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Magazine Chapter 7: Problem 12
Find the general solution to the linear differential equation. $$ y^{\prime \prime}-7 y^{\prime}+12 y=0 $$
Short Answer
Expert verified
The general solution is \( y(t) = C_1e^{4t} + C_2e^{3t} \).
Step by step solution
01
Identifying the Type of Differential Equation
First, recognize that the given equation \( y'' - 7y' + 12y = 0 \) is a second-order linear homogeneous differential equation with constant coefficients.
02
Writing the Characteristic Equation
From the differential equation \( y'' - 7y' + 12y = 0 \), create the characteristic equation by replacing \( y'' \) with \( r^2 \), \( y' \) with \( r \), and \( y \) with 1. The result is \( r^2 - 7r + 12 = 0 \).
03
Solving the Characteristic Equation
To find the roots of the characteristic equation \( r^2 - 7r + 12 = 0 \), use the quadratic formula \( r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), with \( a = 1 \), \( b = -7 \), and \( c = 12 \). Calculate the discriminant: \( (-7)^2 - 4(1)(12) = 49 - 48 = 1 \).
04
Finding the Roots
With the discriminant equal to 1, the roots are real and distinct. Solve using the quadratic formula: \( r = \frac{7 \pm \sqrt{1}}{2} \), resulting in \( r_1 = 4 \) and \( r_2 = 3 \).
05
Writing the General Solution
Since there are two distinct real roots \( r_1 = 4 \) and \( r_2 = 3 \), the general solution to the differential equation is \( y(t) = C_1e^{4t} + C_2e^{3t} \), where \( C_1 \) and \( C_2 \) are arbitrary constants.
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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 involves a second derivative, which means it measures how a quantity changes that itself depends on change. In the equation given, \( y'' - 7y' + 12y = 0 \), the highest derivative is \( y'' \), indicating it is second-order. Such equations are common in physics and engineering for modeling systems with acceleration.
- Notation: The term \( y'' \) represents the second derivative of \( y \) with respect to its variable, often time \( t \).
- Purpose: These equations are used to model various dynamic systems, like oscillations in springs, motion of planets, or circuits.
Characteristic Equation
The characteristic equation is a crucial step in finding general solutions to linear differential equations. By converting the differential equation \( y'' - 7y' + 12y = 0 \) into its characteristic equation, we simplify the problem into a form of algebra.
- Transformation Process: Replace each derivative with powers of \( r \) to derive \( r^2 - 7r + 12 = 0 \).
- Why It Matters: Solving this quadratic equation provides the roots, which determine the form of the solution of the differential equation.
Homogeneous Differential Equations
Homogeneous differential equations, like \( y'' - 7y' + 12y = 0 \), have terms that are all dependent on the function and its derivatives. The distinction 'homogeneous' means every term involves \( y \) or its derivatives, there is no standalone source term.
- Solution Implications: Their solutions involve functions that translate through exponential, sinusoidal, or polynomial functions, often driven by constant coefficients.
- Example Concept: The term \( 0 \) on the right side signifies that the equation models a system without external forces, impacting how the system evolves naturally.
Constant Coefficients
Constant coefficients in differential equations like \( y'' - 7y' + 12y = 0 \) mean the coefficients of \( y \), \( y' \), and \( y'' \) are constants, not functions of the variable. This constancy simplifies the processes used to solve the equation.
- Simplification: Having constant coefficients simplifies determining the characteristic equation and ensures solutions can be expressed in terms of exponential functions.
- Analysis Benefits: Constant coefficients allow us to predict the type of general solution鈥攆iguring out if it's exponential, oscillatory, or involves growth/decay.
