Question

(a) Obtain the general solution of the differential equation. (b) Impose the initial conditions to obtain the unique solution of the initial value problem. (c) Describe the behavior of the solution as \(t \rightarrow-\infty\) and \(t \rightarrow \infty\). In each case, does \(y(t)\) approach \(-\infty,+\infty\), or a finite limit? $$ y^{\prime \prime}+2 y^{\prime}+y=0, \quad y(1)=1, \quad y^{\prime}(1)=0 $$

Short answer

Answer: As \(t \rightarrow \infty\), the solution approaches a finite limit of 0.

Step-by-step solution

Determine the auxiliary equation

We are given the second-order, homogeneous linear differential equation: $$y^{\prime \prime} + 2y^{\prime} + y = 0$$ To solve this equation, we write down the auxiliary (or characteristic) equation, which is a quadratic equation for the corresponding constant coefficients: $$m^2 + 2m + 1 = 0$$

Solve the auxiliary equation

Now we solve the auxiliary equation to find its roots: $$(m + 1)(m + 1) = 0$$ So, the root is \(m = -1\) with multiplicity 2.

Write down the general solution

Since we have a repeated root \(m = -1\), the general solution for the given differential equation is: $$y(t) = C_1 e^{-t} + C_2 te^{-t}$$

Apply the initial conditions to find the unique solution

We have two initial conditions: \(y(1) = 1\) and \(y^{\prime}(1) = 0\). First, we will apply the initial condition \(y(1) = 1\) to find one of the constants. Substitute \(t = 1\) into the general solution: $$1 = C_1 e^{-1} + C_2 e^{-1}$$ Now, we will apply the second initial condition \(y^{\prime}(1) = 0\). To do this, we need to find the derivative of the general solution and substitute \(t = 1\): $$y^{\prime}(t) = -C_1 e^{-t} + C_2 e^{-t} - C_2 te^{-t}$$ $$0 = -C_1 e^{-1} + C_2 e^{-1} - C_2 e^{-1}$$ $$C_1 = C_2$$ Looking at these equations, we can see that \(C_1 = C_2\) and \(C_1 + C_2 = e^{-1}\). Thus, both \(C_1\) and \(C_2\) are equal to \(\frac{e^{-1}}{2}\).

Write down the unique solution

Substitute the values of \(C_1\) and \(C_2\) into the general solution: $$y(t) = \frac{e^{-1}}{2}(e^{-t} + te^{-t})$$ This is the unique solution to the initial value problem.

Analyze the behavior of the solution as \(t \rightarrow -\infty\) and \(t \rightarrow \infty\)

As \(t \rightarrow -\infty\), the negative exponentials will grow, but since we have a \(t\) multiplying one of them, we can't determine the exact limit. The solution does not approach a specific value of \(-\infty\) or \(+\infty\) in this case. As \(t \rightarrow \infty\), the negative exponentials will go to 0, making the entire solution approach 0. So the solution approaches a finite limit of 0 as \(t \rightarrow \infty\).

Key concepts

Differential Equations

Differential equations are mathematical equations that describe the relationship between a function and its derivatives. These equations often model real-world phenomena, such as the growth of populations, the cooling of an object, or the oscillation of a spring.

In the given exercise, we are looking at a second-order linear homogeneous differential equation, which is an equation involving the second derivative of a function that equals zero. This type of equation is crucial because it provides a mathematical framework for understanding dynamic systems and predicting their behavior over time. The solution to such an equation, which we call the general solution, incorporates all possible solutions to the differential equation.

Characteristic Equation

The characteristic equation is a pivotal concept in solving linear homogeneous differential equations with constant coefficients. It is an algebraic equation obtained by substituting a trial solution into the differential equation. For example, in the provided exercise, the trial solution involves an exponential function, which leads us to the characteristic equation \(m^2 + 2m + 1 = 0\).

This characteristic equation determines the nature of the roots which will, in turn, shape the general solution. The roots may be real and distinct, real and repeated, or complex, and each scenario leads to a different form of the general solution. As seen in the solution, solving \(m + 1)^2 = 0\) gave us a repeated root, informing us about the structure of the response to our physical or geometric problem.

General Solution

Once we have the roots from the characteristic equation, we can express the general solution of a differential equation in terms of those roots. If we have distinct real roots, our solution will consist of exponential functions based on those roots. However, when we have repeated roots, as in this exercise, the general solution includes \(e^{mt}\) and \(te^{mt}\) where \(m\) is the repeated root.

Specifically, with a repeated root of \(m = -1\), the general solution is \(y(t) = C_1 e^{-t} + C_2 te^{-t}\), where \(C_1\) and \(C_2\) are constants to be determined by initial conditions. It's important to note that the general solution encompasses all possible specific solutions to the differential equation.

Behavior of the Solution

Understanding the behavior of the solution to a differential equation as time progresses, both towards positive and negative infinity, is essential for comprehending the long-term behavior of the modeled system.

As \(t \rightarrow \infty\), the solution's terms that involve exponential functions with negative exponents approach zero, which means the solution approaches a finite limit. In the case of the specific exercise, as \(t \rightarrow \infty\), the solution \(y(t)\) approaches zero. However, as \(t \rightarrow -\infty\), the solution's behavior is more complex due to the term \(te^{-t}\), and it does not settle towards a specific finite or infinite limit. Analyzing the limits at both extremes of time provides insights into the stability and ultimate fate of the system we are studying.