Question

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

Short answer

In conclusion, the solution to the given second-order linear homogeneous differential equation is $$y(t) = \frac{1}{e^2}e^{-2t} + \frac{2}{e^2}te^{-2t}.$$ The function approaches zero as the value of \(t\) increases, and its graph is an exponentially decaying curve that starts from the point \((-1, 2)\) and approaches 0 for increasing values of \(t\).

Step-by-step solution

Write the characteristic equation

We are given the second-order linear homogeneous differential equation: \(y^{\prime \prime}+4 y^{\prime}+4 y=0\). To convert this to a characteristic equation, we replace \(y^{\prime \prime}\) with \(m^2\), \(y^{\prime}\) with \(m\), and \(y\) with 1. So, the characteristic equation becomes: $$m^2 + 4m + 4 = 0.$$

Solve the characteristic equation

Now, we solve this characteristic equation for \(m\). This is a quadratic equation, which we can either factor or use the quadratic formula. Here, it can be factored: $$(m + 2)^2 = 0.$$ Solving this, we find that both roots are equal and equal to \(m = -2.\) Since we have a repeated root, let's call it \(r = -2.\)

Find the complementary function

Since the characteristic equation has a repeated root (\(r = -2\)), we know the complementary function for the given differential equation is in the form of: $$y_c(t) = C_1 e^{rt} + C_2te^{rt}.$$ Substituting \( r = -2\), we get: $$y_c(t) = C_1 e^{-2t} + C_2te^{-2t}.$$

Use initial conditions to find the particular solution

We are given initial conditions \(y(-1) = 2\) and \(y^{\prime}(-1) = 1\). First, we find the derivative of the complementary function: $$y_c^{\prime}(t) = -2C_1e^{-2t} - 2C_2e^{-2t} + 2C_2te^{-2t}.$$ Now, we use the initial conditions to find the constants \(C_1\) and \(C_2\): $$y(-1) = C_1e^2 + (-1)C_2e^2 = 2.$$ $$y^{\prime}(-1) = -2C_1e^2 - 2C_2e^2 + 2C_2e^2 = 1.$$ Solving these simultaneous equations, we find \(C_1 = \frac{1}{e^2}\) and \(C_2 = \frac{2}{e^2}\). Thus, the particular solution is: $$y(t) = \frac{1}{e^2}e^{-2t} + \frac{2}{e^2}te^{-2t}.$$

Analyze the behavior of the solution for increasing t

When \(t\) increases, both exponential terms \(e^{-2t}\) and \(te^{-2t}\) tend to 0. Therefore, the function \(y(t)\) approaches zero as \(t\) increases.

Sketch the graph of the solution

The graph of the solution \(y(t) = \frac{1}{e^2}e^{-2t} + \frac{2}{e^2}te^{-2t}\) is an exponentially decaying curve. It starts from the point \((-1, 2)\) at the initial condition and approaches 0 for an increasing value of \(t\). The derivative is positive when \(t=-1\), causing the function to initially increase as \(t\) increases from \(-1\). Since both terms are exponentially decaying, the graph will decrease and approach zero as \(t\) goes to infinity. To draw this graph, you can plot the given points and visualize the described behavior.

Key concepts

Characteristic Equation

In differential equations, the characteristic equation is a stepping stone to finding the solution. It's obtained from a given second-order linear homogeneous differential equation. In our problem, we start with an equation of the form:
\[ y^{\prime\prime} + 4y^{\prime} + 4y = 0 \]
To derive the characteristic equation, we replace each derivative with a power of \(m\), like this:

• Replace \(y^{\prime\prime}\) with \(m^2\)
• Replace \(y^{\prime}\) with \(m \)
• Replace \(y\) with 1

This transforms the differential equation into a quadratic: \(m^2 + 4m + 4 = 0\). Solving this equation reveals the nature of the solution's components. The solutions to the roots of this equation greatly influence the type of solutions we get for the original differential equation.

Complementary Function

The complementary function involves finding a general solution to the differential equation using the roots of the characteristic equation. For the differential equation in our problem, the characteristic equation \((m+2)^2=0\) results in a repeated root \(r = -2\). This indicates that there is a specific form the general solution must take because of these repeated roots.
The complementary function associated with it is:\[ y_c(t) = C_1 e^{rt} + C_2 te^{rt} \]Substituting \(r = -2\), we have:\[ y_c(t) = C_1 e^{-2t} + C_2 te^{-2t} \]
Here:

• \(C_1\) and \(C_2\) are constants determined by initial conditions.
• The term \(C_2 te^{-2t}\) arises specifically due to the repeated root, extending the solution beyond just the exponential term.

This formula lays the groundwork upon which we build more specific solutions to meet initial conditions.

Repeated Roots

In the context of differential equations, repeated roots occur when the characteristic equation provides the same root more than once. In our exercise, the characteristic equation \(m^2 + 4m + 4 = 0\) gives a double root at \(m = -2\).
Repeated roots indicate that a simple exponential solution is insufficient to provide a full description of the system’s behavior.
To address this, the general solution incorporates a term involving \(t\) to distinguish the solutions tied to each occurrence of the root. Thus, we incorporate:

• \(e^{rt}\)
• \(te^{rt}\)

When a root has multiplicity \(k\), the terms \(t^ne^{rt}\) for \(n\) from 0 to \(k-1\) are used. Here, \(k=2\), leading to the complementary function \(y_c(t) = C_1 e^{-2t} + C_2 te^{-2t}\).
These additional terms ensure that the complementary function adequately addresses all characteristics of the original differential equation.

Initial Value Problem

Solving an initial value problem requires not only finding the general solution but also using specific conditions to find the exact solution. In our problem, this involves using the given values:

• \(y(-1) = 2\)
• \(y^{\prime}(-1) = 1\)

These initial conditions enable us to determine the constants \(C_1\) and \(C_2\) in the complementary function.
By substituting \(-1\) for \(t\) in both \(y(t)\) and its derivative \(y^{\prime}(t)\), and equating them to the initial conditions, we create equations to solve for \(C_1\) and \(C_2\).
In our example, this results in:

• \(C_1 = \frac{1}{e^2}\)
• \(C_2 = \frac{2}{e^2}\)

Applying these constants yields a particular solution that fits both the differential equation and the given initial conditions, allowing the entire behavior across \(t\) to be understood thoroughly.