Question

(a) Find 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 \(y(t)\) as \(t \rightarrow-\infty\) and as \(t \rightarrow \infty\). Does \(y(t)\) approach \(-\infty,+\infty\), or a finite limit? $$2 y^{\prime \prime}-3 y^{\prime}=0, \quad y(-2)=3, \quad y^{\prime}(-2)=0$$

Short answer

Answer: As \(t\) approaches negative and positive infinity, the specific solution \(y(t) = 3\) approaches a finite limit of 3.

Step-by-step solution

Solve the homogeneous equation

The given equation is a homogeneous second-order linear differential equation: $$2 y^{\prime \prime}-3 y^{\prime}=0$$ Divide by 2 to simplify: $$y^{\prime \prime}-\frac{3}{2} y^{\prime}=0$$ This is a linear constant coefficient equation, and its characteristic equation is given by: $$r^2 - \frac{3}{2}r = 0$$ We can factor this quadratic as \(r(r-\frac{3}{2})=0\). Thus, we have two roots: \(r_1 = 0\) and \(r_2 = \frac{3}{2}\) Given the roots, the general solution of the homogeneous equation is: $$y(t) = C_1 e^{0t} + C_2 e^{\frac{3}{2}t}$$ Simplifying, we have: $$y(t) = C_1 + C_2 e^{\frac{3}{2}t}$$

Apply the initial conditions

Now that we have the general solution, let's find the constants \(C_1\) and \(C_2\) using the given initial conditions: $$y(-2) = 3, \quad y^{\prime}(-2) = 0$$ For the first condition: $$y(-2) = C_1 + C_2 e^{\frac{3}{2}(-2)} = 3$$ For the second condition, we need to find \(y^{\prime}(t)\). Differentiating \(y(t)\) with respect to \(t\): $$y^{\prime}(t) = 0 + \frac{3}{2}C_2 e^{\frac{3}{2}t}$$ So, $$y^{\prime}(-2) = \frac{3}{2}C_2 e^{\frac{3}{2}(-2)} = 0$$ Let's solve the system of equations for \(C_1\) and \(C_2\): $$C_1 + C_2 e^{-3} = 3$$ $$\frac{3}{2}C_2 e^{-3} = 0$$ Since \(e^{-3}\) is not equal to 0, we can conclude that \(C_2 = 0\). Substituting this into the first equation, we get: $$C_1 = 3$$ Thus, our specific solution is: $$y(t) = 3$$

Analyze the behavior of the solution

Now that we have our specific solution \(y(t) = 3\), let's analyze its behavior as \(t \rightarrow -\infty\) and \(t \rightarrow \infty\). As \(t \rightarrow -\infty\), \(y(t) = 3\). As \(t \rightarrow \infty\), \(y(t) = 3\). In both cases, the solution approaches a finite limit, which is 3. Therefore, \(y(t)\) does not approach \(-\infty\) or \(+\infty\), but rather a finite limit.

Key concepts

General Solution of Differential Equation

Understanding the general solution of a differential equation is a foundational concept in differential equations. The general solution encompasses all the possible solutions to a differential equation without any particular initial conditions applied. For second-order linear homogeneous differential equations, the general solution typically takes the form of a combination of exponential functions based on the roots of the characteristic equation. Here's an illustration using the equation from the exercise:

First, identify the characteristic equation of the homogeneous differential equation, which is obtained by replacing the derivatives of the function with powers of an auxiliary variable, usually denoted by 'r'. Once we factor and find the roots of this characteristic equation, the general solution can be constructed. The solution is a linear combination of terms involving exponential functions raised to the power of each root. If we have distinct roots, say, \( r_1 \) and \( r_2 \), the general solution looks like:
\[ y(t) = C_1 e^{r_1t} + C_2 e^{r_2t} \]
where \( C_1 \) and \( C_2 \) are constants to be determined by initial conditions. This form provides a wide range of solutions encompassing all possible scenarios described by the differential equation at hand.

Initial Value Problem

Initial value problems (IVP) are a class of problems in differential equations where the solution to the differential equation is required to satisfy given conditions at a specific point. These conditions, known as initial values, anchor the general solution to a particular instance, making it unique. IVPs are crucial in applications where the state of a system at a given time dictates the future behavior of the system.

For instance, in the exercise, after finding the general solution, the initial values of \( y(-2) = 3 \) and \( y'(-2) = 0 \) are used to find the specific constants \( C_1 \) and \( C_2 \). We plug the initial values into the general solution and its derivative and solve the consequent system of equations. Through this method, we determine the unique constants that satisfy the given conditions. The outcome is a single, unique solution to the differential equation that complies with the specified initial values.

Behavior of Solution

When analyzing the behavior of a solution to a differential equation, we are interested in understanding how the function behaves as the independent variable, often time, approaches infinity or negative infinity. This can provide insights into the stability and long-term trends of the system described by the equation.

In our problem, once the unique solution \( y(t) = 3 \) is found, we can easily determine its behavior for extreme values of time. Since our solution is a constant, it doesn't change with time. Therefore, whether \( t \) heads towards negative or positive infinity, the solution remains the same, indicating a steady state. In more complex scenarios, a solution might grow without bounds, oscillate, or approach other finite limits, which would lead to different interpretations on the system's behavior over time.

Homogeneous Linear Differential Equation

A homogeneous linear differential equation is one in which each term is a constant coefficient multiplied by the function or its derivatives, and there is no free term (a term without the function or its derivatives). The differential equation is considered homogeneous if the right-hand side is zero.

The exercise presents a homogeneous linear differential equation with constant coefficients: \( 2 y'' - 3 y' = 0 \). To solve it, we find a characteristic equation that reflects the structure of the differential equation. This characteristic equation helps determine the nature of the solution to the original equation. Typically, these equations are amenable to explicit solutions, which are invaluable for predicting the behavior of systems modeled by such equations. As in our example, the roots of the characteristic equation, \( r_1 = 0 \) and \( r_2 = \frac{3}{2} \), lead to a general solution composed of exponential functions associated with the roots.