Question

Solve the given initial value problem. Sketch the graph of the solution and describe its behavior for increasing \(t.\) \(y^{\prime \prime}-6 y^{\prime}+9 y=0, \quad y(0)=0, \quad y^{\prime}(0)=2\)

Short answer

Short Answer: The exact solution to the given initial value problem (IVP) is \(y(t) = 2te^{3t}\), which represents an increasing exponential curve as \(t\) increases.

Step-by-step solution

Write down the given information

We are given the second-order linear homogeneous differential equation: \(y^{\prime \prime}-6 y^{\prime}+9 y=0\) With the initial conditions: \(y(0)=0\) and \(y^{\prime}(0)=2\)

Find the characteristic equation and its roots

Write the characteristic equation associated with the differential equation: \(r^2 - 6r + 9 = 0\) This is a quadratic equation. To solve it, we can use the quadratic formula: \(r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) In this case, \(a = 1\), \(b = -6\), and \(c = 9\). Plug these values into the quadratic formula: \(r = \frac{6 \pm \sqrt{(-6)^2 - 4(1)(9)}}{2(1)}\) \(r = \frac{6 \pm \sqrt{36 - 36}}{2}\) \(r = \frac{6 \pm 0}{2}\) \(r = 3\) Since we only found one root, it is a real and repeated root.

Determine the general solution

Since we have a real and repeated root, \(r=3\), the general solution for the differential equation is of the form: \(y(t) = (A + Bt)e^{rt}\) In our case, \(r = 3\). So we have: \(y(t) = (A + Bt)e^{3t}\)

Use the initial conditions to find the constants A and B

Given the initial condition \(y(0)=0\), we can write: \(0 = (A + B(0))e^{3(0)} = A\) So, A = 0. Now, use the second initial condition \(y^{\prime}(0)=2\). Find the first derivative of our general solution: \(y^{\prime}(t) = (3A + (3Bt + B))e^{3t} = (3Bt + B)e^{3t}\) Evaluate the first derivative at \(t=0\): \(2 = (3B(0) + B)e^{3(0)} = B\) So, B = 2.

Write down the exact solution

We now have the values for A and B. The exact solution to the given IVP is: \(y(t) = (0 + 2t)e^{3t}\) Or simply, \(y(t) = 2te^{3t}\)

Sketch the graph of the solution and describe its behavior

By looking at the solution, we can see that as \(t\) increases, both the exponential term \(e^{3t}\) and the linear term \(2t\) increase as well. So, the product of these two terms, \(2te^{3t}\), will also increase for increasing \(t\) values. Ultimately, the graph of the solution will be an increasing exponential curve as \(t\) increases.

Key concepts

Initial Value Problem

An Initial Value Problem (IVP) is a type of differential equation along with specific conditions called initial conditions. These conditions are given at a certain point, usually at the beginning, such as at time zero, and they are used to find a unique solution to the problem. In our exercise, we have the differential equation: \( y^{\prime \prime}-6 y^{\prime}+9 y=0 \), with the initial conditions \( y(0)=0 \) and \( y^{\prime}(0)=2 \).
These initial conditions allow us to find a specific solution that not only satisfies the differential equation but also fits the conditions at the starting point. Without these initial conditions, we would only find a general solution, which represents a family of potential solutions.

Characteristic Equation

In solving differential equations, the characteristic equation plays a crucial role in finding solutions, especially for linear homogeneous equations. The characteristic equation is derived from the coefficients of the differential equation. For the equation \( y^{\prime \prime}-6 y^{\prime}+9 y=0 \), the characteristic equation is \( r^2 - 6r + 9 = 0 \), which was formed by considering the differential operator as an algebraic expression with roots \( r \).
By solving the characteristic equation, we find the roots, which indicate the form of the solution to the differential equation. Quadratic equations like this one can have different types of roots: real and distinct, real and repeated, or complex. Each scenario influences the form of the solution.

Homogeneous Differential Equation

A Homogeneous Differential Equation is one where every term is a function of the dependent variable and its derivatives, and there are no free-standing constant terms. This is important because it often allows for solutions to be expressed in terms of exponential functions.
In our case, the equation \( y^{\prime \prime}-6 y^{\prime}+9 y=0 \) is homogeneous. The solution involves using the characteristic equation's roots to form a solution using exponential functions. Because the root was real and repeated, the general solution took the form \( y(t) = (A + Bt)e^{3t} \). Homogeneous equations like this often involve procedures like finding characteristic equations and solving for constants using initial conditions.

Exponential Growth

Exponential Growth describes how a quantity increases rapidly over time, and is a behavior exhibited by functions of the form \( Ce^{kt} \) where \( k > 0 \). The solution to our initial value problem, \( y(t) = 2te^{3t} \), is an example of exponential growth. Here, both the exponential function \( e^{3t} \) and the linear term \( 2t \) contribute to the rate of increase.
As \( t \) increases, \( e^{3t} \) grows very quickly, dominating the behavior of the function. Thus, the graph of this solution would show a steep, upward-curving line, demonstrating the rapid increase characteristic of exponential growth. This kind of growth is common in phenomena such as population growth and chemical reactions.