Chapter 17: Problem 4
Solve the differential equation. \(y^{\prime \prime}-8 y^{\prime}+12 y=0\)
Short Answer
Expert verified
The solution is \(y(t) = C_1 e^{6t} + C_2 e^{2t}\).
Step by step solution
01
Identify the Characteristic Equation
The first step in solving this second-order linear homogeneous differential equation is to find the characteristic equation. For an equation of the form \(a y'' + b y' + c y = 0\), the characteristic equation is \(ar^2 + br + c = 0\). Here, we have \(y'' - 8y' + 12y = 0\), so the characteristic equation is \(r^2 - 8r + 12 = 0\).
02
Solve the Characteristic Equation
Next, we need to solve the quadratic characteristic equation \(r^2 - 8r + 12 = 0\) to find the roots. We can factor it as \((r - 6)(r - 2) = 0\), which gives us the roots \(r_1 = 6\) and \(r_2 = 2\).
03
Write the General Solution
Since the roots \(r_1\) and \(r_2\) are real and distinct, the general solution to the differential equation is given by \(y(t) = C_1 e^{r_1 t} + C_2 e^{r_2 t}\). Plugging the roots into this formula, we get \(y(t) = C_1 e^{6t} + C_2 e^{2t}\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Characteristic Equation
The characteristic equation plays a crucial role in solving second-order linear differential equations. To understand it, consider the standard form of a second-order linear homogeneous differential equation: \[ a y'' + b y' + c y = 0 \] Here, a characteristic equation is associated with this differential equation and is of the quadratic form: \[ ar^2 + br + c = 0 \] This equation helps us find the values of r that will be used to form the differential equation's solution. In our exercise, the given differential equation is \(y'' - 8y' + 12y = 0\). By comparing it to the standard form, we can construct the characteristic equation as \[ r^2 - 8r + 12 = 0 \] Solving this quadratic equation provides us with the roots (or values of \(r\)) that are fundamental to building the general solution.
Second-Order Linear Differential Equations
Differential equations that involve second derivatives are called second-order differential equations. Specifically, those in the form of \[ a y'' + b y' + c y = 0 \] are referred to as second-order linear homogeneous differential equations. The term "linear" indicates that the equation concerns linear terms with respect to \(y'\), \(y''\), and \(y\), without involving any multiplicative interaction between these terms.
- Homogeneous: This means the right-hand side equals zero, ensuring the function behaves consistently, simplifying analysis and solution derivation.
- Structure: The equation consists of the sum of its derivatives multiplied by constant coefficients, which facilitates the solving process.
General Solution of Differential Equations
The general solution of a differential equation provides a comprehensive form that includes all possible solutions. Upon solving the characteristic equation and finding its roots, the nature of these roots determines the structure of the general solution.
- General Solution Formula: For real and distinct roots \(r_1\) and \(r_2\), the general solution can be expressed as: \[ y(t) = C_1 e^{r_1 t} + C_2 e^{r_2 t} \] Here, \(C_1\) and \(C_2\) are arbitrary constants determined by initial conditions or boundary values.
- Application: In our example, the roots found were \(r_1 = 6\) and \(r_2 = 2\). Therefore, the general solution becomes: \[ y(t) = C_1 e^{6t} + C_2 e^{2t} \] This solution comprises two parts, each part representing an exponential function of time, governed by their respective roots.