Question

Find the solution of the given initial value problem. Sketch the graph of the solution and describe its behavior as \(t\) increases. $$ y^{\prime \prime}+4 y^{\prime}+3 y=0, \quad y(0)=2, \quad y^{\prime}(0)=-1 $$

Short answer

Answer: The specific solution to the initial value problem is \(y(t) = 5 e^{-t} - 3 e^{-3t}\).

Step-by-step solution

Write down the initial value problem

The given initial value problem is: $$ y^{\prime \prime}+4 y^{\prime}+3 y=0, \quad y(0)=2, \quad y^{\prime}(0)=-1 $$

Find the characteristic equation

The characteristic equation associated with the given homogeneous differential equation is: $$ r^2 + 4r + 3 = 0 $$

Solve the characteristic equation

Solve the quadratic equation by factoring it: $$ (r + 3)(r + 1) = 0 $$ The roots of the equation are \(r_1 = -1\) and \(r_2 = -3\).

Construct the general solution

Using the roots, we can construct the general solution of the homogeneous differential equation: $$ y(t) = C_1 e^{-t} + C_2 e^{-3t} $$

Apply the initial conditions

Next, we apply the initial conditions to find the specific solution. First, applying the initial condition \(y(0) = 2\), we get: $$ 2 = C_1 e^{0} + C_2 e^{0} \Rightarrow 2 = C_1 + C_2 $$ Now, applying \(y^{\prime}(0)=-1\), we differentiate the general solution and evaluate it at \(t=0\): $$ y^{\prime}(t) = -C_1 e^{-t} - 3C_2 e^{-3t} $$ $$ -1 = -C_1 e^{0} - 3C_2 e^{0} \Rightarrow -1 = -C_1 - 3C_2 $$ Solve the system of linear equations to find \(C_1\) and \(C_2\): $$ C_1 = 5, \quad C_2 = -3 $$ Now, the specific solution is: $$ y(t) = 5 e^{-t} - 3 e^{-3t} $$

Describe the behavior of the solution as \(t\) increases

As \(t\) increases, both terms in the solution decrease due to the exponential decay of the terms. The term with the faster decay is the one with \(e^{-3t}\), so it will approach zero more quickly than the term with \(e^{-t}\). As a result, the behavior of the solution will be dominated by the slower decaying term (\(5e^{-t}\)) as \(t\) increases. Overall, this solution represents a decaying function that approaches zero as \(t\) goes to infinity.

Key concepts

Homogeneous Differential Equation

A homogeneous differential equation is a type of differential equation where all terms depend on the function and its derivatives. The equation given in the exercise, \(y'' + 4y' + 3y = 0\), is an example of a second-order linear homogeneous differential equation. It is called "homogeneous" because it equates to zero. This term signifies there are no external influences or forcing functions acting on the system, only intrinsic properties. Homogeneous equations are commonly encountered in physics and engineering because they describe natural systems where internal factors dictate behavior without external input. Understanding these equations involves finding functions that describe the evolution of the system's state over time.

Characteristic Equation

The characteristic equation is a crucial part of solving linear homogeneous differential equations with constant coefficients. It's derived from the differential equation by assuming a solution of the form \(y(t) = e^{rt}\), where \(r\) is a constant. Substituting this assumed solution into the differential equation transforms it into a polynomial equation, specifically in terms of \(r\). For the exercise's equation, \(y'' + 4y' + 3y = 0\), the characteristic equation becomes \(r^2 + 4r + 3 = 0\). Solving this quadratic equation gives us the roots, which are fundamental in determining the solution's behavior. In our case, \(r_1 = -1\) and \(r_2 = -3\) indicate the system's natural response.

General Solution

The general solution to a homogeneous differential equation involves using the roots of the characteristic equation. With roots \(r_1\) and \(r_2\), the form of the general solution will be \(y(t) = C_1 e^{r_1 t} + C_2 e^{r_2 t}\). For the given exercise, these roots are \(-1\) and \(-3\), leading to the general solution \(y(t) = C_1 e^{-t} + C_2 e^{-3t}\). The coefficients \(C_1\) and \(C_2\) must be determined from initial conditions, which reveal how the solution fits a specific scenario. By solving simple linear equations, we find \(C_1 = 5\) and \(C_2 = -3\). This gives us the specific solution \(y(t) = 5 e^{-t} - 3 e^{-3t}\), ensuring it meets both initial conditions: \(y(0) = 2\) and \(y'(0) = -1\).

Exponential Decay

Exponential decay occurs when quantities decrease rapidly at first, and then slowly as time progresses. In the context of the solution \(y(t) = 5 e^{-t} - 3 e^{-3t}\), exponential decay is evident in both terms: \(e^{-t}\) and \(e^{-3t}\). The rate at which each term tends towards zero depends on its exponent. The term \(e^{-3t}\) decays much faster than \(e^{-t}\) because the larger exponent makes it diminish quickly. As time goes on, the slower decaying term, \(5 e^{-t}\), becomes the predominant factor influencing the system's behavior. Ultimately, both exponential terms will approach zero, indicating that the system stabilizes at zero as \(t\) approaches infinity. This characteristic is crucial in predicting the long-term behavior of systems depicted by such equations.