applications of differential equations.

2. Finding a Particular Solution

Now that we have the homogeneous solution, the next step in solving the non-homogeneous differential equation y'' + y' + 2y = 2e^x + 1 is to find a particular solution. A particular solution is any specific function that satisfies the original non-homogeneous equation. To find a particular solution, we use the method of undetermined coefficients. This method involves making an educated guess about the form of the particular solution based on the form of the non-homogeneous term (the right-hand side of the equation). In our case, the non-homogeneous term is 2e^x + 1, which is a sum of an exponential function and a constant. Therefore, we assume a particular solution of the form:

y_p(x) = Ae^x + B

where A and B are constants that we need to determine. This assumption is based on the principle that the derivatives of exponential functions are proportional to themselves, and the derivative of a constant is zero. This makes this form a suitable candidate for satisfying the differential equation. To find the values of A and B, we substitute the assumed particular solution into the original non-homogeneous equation. First, we need to find the first and second derivatives of y_p(x):

y_p'(x) = Ae^x y_p''(x) = Ae^x

Now, substitute y_p(x), y_p'(x), and y_p''(x) into the original equation:

(Ae^x) + (Ae^x) + 2(Ae^x + B) = 2e^x + 1

Simplify the equation:

Ae^x + Ae^x + 2Ae^x + 2B = 2e^x + 1 4Ae^x + 2B = 2e^x + 1

For this equation to hold for all values of x, the coefficients of the corresponding terms must be equal. Therefore, we can equate the coefficients of e^x and the constant terms:

4A = 2 2B = 1

Solving these equations for A and B, we get:

A = 2/4 = 1/2 B = 1/2

Now, substitute the values of A and B back into the assumed particular solution:

y_p(x) = (1/2)e^x + 1/2

This is the particular solution to the non-homogeneous differential equation. It represents one specific solution that satisfies the equation, without considering the initial conditions. This particular solution, when combined with the homogeneous solution, will give us the general solution to the differential equation. The method of undetermined coefficients is a powerful technique for finding particular solutions, and it is widely used in solving differential equations. However, it is important to note that the choice of the form of the particular solution depends on the form of the non-homogeneous term, and sometimes adjustments need to be made to the initial guess if it does not lead to a solution.

3. General Solution

Having found both the homogeneous solution y_h(x) and a particular solution y_p(x), we can now construct the general solution to the non-homogeneous differential equation y'' + y' + 2y = 2e^x + 1. The general solution is the sum of the homogeneous solution and a particular solution:

y(x) = y_h(x) + y_p(x)

Substituting the expressions we found for y_h(x) and y_p(x), we get:

y(x) = c₁e^(-x/2)cos((√7/2)x) + c₂e^(-x/2)sin((√7/2)x) + (1/2)e^x + 1/2

This is the general solution to the differential equation. It represents a family of solutions, where each solution corresponds to a different pair of values for the arbitrary constants c₁ and c₂. These constants are determined by the initial conditions, which are typically given as the values of y(x) and y'(x) at a specific point. If initial conditions are provided, we can substitute them into the general solution and its derivative to obtain a system of equations for c₁ and c₂. Solving this system will give us the specific values of the constants, and substituting these values back into the general solution will give us the unique solution that satisfies the given initial conditions. The general solution encompasses all possible solutions to the differential equation, and it is a crucial step in understanding the behavior of the system being modeled. The presence of the arbitrary constants allows the solution to adapt to different initial states of the system. The form of the general solution, with its combination of oscillatory and exponential terms, provides valuable insights into the dynamics of the system and its response to external inputs.

4. Summary of Steps

To recap, here are the steps we followed to solve the non-homogeneous differential equation y'' + y' + 2y = 2e^x + 1:

  1. Find the Homogeneous Solution:

    • Set the right-hand side of the equation to zero to obtain the homogeneous equation: y'' + y' + 2y = 0.
    • Assume a solution of the form y = e^(rx) and substitute it into the homogeneous equation.
    • Obtain the characteristic equation: r^2 + r + 2 = 0.
    • Solve the characteristic equation for r using the quadratic formula. In this case, we found complex conjugate roots: r = (-1 ± i√7) / 2.
    • Write the general solution to the homogeneous equation using the complex roots: y_h(x) = c₁e^(-x/2)cos((√7/2)x) + c₂e^(-x/2)sin((√7/2)x), where c₁ and c₂ are arbitrary constants.

  2. Find a Particular Solution:

    • Use the method of undetermined coefficients to find a particular solution. Based on the non-homogeneous term 2e^x + 1, assume a particular solution of the form y_p(x) = Ae^x + B.
    • Find the first and second derivatives of y_p(x): y_p'(x) = Ae^x and y_p''(x) = Ae^x.
    • Substitute y_p(x), y_p'(x), and y_p''(x) into the original non-homogeneous equation.
    • Equate the coefficients of corresponding terms to obtain a system of equations for A and B.
    • Solve the system of equations for A and B. In this case, we found A = 1/2 and B = 1/2.
    • Substitute the values of A and B back into the assumed particular solution: y_p(x) = (1/2)e^x + 1/2.

  3. Form the General Solution:

    • Add the homogeneous solution and the particular solution to obtain the general solution: y(x) = y_h(x) + y_p(x).
    • In this case, the general solution is: y(x) = c₁e^(-x/2)cos((√7/2)x) + c₂e^(-x/2)sin((√7/2)x) + (1/2)e^x + 1/2.

  4. Apply Initial Conditions (if provided):

    • If initial conditions are given, substitute them into the general solution and its derivative to obtain a system of equations for the arbitrary constants c₁ and c₂.
    • Solve the system of equations for c₁ and c₂.
    • Substitute the values of c₁ and c₂ back into the general solution to obtain the unique solution that satisfies the initial conditions.

This step-by-step process provides a systematic approach to solving second-order linear non-homogeneous differential equations with constant coefficients. By mastering these steps, you can confidently tackle a wide range of differential equations and apply them to various problems in science and engineering. The ability to solve differential equations is a fundamental skill in many disciplines, and this guide provides a solid foundation for further exploration and application of these powerful mathematical tools.

5. Conclusion

In conclusion, we have successfully solved the non-homogeneous differential equation y'' + y' + 2y = 2e^x + 1 by following a systematic approach. We first determined the homogeneous solution by solving the characteristic equation and finding the complex conjugate roots. Then, we employed the method of undetermined coefficients to find a particular solution that satisfies the non-homogeneous equation. Finally, we combined the homogeneous and particular solutions to obtain the general solution, which represents a family of solutions parameterized by arbitrary constants. This process highlights the key steps involved in solving second-order linear non-homogeneous differential equations with constant coefficients.

The general solution we obtained, y(x) = c₁e^(-x/2)cos((√7/2)x) + c₂e^(-x/2)sin((√7/2)x) + (1/2)e^x + 1/2, provides valuable insights into the behavior of the system being modeled. The oscillatory terms, cos((√7/2)x) and sin((√7/2)x), indicate the presence of oscillations, while the exponential term e^(-x/2) suggests that these oscillations will decay over time. The term (1/2)e^x represents a forcing function that grows exponentially, and the constant term 1/2 contributes a steady-state component to the solution. The arbitrary constants c₁ and c₂ allow the solution to adapt to different initial conditions, making it a versatile tool for modeling various scenarios.

Understanding and solving differential equations is essential in many fields, including physics, engineering, economics, and biology. These equations provide a powerful framework for describing and analyzing systems that change over time. The techniques presented in this article, such as finding homogeneous solutions and using the method of undetermined coefficients, are fundamental tools in the arsenal of any scientist or engineer. By mastering these techniques, you can gain a deeper understanding of the world around us and develop innovative solutions to complex problems. This guide serves as a stepping stone for further exploration and application of differential equations in diverse areas of study and practice.