Problem 21 - Thomas Calculus

Question

Solve the exponential growth/decay initial value problem for \(y\) as a function of \(t\) thinking of the differential equation as a first-order linear equation with \(P(x)=-k\) and \(Q(x)=0 :\) $$\frac{d y}{d t}=k y \quad(k \text { constant }), \quad y(0)=y_{0}$$

Step-by-Step Solution

Answer

The solution is \(y(t) = y_0e^{kt}\).

Step 1: Understand the Equation

We are given a differential equation \(\frac{dy}{dt} = ky\) with the condition \(y(0) = y_0\). This type of equation represents either exponential growth or decay depending on the sign of \(k\).

Step 2: Identify the Standard Form

The differential equation can be written in the standard form of a first-order linear differential equation: \(\frac{dy}{dt} - ky = 0\). Here, \(P(t) = -k\) and \(Q(t) = 0\).

Step 3: Solve the Homogeneous Equation

Since the term \(Q(t) = 0\), this equation is homogeneous, and its solution can be assumed to be of the form \(y(t) = Ce^{kt}\).

Step 4: Apply the Initial Condition

Using the initial condition \(y(0) = y_0\), we substitute into our solution: \(y(0) = Ce^{k \cdot 0} = C = y_0\). Hence, \(C = y_0\).

Step 5: Write the General Solution

Substitute \(C = y_0\) into the assumed solution: \(y(t) = y_0e^{kt}\). This is the solution of the differential equation with the given initial condition.

Key Concepts

Exponential Growth

Exponential growth is a process where the quantity increases over time in a way that the rate of growth is proportional to the current amount. This means that as time goes on, the quantity grows faster. The term is represented mathematically by the differential equation \(\frac{dy}{dt} = ky\), where \(k\) is a constant. If \(k > 0\), the process is called exponential growth. The positive \(k\) indicates that the function will keep increasing. If \(k < 0\), it is known as exponential decay, representing a decrease over time.
To identify exponential growth or decay:

• Check the sign of \(k\). Positive \(k\) indicates growth; negative \(k\) indicates decay.
• Observe the function form, \(y(t) = y_0e^{kt}\), where \(e\) is the base of the natural logarithm and \(y_0\) is the initial amount.

Exponential functions are common in real-life situations, such as population growth, radioactive decay, and interest calculations. They are powerful because they show how quickly quantities can increase or decrease under constant proportional conditions. Understanding exponential growth helps in forecasting future scenarios in various fields.

Initial Value Problem

An Initial Value Problem (IVP) in the context of differential equations is a problem where the solution is determined by the initial conditions provided. It involves finding a function \(y(t)\) that satisfies a differential equation and also meets the initial condition, \(y(0) = y_0\). The given condition helps in finding the particular solution out of many potential solutions for a differential equation.
Steps involved in solving an IVP include:

• Expressing the differential equation in a recognizable form.
• Solving the differential equation to obtain a general solution.
• Applying the initial condition to find the specific solution that satisfies the problem.

In our exercise, the initial condition \(y(0) = y_0\) allows us to determine the constant \(C\) when we solve the exponential differential equation. This ensures that our solution not only satisfies the equation but also starts at the right initial state. Understanding initial value problems is crucial for applications where you know the starting point and need to model future behavior.

Homogeneous Equations

A homogeneous differential equation is one where all terms are dependent on the function and its derivatives, and there is no term that stands alone (or can be seen as "free-standing"). In the context of linear first-order differential equations, a homogeneous equation takes the form \(\frac{dy}{dt} - ky = 0\), where the function \(Q(t)\) equals zero.
Characteristics of homogeneous equations include:

• The right-hand side of the equation is zero, indicating no external forces or sources.
• Solutions often take the form of exponential functions, as with \(y(t) = Ce^{kt}\).

These equations are simpler to solve since the absence of the \(Q(t)\) term allows us to assume solutions like \(Ce^{kt}\). By focusing purely on the function and its derivative, these equations describe natural processes driven entirely by intrinsic characteristics without external influence. Recognizing homogeneous equations helps in simplifying analysis and in predicting steady-state behaviors where growth or decay is uninfluenced by external factors.