Chapter 31: Problem 40
Solve the given problems. The displacement \(y\) (in \(\mathrm{cm}\) ) of an object at the end of a spring is described by the equation \(d^{2} y / d t^{2}+4 d y / d t+4 y=0,\) where \(t\) is the time (in \(\mathrm{s}\) ) if \(f(0)=0\) and \(f(1)=0.50 \mathrm{cm} .\) Solve for \(y=f(t)\).
Short Answer
Expert verified
The solution is \( y(t) = 0.50 e^{2-2t} t \).
Step by step solution
01
Identify the type of differential equation
This is a second-order linear homogeneous differential equation with constant coefficients, given by \( \frac{d^{2} y}{d t^{2}} + 4 \frac{d y}{d t} + 4 y = 0 \).
02
Write the characteristic equation
To solve the differential equation, we find its characteristic equation by assuming \( y = e^{rt} \). This gives us the characteristic equation: \( r^2 + 4r + 4 = 0 \).
03
Solve the characteristic equation
Solve the quadratic equation \( r^2 + 4r + 4 = 0 \) using the quadratic formula: \( r = \frac{-b \pm \sqrt{b^2-4ac}}{2a} \). Here, \( a=1, b=4, c=4 \), so \( r = \frac{-4 \pm \sqrt{16-16}}{2} = \frac{-4 \pm 0}{2} = -2 \). The roots are equal, \( r_1 = r_2 = -2 \).
04
Form the general solution
Since the characteristic equation has a repeated root, the general solution for \(y\) is \( y(t) = (C_1 + C_2 t) e^{-2t} \), where \( C_1 \) and \( C_2 \) are constants to be determined.
05
Apply the initial conditions
Use the initial conditions to find \( C_1 \) and \( C_2 \). Given \( f(0) = 0 \), substitute \( t = 0 \) into the general solution: \( 0 = (C_1 + C_2 \cdot 0) e^0 = C_1 \). So, \( C_1 = 0 \).
06
Use the second initial condition
With \( C_1 = 0 \), the general solution becomes \( y(t) = C_2 t e^{-2t} \). Using \( f(1) = 0.50 \), \( 0.50 = C_2 \cdot 1 \cdot e^{-2} \). Solving gives \( C_2 = 0.50 e^{2} \).
07
Write the final solution
Substitute the value of \( C_2 \) back into the general solution: \( y(t) = 0.50 e^{2} t e^{-2t} = 0.50 e^{2-2t} t \). This is the solution function \( y(t) \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Second-Order Linear Differential Equations
Second-order linear differential equations are a type of differential equation that involves second derivatives, as seen in terms like \( \frac{d^2y}{dt^2} \). These equations are called "linear" because they don't involve powers or products of the unknown function and its derivatives beyond the first power. Typically, they appear in the form: \[ a \frac{d^2y}{dt^2} + b \frac{dy}{dt} + c y = 0 \]When we discuss second-order linear differential equations, it's important to also consider the coefficients \( a \), \( b \), and \( c \), which remain constant in these cases. Studying the specific solutions of these equations requires solving for the function \( y(t) \) that describes the behavior of a system over time.
This particular dynamic is common in physics, especially in the analysis of mechanical vibrations and electrical circuits. An example of such a system is the displacement of an object attached to a spring, as given in the original problem.
This particular dynamic is common in physics, especially in the analysis of mechanical vibrations and electrical circuits. An example of such a system is the displacement of an object attached to a spring, as given in the original problem.
Homogeneous Differential Equations
Homogeneous differential equations are a subset where every term is a function of the unknown variable and its derivatives and equals zero. The term "homogeneous" implies that the equations have a zero on one side:\[ a \frac{d^2y}{dt^2} + b \frac{dy}{dt} + c y = 0 \]In homogeneous differential equations, the structure ensures that if you have a solution, any multiple of this solution is also a solution. This property makes it easier to find solutions using superposition—the idea that solutions can be added together to find new solutions.
When solving these equations, the concept of a characteristic equation often comes into play. This provides essential insight into finding and verifying solutions.
When solving these equations, the concept of a characteristic equation often comes into play. This provides essential insight into finding and verifying solutions.
Characteristic Equation
The characteristic equation is derived from assuming a solution of a specific form for the differential equation. For our second-order homogeneous differential equation, we assume a solution like \[ y = e^{rt} \]This assumption simplifies the differential equation into an algebraic equation, specific to the problem at hand:\[ r^2 + 4r + 4 = 0 \]Solving this algebraic equation, called the characteristic equation, gives the values of \( r \) that correspond to different valid solutions for \( y(t) \). In our exercise, the solutions are repeated roots (both are \( r = -2 \)), which lead us to a special form of the general solution involving both \( t \) and \( e^{-2t} \). Repeated roots imply the solution takes the form: \[ y(t) = (C_1 + C_2 t) e^{-2t} \]This characteristic method effectively connects complex calculus with more familiar algebraic techniques, making problem-solving manageable.
Initial Conditions
Initial conditions provide crucial information needed to determine specific solutions for differential equations. They tell us the state of a system at specific points in time, allowing us to solve for the constants in the general solution. For our given problem, initial conditions are:- \( f(0) = 0 \)- \( f(1) = 0.50 \, \text{cm} \)Applying these to the general solution helps in finding \( C_1 \) and \( C_2 \):- Substituting \( t=0 \): \( 0 = (C_1 + C_2 \cdot 0) e^0 \) gives \( C_1 = 0 \).- Substituting \( t=1 \): \( 0.50 = C_2 e^{-2} \) helps compute \( C_2 \).Incorporating the initial conditions anchors the mathematical solution to physical reality, providing the specific path the system follows.
General Solution
The general solution of a differential equation encompasses all possible solutions that satisfy the differential equation. For second-order linear differential equations with constant coefficients, the general solution is based on the roots of the characteristic equation. Given a characteristic equation with repeated roots as seen here, \( r_1 = r_2 = -2 \), the general solution is:\[ y(t) = (C_1 + C_2 t) e^{-2t} \]Each combination of \( C_1 \) and \( C_2 \) represents a potential solution, but specific initial conditions narrow this down to a unique path. Once initial conditions are applied, like \( f(0) = 0 \) and \( f(1) = 0.50 \), we determine specific values for these constants, arriving at the specific function:\[ y(t) = 0.50 e^{2} t e^{-2t} \]The general solution thus provides both depth and flexibility, capturing a wide array of possibilities converging to a true representation of the physical system.