General Solution For Y'' - 4y' + 9y = 2x - 3 Differential Equation
Hey there, math enthusiasts! Let's dive into the fascinating world of differential equations and crack the code of a specific one: y'' - 4y' + 9y = 2x - 3. Our mission? To find the general solution. To do this effectively, we'll leverage the power of breaking down the problem into manageable parts: the homogeneous solution (y_h) and the particular solution (y_p). By understanding these components, we can piece together the complete picture. So, grab your mathematical toolkit, and let's get started!
Understanding the Homogeneous Solution (y_h)
To kick things off, we need to tackle the homogeneous equation associated with our original problem. This means setting the right-hand side to zero, giving us y'' - 4y' + 9y = 0. The solutions to this equation, the y_h, form the backbone of our general solution. We find y_h by assuming a solution of the form y = e^(rx), where 'r' is a constant we need to determine. Substituting this into the homogeneous equation, we get the characteristic equation: r² - 4r + 9 = 0. This quadratic equation holds the key to our homogeneous solutions.
Now, let's solve this characteristic equation. Using the quadratic formula, r = [-b ± √(b² - 4ac)] / 2a, where a = 1, b = -4, and c = 9, we find the roots. Plugging in the values, we get r = [4 ± √((-4)² - 4 * 1 * 9)] / 2 * 1 = [4 ± √(-20)] / 2. Aha! We've encountered a negative value under the square root, which means our roots are complex numbers. Specifically, r = 2 ± i√5. These complex roots tell us that our homogeneous solution will involve trigonometric functions. When dealing with complex roots of the form α ± iβ, the homogeneous solution takes the form y_h = e^(αx)(C₁cos(βx) + C₂sin(βx)), where C₁ and C₂ are arbitrary constants. In our case, α = 2 and β = √5, so the homogeneous solution is y_h = e^(2x)(C₁cos(√5x) + C₂sin(√5x)). This is a crucial piece of our puzzle.
The homogeneous solution y_h = e^(2x)(C₁cos(√5x) + C₂sin(√5x)) represents the natural behavior of the system described by the differential equation when there's no external input (the 2x - 3 term in the original equation). The exponential term e^(2x) indicates that the solution's amplitude grows over time, while the trigonometric terms cos(√5x) and sin(√5x) introduce oscillations. The constants C₁ and C₂ allow us to adjust the initial amplitude and phase of these oscillations to match specific initial conditions of the problem. For example, if we knew the value of y and y' at x = 0, we could solve for C₁ and C₂ to get a unique solution. The presence of complex roots in the characteristic equation is a hallmark of systems that exhibit oscillatory behavior, such as damped harmonic oscillators in physics or oscillating circuits in electrical engineering. Understanding the homogeneous solution provides valuable insights into the system's inherent dynamics before we even consider the external forcing function (2x - 3).
Finding the Particular Solution (y_p)
Now that we've conquered y_h, let's turn our attention to the particular solution, y_p. This is a specific solution that satisfies the original non-homogeneous equation, y'' - 4y' + 9y = 2x - 3. To find y_p, we'll use the method of undetermined coefficients. The idea here is to guess a solution that has the same form as the non-homogeneous term (2x - 3) and then determine the coefficients that make it work. Since our non-homogeneous term is a linear function, we'll guess a linear solution of the form y_p = Ax + B, where A and B are constants we need to find. We have chosen this guess because linear functions result in constants and linear terms when differentiated, which can then match the form of 2x-3 on the right side of the equation.
Next, we need to find the first and second derivatives of our guessed solution. The first derivative, y_p', is simply A, and the second derivative, y_p'', is 0. Now, we substitute y_p, y_p', and y_p'' into the original differential equation: 0 - 4(A) + 9(Ax + B) = 2x - 3. This simplifies to 9Ax + (9B - 4A) = 2x - 3. For this equation to hold true for all values of x, the coefficients of the corresponding terms must be equal. This gives us a system of two linear equations: 9A = 2 and 9B - 4A = -3. Solving this system, we find A = 2/9 and B = -19/81. Thus, our particular solution is y_p = (2/9)x - 19/81. This particular solution represents a steady-state response of the system to the external input 2x - 3. It doesn't capture the oscillatory behavior described by y_h; instead, it shows how the system responds to the specific forcing function.
The particular solution y_p = (2/9)x - 19/81 represents a specific response of the system to the non-homogeneous term 2x - 3. Unlike the homogeneous solution, which describes the system's natural behavior, the particular solution shows how the system is forced to behave by the external input. The linear form of y_p reflects the linear nature of the forcing function. The coefficient of x, 2/9, indicates the system's sensitivity to the input's linear term, while the constant term, -19/81, represents a constant offset in the response. This particular solution is crucial because it adds the necessary component to capture the full behavior of the system under the influence of the external force. In many physical systems, the particular solution corresponds to the steady-state response, which is the behavior the system settles into after any initial transients have died out. Understanding y_p allows us to predict how the system will respond to different types of external inputs, which is essential in control systems design and other applications.
The Grand Finale: The General Solution
We've reached the exciting part – combining our homogeneous solution (y_h) and particular solution (y_p) to form the general solution. The principle here is that the general solution is simply the sum of the homogeneous and particular solutions: y = y_h + y_p. This combination captures both the natural behavior of the system (y_h) and its response to the external input (y_p).
So, plugging in our previously found solutions, we get y = e^(2x)(C₁cos(√5x) + C₂sin(√5x)) + (2/9)x - 19/81. Voila! This is the general solution to the differential equation y'' - 4y' + 9y = 2x - 3. This solution contains two arbitrary constants, C₁ and C₂, which means it represents a family of solutions. To find a unique solution, we would need additional information, such as initial conditions (the values of y and y' at a specific point). The general solution encompasses all possible behaviors of the system, allowing us to model a wide range of scenarios.
The general solution y = e^(2x)(C₁cos(√5x) + C₂sin(√5x)) + (2/9)x - 19/81 is the complete picture of the system's behavior. It tells us everything we need to know about how the system will evolve over time, given any initial conditions. The homogeneous part, e^(2x)(C₁cos(√5x) + C₂sin(√5x)), describes the transient response, which is the system's initial reaction to a disturbance or change in conditions. This part of the solution decays over time if the exponential term has a negative exponent (which it doesn't in this case, indicating an increasing oscillation amplitude), and it oscillates due to the trigonometric functions. The particular part, (2/9)x - 19/81, represents the steady-state response, which is the behavior the system settles into after the transient response has died out. The constants C₁ and C₂ are determined by the initial conditions of the problem, such as the initial position and velocity of a mass-spring system or the initial voltage and current in an electrical circuit. By adjusting these constants, we can tailor the general solution to fit any specific situation. Understanding the general solution is the ultimate goal in analyzing differential equations because it provides a comprehensive understanding of the system's dynamics.
In essence, the general solution is a powerful tool. It not only satisfies the differential equation but also provides a framework for understanding the behavior of the system being modeled. Whether it's the motion of a spring, the flow of current in a circuit, or the spread of a disease, differential equations and their general solutions offer invaluable insights. So, the next time you encounter a differential equation, remember the power of breaking it down into homogeneous and particular solutions – you'll be well on your way to finding the answer!
