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? $$y^{\prime \prime}-4 y=0, \quad y(3)=0, \quad y^{\prime}(3)=0$$

Short answer

**Answer:** The unique solution to the given differential equation is the constant function \(y(t) = 0\). The behavior of this solution as \(t\) approaches positive or negative infinity remains the same: it approaches a finite limit of 0.

Step-by-step solution

Write the characteristic equation for the given differential equation

Rewrite the given differential equation as a characteristic equation, as follows: $$m^2 - 4 = 0$$

Solve the characteristic equation

To find the roots of the characteristic equation, solve the following quadratic equation: $$m^2 - 4 = (m + 2)(m - 2) = 0$$ This equation has roots \(m_1 = -2\) and \(m_2 = 2\).

Write the general solution of the differential equation

Given that the roots are distinct real numbers, the general solution will take the form: $$y(t) = A e^{m_1 t} + B e^{m_2 t}$$ Substitute the roots found in Step 2 to get: $$y(t) = A e^{-2t} + B e^{2t}$$

Apply the initial conditions to obtain the unique solution

Apply the initial condition \(y(3) = 0\): $$0 = A e^{-2 \cdot 3} + B e^{2 \cdot 3}$$ Next, find the derivative of \(y(t)\): $$y'(t) = -2A e^{-2t} + 2B e^{2t}$$ Apply the initial condition \(y'(3) = 0\): $$0 = -2A e^{-6} + 2B e^6$$

Solve for A and B

From the equations obtained in Step 4: $$A e^{-6} + B e^6 = 0 \quad (1)$$ $$-A e^{-6} + B e^6 = 0 \quad (2)$$ Now, add \((1)\) and \((2)\) together to solve for A and B: $$2B e^6 = 0$$ Which implies: $$B = 0$$ Now substitute B = 0 into \((1)\): $$A e^{-6} = 0$$ Which implies: $$A = 0$$

Write the unique solution

Substituting A = 0 and B = 0 in our general solution, we get: $$y(t) = 0$$

Describe the behavior of the solution as t approaches infinity or negative infinity

Since the unique solution is a constant function \(y(t) = 0\), its behavior as \(t\) approaches negative or positive infinity does not change: As \(t \rightarrow -\infty\), \(y(t) = 0\). As \(t \rightarrow +\infty\), \(y(t) = 0\). The function \(y(t)\) approaches a finite limit of 0 as t goes to either positive or negative infinity.

Key concepts

Initial Value Problem

An initial value problem in differential equations refers to finding a specific solution that satisfies both a differential equation and a set of initial conditions. These conditions specify the value of the solution and possibly its derivatives at a particular point. They are crucial because a differential equation alone has infinitely many solutions. To tackle an initial value problem, follow these steps:

• First, determine the general solution of the differential equation.
• Then, apply the given initial conditions to pinpoint the unique solution that fits the criteria.

In the exercise provided, the initial values are given as the values of the function and its derivative at a specific point: \(y(3) = 0\) and \(y'(3) = 0\). This information constrains the general solution, helping to determine the specific constants that fulfill these conditions. In this case, both constants turned out to be zero, resulting in the solution \(y(t) = 0\).

Characteristic Equation

The characteristic equation is a key concept when solving linear homogeneous differential equations with constant coefficients. It allows us to determine the form of the general solution to such equations.For a second-order differential equation like \(y'' - 4y = 0\), we form the characteristic equation by replacing derivatives with powers of a variable, usually \(m\). This substitution simplifies the process because the equation becomes algebraic, specifically quadratic in nature.In our example:

• The characteristic equation is \(m^2 - 4 = 0\).
• Solving this gives the roots \(m_1 = -2\) and \(m_2 = 2\), indicating real and distinct solutions.

The roots indicate exponential solutions for the differential equation, leading to the general form \(y(t) = Ae^{-2t} + Be^{2t}\). The characteristic equation thus plays a crucial role in shaping the possible solutions of our differential equation.

General Solution

The general solution to a differential equation encompasses all potential solutions. For a second-order homogeneous linear differential equation with constant coefficients, the general solution takes the form of a combination of exponential functions. In mathematical terms, for the equation \(y'' - 4y = 0\), we use the roots of the characteristic equation. Given that the roots \(m_1 = -2\) and \(m_2 = 2\) are real and distinct, the general solution is:

• \(y(t) = A e^{-2t} + B e^{2t}\)

Here, \(A\) and \(B\) are arbitrary constants that are determined when initial conditions are introduced.The general solution acts as the underlying formula from which all specific solutions, for different initial conditions, stem. This is why knowing it is vital; it provides the framework for understanding the possible behaviors of the differential equation under various scenarios.

Behavior of Solutions

Understanding the behavior of solutions means analyzing how they act as time progresses to infinity or negative infinity. For differential equations like the one given, behavior can often be inferred partially from the structure of the solution.In our exercise, the derived solution is simply \(y(t) = 0\). Hence, no matter what value \(t\) takes, \(y(t)\) remains constant. This is a special case of solutions where all terms vanish due to initial conditions leading to zero constants.Regarding behavior:

• As \(t \rightarrow -\infty\), \(y(t) = 0\).
• As \(t \rightarrow +\infty\), \(y(t) = 0\).

This demonstrates that the solution tends to a stable, finite value. The behavior of solutions is crucial in both theoretical and practical applications, as it helps predict long-term outcomes based on the differential equation's structure and initial conditions.