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Magazine Chapter 4: Problem 10
In Problems 1-36 find the general solution of the given differential equation. \(\frac{d^{2} y}{d x^{2}}-10 \frac{d y}{d x}+25 y=0\)
Short Answer
Expert verified
The general solution is \( y(x) = C_1 e^{5x} + C_2 x e^{5x} \).
Step by step solution
01
Identify the Type of Differential Equation
The given differential equation is \( \frac{d^2 y}{dx^2} - 10 \frac{dy}{dx} + 25y = 0 \). This is a second-order linear homogeneous differential equation with constant coefficients.
02
Write the Characteristic Equation
For a second-order linear differential equation of the form \( a \frac{d^2 y}{dx^2} + b \frac{dy}{dx} + c y = 0 \), the characteristic equation is obtained by replacing \( \frac{d^2 y}{dx^2} \) with \( r^2 \), \( \frac{dy}{dx} \) with \( r \), and \( y \) with 1. Thus, the characteristic equation becomes \( r^2 - 10r + 25 = 0 \).
03
Solve the Characteristic Equation
The characteristic equation \( r^2 - 10r + 25 = 0 \) is a quadratic equation. To solve it, we can factor it: \((r - 5)(r - 5) = r^2 - 10r + 25 = 0\). Thus, \( r = 5 \) is a repeated root.
04
Write the General Solution
Since the characteristic equation has a repeated root \( r = 5 \), the general solution of the differential equation is given by \( y(x) = C_1 e^{5x} + C_2 x e^{5x} \), where \( C_1 \) and \( C_2 \) are constants determined by initial conditions or boundary conditions if provided.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Second-order differential equations
A second-order differential equation is an equation involving a function and its derivatives up to the second order. This type of equation is fundamental in describing a variety of physical systems where accelerations (second derivative of position) are involved. For example, the motion of a pendulum, vibrations of mechanical structures, or electrical circuits use second-order differential equations to model their behaviors.
These equations typically have the form: \[ a \frac{d^2 y}{dx^2} + b \frac{dy}{dx} + c y = f(x) \] where:
These equations typically have the form: \[ a \frac{d^2 y}{dx^2} + b \frac{dy}{dx} + c y = f(x) \] where:
- \(a\), \(b\), and \(c\) are constants.
- \(f(x)\) is a function of \(x\).
Homogeneous equations
A homogeneous equation is a specific type of differential equation where the right-hand side is zero, that means \( f(x) = 0 \). In the given exercise, we see an example of a homogeneous equation: \[ \frac{d^2 y}{dx^2} - 10 \frac{dy}{dx} + 25y = 0 \] The term homogeneous comes because all terms involve the function \(y\) or its derivatives.
Key characteristics of homogeneous equations include:
Key characteristics of homogeneous equations include:
- They simplify to the form \(L(y) = 0\), where \(L(y)\) is a linear differential operator.
- Solutions to homogeneous equations can often be superimposed; that is, if \(y_1(x)\) and \(y_2(x)\) are solutions, then \(C_1 y_1(x) + C_2 y_2(x)\) is also a solution, where \(C_1\) and \(C_2\) are constants.
Characteristic equation
The characteristic equation is central to solving linear homogeneous differential equations with constant coefficients. To derive it, replace each derivative in the differential equation with a power of \(r\), a variable typically used to denote roots.
For example, for the equation \( \frac{d^2 y}{dx^2} - 10 \frac{dy}{dx} + 25y = 0 \), replace \( \frac{d^2 y}{dx^2} \) with \(r^2\), \( \frac{dy}{dx} \) with \(r\), and \(y\) with 1, resulting in: \[ r^2 - 10r + 25 = 0 \] Solving the characteristic equation usually involves finding roots which indicate solutions to the differential equation.
For example, for the equation \( \frac{d^2 y}{dx^2} - 10 \frac{dy}{dx} + 25y = 0 \), replace \( \frac{d^2 y}{dx^2} \) with \(r^2\), \( \frac{dy}{dx} \) with \(r\), and \(y\) with 1, resulting in: \[ r^2 - 10r + 25 = 0 \] Solving the characteristic equation usually involves finding roots which indicate solutions to the differential equation.
- These roots can either be real and distinct, repeated, or complex.
- The nature of the roots determines the form of the general solution.
General solution of differential equations
The general solution of a differential equation provides a family of solutions for the given equation. It encompasses all possible solutions based on the initial or boundary conditions.
For second-order linear homogeneous differential equations, the general solution depends on the roots of the characteristic equation:
For second-order linear homogeneous differential equations, the general solution depends on the roots of the characteristic equation:
- **Real and distinct roots**: If the characteristic equation has two real and distinct roots \( r_1 \) and \( r_2 \), the general solution is \( y(x) = C_1 e^{r_1 x} + C_2 e^{r_2 x} \).
- **Repeated roots**: If there is a repeated real root \( r \), the general solution becomes \( y(x) = C_1 e^{rx} + C_2 x e^{rx} \).
- **Complex roots**: If the roots are complex, \( \alpha \pm \beta i \), the solution is \( y(x) = e^{\alpha x}(C_1 \cos(\beta x) + C_2 \sin(\beta x)) \).
