Question

Solve the given initial value problem. Sketch the graph of the solution and describe its behavior for increasing \(t.\) \(9 y^{\prime \prime}-12 y^{\prime}+4 y=0, \quad y(0)=2, \quad y^{\prime}(0)=-1\)

Short answer

Answer: The particular solution for the given initial value problem is \(y(t) = 2 e^{\frac{2}{3}t} - \frac{5}{2} t e^{\frac{2}{3}t}\). The graph of the solution starts at \((0, 2)\), initially decreases, and then turns concave up as it eventually grows exponentially for increasing \(t\).

Step-by-step solution

Find the General Solution

To find the general solution of the given second-order linear homogeneous differential equation, let's solve its characteristic equation: \(9m^2 - 12m + 4 = 0.\) We can factor this quadratic equation as follows: \((3m - 2)^2 = 0.\) So we have a real, repeated root \(m = \frac{2}{3}.\) Therefore, the general solution of the given differential equation is \(y(t) = c_1 e^{\frac{2}{3}t} + c_2 t e^{\frac{2}{3}t}.\)

Apply the Initial Conditions

Now, let's apply the initial conditions. First, the condition \(y(0) = 2\): \(2 = c_1 e^0 + c_2(0) e^0 \implies c_1 = 2.\) Next, apply the condition \(y'(0) = -1.\) To do this, differentiate the expression for \(y(t)\) with respect to \(t,\) which gives: \(y'(t) = \frac{2}{3} c_1 e^{\frac{2}{3}t} + \frac{2}{3} c_2 e^{\frac{2}{3}t} + c_2 t e^{\frac{2}{3}t}.\) Now, plug in \(t = 0\) and set \(y'(0) = -1\): \(-1 = \frac{2}{3} (2) e^0 + \frac{2}{3} c_2 e^0 \implies c_2 = -\frac{5}{2}.\)

Determine the Particular Solution

With the constants \(c_1 = 2\) and \(c_2 = -\frac{5}{2},\) we can find the particular solution: \(y(t) = 2 e^{\frac{2}{3}t} - \frac{5}{2} t e^{\frac{2}{3}t}.\)

Sketch the Graph and Describe Behavior

The graph of the particular solution will have the following characteristics: - It starts at the point \((0, 2)\). - The initial slope is negative, which means the graph will begin decreasing. - As \(t\) increases, the exponential term \(e^{\frac{2}{3}t}\) will dominate the function and will make it grow at a rate proportional to \(\frac{2}{3}\). - The function will turn concave up as \(t\) increases. In summary, the graph of the solution will start at \((0, 2),\) initially decreasing and then turning concave up as it eventually grows exponentially for increasing \(t.\)

Key concepts

Differential Equations

Differential equations are equations that relate a function with its derivatives. They show how a function changes over time or space and are essential in fields like physics, engineering, and biology. A differential equation can be simple, involving only first derivatives, or complex, involving higher-order derivatives.
In solving them, one of the main goals is to find a function or a family of functions that satisfies the relationship defined by the equation. For instance, if we are given a differential equation related to the motion of a particle, solving it would tell us how the position of the particle changes over time.
In our specific problem, we dealt with a second-order linear differential equation. This means it involved the second derivative of a function, representing how the rate of change of the rate of change is behaving.

Homogeneous Equations

Homogeneous differential equations are a specific type of differential equation where every term is a multiple of the dependent variable or its derivatives. Simply put, they are equations that can usually be solved using standard procedures such as finding the characteristic equation. These equations are quite symmetric in nature, and because of their specific structure, they allow certain solution methods.
The equation in our problem, \(9 y'' - 12 y' + 4 y = 0\), is homogeneous. Surprisingly, despite the complexity that may seem to arise due to the second derivative, such equations often have elegant solutions.
Solving a homogeneous equation involves finding a characteristic equation, which helps determine the nature of the solutions whether they are real or complex, repeated, or distinct.

Characteristic Equation

A characteristic equation is a polynomial equation derived from a linear differential equation used to find its solutions. For a differential equation of the form \(ay'' + by' + cy = 0\), the characteristic equation would be a quadratic or higher-order equation derived by substituting \(y = e^{mt}\) into the differential equation. This substitution turns differentiation into multiplication by \(m\), transforming the differential equation into a polynomial equation.
In our problem, the characteristic equation was \(9m^2 - 12m + 4 = 0\), which simplifies to \((3m - 2)^2 = 0\) indicating a repeated root at \(m = \frac{2}{3}\). The repeated root suggests a solution that involves an exponential function combined with a term linear in \(t\), leading to solutions of the form \(c_1 e^{mt} + c_2 t e^{mt}\).
This approach provides systematic access to solutions that may depict complex behaviors in real-world applications.

Exponential Growth

Exponential growth describes a process where the rate of change of a quantity is proportional to the current amount, leading to growth that accelerates over time. In mathematics, this type of growth is typically expressed in the form \(e^{rt}\), where \(e\) represents Euler's number, an important constant approximately equal to 2.71828, and \(r\) is a constant rate of growth.
In our problem's solution, we encounter an exponential term \(e^{\frac{2}{3}t}\). This indicates that for large \(t\), the solution grows rapidly because the exponential term dominates the behavior. Initially, the behavior of the solution might be influenced by other components, like the linear term in \(t\), but eventually, the exponential growth defines the function's long-term behavior.
Exponential growth models are widespread and can describe various natural phenomena, including population growth, biological processes, and financial growth scenarios.