Homogeneous linear second order differential equations

0/2
?
Intros
Lessons
  1. What are Homogeneous Linear Second Order Differential Equations? And what are some sorts of solutions to them?
  2. What is the Characteristic Equation?
0/2
?
Examples
Lessons
  1. Solving Homogeneous Linear Second Order Differential Equations
    Find some general solutions to the following constant coefficient homogeneous linear second order differential equation:

    2y+5y3y=02y''+5y'-3y=0
    1. Using the initial conditions find a particular solution to the following differential equation:

      y+3y4y=0y''+3y'-4y=0
      y(0)=3,y(0)=2y(0)=3, y' (0)=-2
      Topic Notes
      ?

      Introduction to Homogeneous Linear Second Order Differential Equations

      Welcome to the fascinating world of homogeneous linear second order differential equations! These equations are a crucial part of advanced mathematics and have numerous applications in physics, engineering, and other sciences. In our introduction video, we'll explore the fundamental concepts and characteristics of these equations. You'll learn how to identify a homogeneous linear second order differential equation and understand its general form. We'll discuss the importance of the coefficients and how they influence the solution. The video will also cover the process of finding the general solution using various methods, such as the characteristic equation approach. Don't worry if it seems complex at first we'll break it down step-by-step, making it easier to grasp. By the end of this introduction, you'll have a solid foundation to tackle more advanced topics in differential equations. Remember, mastering these equations opens doors to understanding complex systems in nature and technology. Let's embark on this mathematical journey together!

      Understanding the Structure of Homogeneous Linear Second Order Differential Equations

      Homogeneous linear second order differential equations are a fundamental concept in mathematics, particularly in the study of calculus and differential equations. These equations play a crucial role in modeling various physical phenomena and are widely used in engineering, physics, and other scientific fields. Let's dive into the general form of these equations and explore their components.

      The general form of a homogeneous linear second order differential equation with constant coefficients is:

      ay'' + by' + cy = 0

      Here, y represents the dependent variable, and the prime notation (y' and y'') denotes the first and second derivatives of y with respect to the independent variable, typically x. The coefficients a, b, and c are constants, and their values play a significant role in determining the behavior of the equation's solutions.

      Let's break down the components of this equation:

      • a: This coefficient is associated with the second derivative (y''). It represents the acceleration or rate of change of the rate of change in the system being modeled.
      • b: This coefficient is linked to the first derivative (y'). It often represents damping or resistance in physical systems.
      • c: This coefficient is associated with y itself and typically represents a restoring force or spring constant in mechanical systems.

      The term "homogeneous" in this context means that the right-hand side of the equation is equal to zero. This property is crucial as it allows for specific solution methods and behaviors.

      To illustrate the concept, let's consider a few examples:

      1. y'' + 4y' + 4y = 0: This equation might represent a damped harmonic oscillator in physics.
      2. 2y'' - 3y' + y = 0: This could model a more complex system with different rates of damping and restoration.
      3. y'' - y = 0: A simple case that might represent an undamped oscillator or exponential functions growth/decay.

      The solutions to these equations can take various forms depending on the values of a, b, and c. They might be exponential functions, trigonometric functions, or a combination of both. The nature of the solutions is determined by the roots of the characteristic equation associated with the differential equation.

      Understanding the general form of homogeneous linear second order differential equations is crucial for several reasons:

      • It provides a framework for modeling many real-world phenomena.
      • It allows for the development of systematic solution methods.
      • It helps in predicting the behavior of systems without needing to solve the equation explicitly.

      As you delve deeper into the study of differential equations, you'll encounter various methods for solving these equations, such as the characteristic equation method, the method of undetermined coefficients, and variation of parameters. Each of these methods builds upon the understanding of the general form we've discussed here.

      In conclusion, the general form ay'' + by' + cy = 0 of homogeneous linear second order differential equations with constant coefficients is a powerful tool in mathematics and science. By understanding the roles of the coefficients a, b, and c, and recognizing the homogeneous nature of the equation, you'll be well-equipped to analyze and solve a wide range of problems in various fields. Remember, practice and application are key to mastering these concepts, so don't hesitate to work through numerous examples and real-world applications.

      The Characteristic Equation: Key to Solving Homogeneous Linear Second Order Differential Equations

      The characteristic equation is a fundamental concept in the study of differential equations, particularly when dealing with homogeneous linear second-order differential equations. This powerful tool provides a systematic approach to finding solutions, making it an essential technique for mathematicians, engineers, and scientists alike. In this section, we'll explore the characteristic equation, its derivation, and its crucial role in solving differential equations.

      To understand the characteristic equation, we must first consider the general form of a homogeneous linear second-order differential equation:

      ay'' + by' + cy = 0

      where a, b, and c are constants, and y is a function of x. The characteristic equation is derived from this differential equation by assuming a solution of the form y = e^(rx), where r is a constant. This assumption is based on the fact that exponential functions are eigenfunctions of the differential operator.

      To form the characteristic equation, we follow these steps:

      1. Substitute y = e^(rx) into the original differential equation.
      2. Calculate the first and second derivatives: y' = re^(rx) and y'' = r^2e^(rx).
      3. Substitute these expressions into the differential equation.
      4. Factor out e^(rx), as it is non-zero for all x.
      5. Set the resulting expression equal to zero.

      The result is the characteristic equation:

      ar^2 + br + c = 0

      This quadratic equation in r is the key to finding solutions to the original differential equation. The roots of this equation, often denoted as r1 and r2, determine the nature of the solutions.

      To solve the characteristic equation, we use the quadratic formula:

      r = [-b ± (b^2 - 4ac)] / (2a)

      The discriminant, b^2 - 4ac, plays a crucial role in determining the nature of the roots:

      Once we have the roots, we can construct the general solution to the differential equation:

      The importance of the characteristic equation cannot be overstated. It provides a direct link between the coefficients of the differential equation and its solutions. By analyzing the roots of the characteristic equation, we can predict the behavior of the solutions without explicitly solving the differential equation.

      Moreover, the characteristic equation method extends to higher-order linear differential equations with constant coefficients. For an nth-order equation, the characteristic equation will be an nth-degree polynomial, and its roots will determine the form of the general solution.

      In practical applications, the characteristic equation is invaluable for modeling physical systems, such as mechanical vibrations, electrical circuits, and population dynamics. The nature of the roots provides insights into the system's stability and long-term behavior. For instance, negative real parts of complex roots indicate damped oscillations, while positive real parts suggest unstable, growing solutions.

      In conclusion, mastering the characteristic equation is essential for anyone working with differential equations. Its elegance lies in transforming a complex differential equation into a more manageable algebraic problem. By understanding how to

      Types of Solutions: Real, Complex, and Repeated Roots

      When solving second-order linear homogeneous differential equations, the characteristic equation plays a crucial role in determining the nature of the solution. The roots of this equation can fall into three distinct categories: distinct real roots, complex conjugate roots, and repeated roots. Each type of root leads to a different form of the general solution, significantly impacting how we approach and interpret the differential equation's behavior.

      1. Distinct Real Roots:

      When the characteristic equation yields two different distinct real roots, say r and r, the general solution takes the form y = ce^(rx) + ce^(rx), where c and c are arbitrary constants. This scenario represents a combination of two exponential functions, each growing or decaying at its own rate. For example, consider the differential equation y'' - 5y' + 6y = 0. Its characteristic equation r² - 5r + 6 = 0 has roots r = 2 and r = 3, resulting in the general solution y = ce^(2x) + ce^(3x). This solution describes a system where two independent exponential functions coexist, potentially leading to interesting long-term dynamics depending on the initial conditions.

      2. Complex Roots:

      When the characteristic equation produces complex conjugate roots, typically in the form a ± bi, the general solution becomes y = e^(ax)(ccos(bx) + csin(bx)). This solution represents oscillatory behavior modulated by an exponential envelope. The real part 'a' determines whether the oscillations grow (a > 0), decay (a < 0), or maintain constant amplitude (a = 0). The imaginary part 'b' influences the frequency of oscillations. For instance, the equation y'' + 4y' + 13y = 0 has the characteristic equation r² + 4r + 13 = 0, yielding roots r = -2 ± 3i. The general solution is y = e^(-2x)(ccos(3x) + csin(3x)), describing damped oscillations with decreasing amplitude over time.

      3. Repeated Roots:

      In some cases, the characteristic equation may have a single repeated root, r. When this occurs, the general solution takes the form y = (c + cx)e^(rx). This solution combines exponential behavior with linear growth, often resulting in solutions that exhibit more dramatic long-term behavior compared to the distinct real roots case. An example of this is the differential equation y'' + 4y' + 4y = 0, where the characteristic equation r² + 4r + 4 = 0 has the repeated root r = -2. The general solution is y = (c + cx)e^(-2x), representing a decaying exponential modulated by a linear term.

      Understanding these three types of roots is essential for analyzing the behavior of systems modeled by second-order differential equations. Distinct real roots often appear in systems with overdamped behavior, where the solution approaches equilibrium without oscillation. Complex roots are common in oscillatory systems, such as spring-mass systems or electrical circuits with inductance and capacitance. Repeated roots, while less common in natural systems, can arise in carefully tuned or idealized scenarios and often represent a transition point between overdamped and oscillatory behavior.

      The nature of the roots not only determines the form of the general solution but also provides insights into the system's stability and long-term behavior. Real roots indicate non-oscillatory behavior, with negative roots leading to decay and positive roots to growth. Complex roots with negative real parts result in damped oscillations, while those with positive real parts lead to growing oscillations. Repeated roots can produce solutions that grow more rapidly than simple exponentials, even when the root is negative.

      In practical applications, engineers and scientists often analyze these root types to design systems with desired behaviors. For instance, in control systems, one might aim for distinct negative real roots to ensure quick, non-oscillatory convergence to a setpoint. In contrast, some audio systems might be designed to have complex roots.

      Forming the General Solution for Homogeneous Linear Second Order Differential Equations

      The general solution of a homogeneous linear second-order differential equation is a crucial concept in mathematics and engineering. It provides a comprehensive description of all possible solutions to the equation. To form this general solution, we rely on the roots of the characteristic equation, which are fundamental to understanding the behavior of the differential equation.

      The process begins by solving the characteristic equation, which is derived from the original differential equation. Once we have the roots, we can construct the general solution using exponential functions. The form of this solution depends on the nature of the roots obtained.

      When we have two distinct real roots, r and r, the general solution takes the form y = ce^(rx) + ce^(rx), where c and c are arbitrary constants. For example, if the roots are 2 and -3, the general solution would be y = ce^(2x) + ce^(-3x).

      In the case of a repeated real root r, the general solution becomes y = (c + cx)e^(rx). This form ensures we have two linearly independent solutions, which is crucial for the completeness of the general solution. An example of this would be y = (c + cx)e^(4x) for a repeated root of 4.

      When we encounter complex conjugate roots a ± bi, the general solution takes the form y = e^(ax)(ccos(bx) + csin(bx)). This representation uses Euler's formula to express complex exponentials in terms of real-valued functions. For instance, if the roots are 1 ± 2i, the general solution would be y = e^x(ccos(2x) + csin(2x)).

      The importance of having two linearly independent solutions cannot be overstated. Linear independence ensures that the general solution spans the entire solution space of the differential equation. This means that any particular solution can be expressed as a linear combination of these two independent solutions. Without linear independence, we would not be able to represent all possible solutions to the equation.

      To verify linear independence, we can use the Wronskian determinant. If the Wronskian is non-zero, the solutions are linearly independent. This check is particularly important when dealing with repeated roots, where the second solution involves multiplication by x to ensure independence from the first solution.

      The constants c and c in the general solution play a crucial role. They allow us to tailor the solution to specific initial conditions or boundary values. By adjusting these constants, we can generate an infinite number of particular solutions that satisfy the differential equation.

      It's worth noting that the method of undetermined coefficients or variation of parameters can be used to find particular solutions for non-homogeneous equations. These methods build upon the general solution of the homogeneous equation.

      In practical applications, the general solution provides insights into the system's behavior. For instance, in physics, it might describe the motion of an oscillating system, while in engineering, it could model the response of an electrical circuit. The exponential functions in the solution often represent growth or decay processes, while trigonometric functions indicate oscillatory behavior.

      Understanding how to form and interpret the general solution is essential for analyzing differential equations in various fields. It provides a powerful tool for predicting system behavior, designing control systems, and solving complex real-world problems. By mastering this concept, students and professionals alike can tackle a wide range of mathematical and scientific challenges with confidence.

      Applying Initial Conditions to Find Particular Solutions

      Understanding how to use initial conditions to find particular solutions from the general solution is a crucial skill in mathematics and its practical applications. This process allows us to transform a broad, general solution into a specific solution that fits a particular scenario or problem. Let's explore this concept step-by-step and examine its importance in real-world situations.

      The general solution of a differential equation typically contains arbitrary constants. These constants represent the infinite number of possible solutions that satisfy the equation. However, in practical problems, we often need a specific solution that meets certain initial conditions. This is where the process of finding particular solutions comes into play.

      Step 1: Obtain the general solution First, solve the differential equation to find its general solution. This solution will contain arbitrary constants, usually denoted as C or C1, C2, etc.

      Step 2: Identify the initial conditions Initial conditions are specific values of the dependent variable and its derivatives at a given point, usually at t = 0. These conditions are typically provided in the problem statement.

      Step 3: Substitute the initial conditions Replace the variables in the general solution with the given initial conditions. This will create a system of equations involving the arbitrary constants.

      Step 4: Solve for the constants Use algebraic methods to solve the system of equations and determine the values of the arbitrary constants.

      Step 5: Substitute the constants back into the general solution Once you have found the values of the constants, substitute them back into the general solution. This will give you the particular solution that satisfies the initial conditions.

      Let's consider an example to illustrate this process: Suppose we have the general solution y = C1e^(2t) + C2e^(-3t), and we're given the initial conditions y(0) = 4 and y'(0) = 5.

      Step 1: We already have the general solution. Step 2: Initial conditions are y(0) = 4 and y'(0) = 5. Step 3: Substitute t = 0 into the general solution and its derivative: 4 = C1 + C2 5 = 2C1 - 3C2 Step 4: Solve this system of equations to find C1 = 3.5 and C2 = 0.5. Step 5: Substitute these values back into the general solution: y = 3.5e^(2t) + 0.5e^(-3t)

      This process is crucial in various practical applications. In physics, it allows us to model the motion of objects given their initial position and velocity. In engineering, it helps in designing control systems by specifying their initial states. In biology, it aids in predicting population growth based on initial population sizes.

      The importance of finding particular solutions extends to fields like economics, where initial conditions might represent starting capital or market conditions. In environmental science, it helps model the spread of pollutants given initial concentrations. Even in computer graphics, initial conditions can determine the starting point of animations or simulations.

      By mastering the process of applying initial conditions to find particular solutions, you gain the ability to bridge the gap between theoretical mathematics and real-world problem-solving. This skill allows you to take a general mathematical model and tailor it to specific scenarios, making it an invaluable tool in scientific research, engineering design, and data analysis.

      In conclusion, the process of using initial conditions to find particular solutions is a fundamental technique in applied mathematics. It transforms abstract general solutions into concrete, applicable results. By following the steps outlined above and practicing with various examples, you can develop proficiency in this essential skill, enabling you to tackle a wide range of practical problems across numerous fields of study and industry.

      Conclusion: Mastering Homogeneous Linear Second Order Differential Equations

      In summary, we've explored the essential concepts of homogeneous linear second order differential equations. We've covered the characteristic equation, its roots, and how they determine the general solution. Understanding these fundamental principles is crucial for tackling more complex differential equations in advanced mathematics and various scientific fields. The importance of mastering these concepts cannot be overstated, as they form the foundation for solving a wide range of real-world problems in physics, engineering, and other disciplines. To truly grasp these concepts, consistent practice is key. We encourage you to work through additional problems, explore different scenarios, and challenge yourself with more advanced applications. Further study in this area will not only enhance your mathematical skills but also open doors to deeper understanding in related fields. Remember, each equation you solve strengthens your problem-solving abilities and prepares you for more sophisticated mathematical challenges ahead.

      Homogeneous Linear Second Order Differential Equations

      What are Homogeneous Linear Second Order Differential Equations? And what are some sorts of solutions to them?

      Step 1: Understanding the Order of Differential Equations

      To begin with, we need to understand the concept of the order of a differential equation. The order of a differential equation is determined by the highest derivative present in the equation. For instance, if the highest derivative is the second derivative, then it is a second order differential equation. In our case, we are dealing with second order differential equations, which means the highest derivative is the second derivative of the function y with respect to x, denoted as d2ydx2 \frac{d^2y}{dx^2} or y y'' .

      Step 2: Defining Linear Second Order Differential Equations

      A linear second order differential equation is one where the function y and its derivatives appear linearly. This means that the equation can be written in the form:

      a(x)d2ydx2+b(x)dydx+c(x)y=0 a(x) \frac{d^2y}{dx^2} + b(x) \frac{dy}{dx} + c(x)y = 0

      Here, a(x) a(x) , b(x) b(x) , and c(x) c(x) are functions of x. These functions can be constants or more complex functions of x. The key point is that y and its derivatives are not raised to any power other than one and are not multiplied together.

      Step 3: Homogeneous Differential Equations

      A differential equation is said to be homogeneous if all the terms involve the function y or its derivatives. In other words, there are no terms that are purely functions of x or constants. The general form of a homogeneous linear second order differential equation is:

      a(x)d2ydx2+b(x)dydx+c(x)y=0 a(x) \frac{d^2y}{dx^2} + b(x) \frac{dy}{dx} + c(x)y = 0

      In this form, the right-hand side of the equation is zero, indicating that the equation is homogeneous.

      Step 4: Constant Coefficient Homogeneous Equations

      We often deal with a special case of homogeneous linear second order differential equations where the coefficients a a , b b , and c c are constants. This simplifies the equation to:

      ad2ydx2+bdydx+cy=0 a \frac{d^2y}{dx^2} + b \frac{dy}{dx} + c y = 0

      For example, if a=4 a = 4 , b=3 b = -3 , and c=2 c = 2 , the equation becomes:

      4d2ydx23dydx+2y=0 4 \frac{d^2y}{dx^2} - 3 \frac{dy}{dx} + 2 y = 0

      Step 5: Finding Solutions

      To find solutions to these equations, we assume a solution of the form y=erx y = e^{rx} . By substituting this assumed solution into the differential equation, we can derive a characteristic equation in terms of r. Solving this characteristic equation gives us the values of r, which in turn provide us with the general solution to the differential equation.

      For example, substituting y=erx y = e^{rx} into 4d2ydx23dydx+2y=0 4 \frac{d^2y}{dx^2} - 3 \frac{dy}{dx} + 2 y = 0 results in the characteristic equation:

      4r23r+2=0 4r^2 - 3r + 2 = 0

      Solving this quadratic equation for r gives us the roots, which can be real and distinct, real and repeated, or complex conjugates. Each case leads to a different form of the general solution.

      Step 6: General Solution for Real and Distinct Roots

      If the characteristic equation has real and distinct roots r1 r_1 and r2 r_2 , the general solution to the differential equation is:

      y=C1er1x+C2er2x y = C_1 e^{r_1 x} + C_2 e^{r_2 x}

      where C1 C_1 and C2 C_2 are arbitrary constants determined by initial conditions.

      Step 7: General Solution for Real and Repeated Roots

      If the characteristic equation has real and repeated roots r r , the general solution is:

      y=(C1+C2x)erx y = (C_1 + C_2 x) e^{r x}

      Here, C1 C_1 and C2 C_2 are arbitrary constants.

      Step 8: General Solution for Complex Conjugate Roots

      If the characteristic equation has complex conjugate roots α±βi \alpha \pm \beta i , the general solution is:

      y=eαx(C1cos(βx)+C2sin(βx)) y = e^{\alpha x} (C_1 \cos(\beta x) + C_2 \sin(\beta x))

      Again, C1 C_1 and C2 C_2 are arbitrary constants.

      Step 9: Superposition Principle

      One important property of linear homogeneous differential equations is the superposition principle. If y1 y_1 and y2 y_2 are solutions to the differential equation, then any linear combination of these solutions is also a solution. This means that if y1 y_1 and y2 y_2 are solutions, then:

      y=C1y1+C2y2 y = C_1 y_1 + C_2 y_2

      is also a solution, where C1 C_1 and C2 C_2 are constants.

      Step 10: Conclusion

      In summary, homogeneous linear second order differential equations are equations involving the second derivative of a function, where the function and its derivatives appear linearly and the equation is set to zero. Solutions to these equations can be found by solving the characteristic equation, and the general solution depends on the nature of the roots of this characteristic equation. The superposition principle allows us to combine solutions to form new solutions, providing a powerful tool for solving these types of differential equations.

      FAQs

      Here are some frequently asked questions about homogeneous linear second order differential equations:

      1. What is a homogeneous linear second order differential equation?

        A homogeneous linear second order differential equation is an equation of the form ay'' + by' + cy = 0, where a, b, and c are constants, and y is a function of an independent variable (usually x). The equation is called homogeneous because all terms contain y or its derivatives, and the right-hand side is zero.

      2. How do you solve a homogeneous linear second order differential equation?

        The general approach involves finding the characteristic equation (ar² + br + c = 0), solving for its roots, and then using these roots to construct the general solution. The form of the solution depends on whether the roots are real and distinct, real and repeated, or complex conjugates.

      3. What is the significance of the characteristic equation?

        The characteristic equation is crucial because its roots determine the form of the general solution. Real distinct roots lead to solutions with exponential functions, repeated roots involve a product of exponential and polynomial functions, and complex roots result in solutions with sine and cosine functions.

      4. How do initial conditions affect the solution?

        Initial conditions are used to find specific values for the arbitrary constants in the general solution, resulting in a particular solution. Typically, two conditions are needed for a second order equation, often specifying the value of y and y' at a particular point.

      5. What are some real-world applications of these equations?

        Homogeneous linear second order differential equations are used to model various physical phenomena, including simple harmonic motion, damped oscillations in mechanical and electrical systems, population dynamics, and heat transfer problems. They are fundamental in fields such as physics, engineering, and applied mathematics.

      Prerequisite Topics for Homogeneous Linear Second Order Differential Equations

      Understanding homogeneous linear second order differential equations requires a solid foundation in several key mathematical concepts. These prerequisite topics are crucial for grasping the intricacies of this advanced subject and solving complex problems effectively.

      One of the fundamental skills needed is the ability to work with derivative of exponential functions. This concept is essential when dealing with equations that involve exponential terms, which are common in many real-world applications of differential equations.

      Another critical prerequisite is understanding the characteristic equation with complex roots. This topic is directly related to solving homogeneous linear second order differential equations, as the roots of the characteristic equation determine the form of the solution.

      The nature of roots of quadratic equations: The discriminant is also crucial. This concept helps in determining the type of roots (real, complex, or repeated) of the characteristic equation, which in turn affects the solution method.

      While not directly related, solving differential equations with step functions can provide valuable insights into more complex problem-solving techniques used in homogeneous linear second order differential equations.

      Understanding position velocity acceleration relationships is beneficial, as many real-world applications of these equations involve motion and physics concepts.

      The concept of linear independence is crucial when working with the general solution of homogeneous equations, as it helps in determining whether two solutions form a fundamental set.

      Graphing from slope-intercept form y=mx+b provides a strong foundation for visualizing and interpreting solutions to differential equations.

      Knowledge of the derivative of inverse trigonometric functions can be helpful when dealing with certain types of differential equations that involve these functions.

      Determining number of solutions to linear equations is a skill that translates well to understanding the nature of solutions in differential equations.

      Lastly, familiarity with the characteristic equation with real distinct roots is essential, as it forms the basis for solving many homogeneous linear second order differential equations.

      By mastering these prerequisite topics, students will be well-prepared to tackle the challenges of homogeneous linear second order differential equations, enabling them to solve complex problems and apply their knowledge to real-world situations with confidence.

      A Linear Second Order Differential Equation is of the form:

      a(x)d2ydx2+b(x)dydx+c(x)y=d(x)a(x) \frac{d^2 y}{dx^2} +b(x) \frac{dy}{dx}+c(x)y=d(x)

      Or equivalently,

      a(x)y+b(x)y+c(x)y=d(x)a(x) y''+b(x) y'+c(x)y=d(x)

      Where all of a(x),b(x),c(x),d(x)a(x),b(x),c(x),d(x) are functions of xx.

      A Linear Second Order Differential Equation is called homogeneous if d(x)=0d(x)=0. So,

      a(x)y+b(x)y+c(x)y=0a(x) y''+b(x) y'+c(x)y=0

      And a general constant coefficient linear homogeneous, second order differential equation looks like this:

      Ay+By+Cy=0Ay''+By'+Cy=0

      Let's suppose that both f(x) and g(x) are solutions to the above differential equations, then so is

      y(x)=c1f(x)+c2g(x)y(x)=c_1 f(x)+c_2 g(x)

      Where c1c_1 and c2c_2 are constants

      Characteristic Equation
      The general solution to the differential equation:

      Ay+By+Cy=0Ay''+By'+Cy=0

      Will be of the form: y(x)=erxy(x)=e^{rx}
      Taking the derivatives:
      y(x)=rerxy' (x)=re^{rx}
      y(x)=r2erxy'' (x)=r^2 e^{rx}

      And inputting them into the above equation:

      Ar2erx+Brerx+Cerx=0Ar^2 e^{rx}+Bre^{rx}+Ce^{rx}=0

      So we will have: erx(Ar2+Br+C)=0e^{rx} (Ar^2+Br+C)=0

      The equation Ar2+Br+C=0Ar^2+Br+C=0 is called the Characteristic Equation. And is used to solve these sorts of questions.

      Solving the quadratic we will get some values for rr:

      Real Roots: r1r2r_1 \neq r_2
      Complex Roots: r1,r2=λ±μir_1,r_2=\lambda \pm \mu i
      Repeated Real Roots: r1=r2r_1=r_2

      Though let's deal with only real roots for now

      As we will get two solutions to the characteristic equation:

      y1(x)=er1xy_1 (x)=e^{r_1 x}
      y2(x)=er2xy_2 (x)=e^{r_2 x}

      So all solutions will be of the form:

      y(x)=c1erx+c2erxy(x)=c_1 e^{rx}+c_2 e^{rx}