Question

In case you haven't studied any differential equations before, I shall be introducing the necessary ideas as needed. Here is a simple excercise to get you started: Find the general solution of the firstorder equation \(d f / d t=f\) for an unknown function \(f(t) .\) [There are several ways to do this. One is to rewrite the equation as \(d f / f=d t\) and then integrate both sides.] How many arbitrary constants does the general solution contain? [Your answer should illustrate the important general theorem that the solution to any \(n\) th-order differential equation (in a very large class of "reasonable" equations) contains \(n\) arbitrary constants.]

Short answer

The general solution is \( f(t) = C_1 e^t \) and it contains one arbitrary constant.

Step-by-step solution

Rewrite the Differential Equation

The given differential equation is \( \frac{df}{dt} = f \). A first step is to separate the variables so we can integrate. We rewrite it as \( \frac{df}{f} = dt \). This step involves manipulating the equation to enable integration.

Integrate Both Sides

Next, we integrate both sides of the equation. The left-hand side is \( \int \frac{1}{f} \, df \) and the right-hand side is \( \int dt \). This gives us \( \ln |f| = t + C \), where \( C \) is the integration constant. Integration introduces the arbitrary constant because it represents an indefinite integral.

Solve for the Unknown Function

To solve for \( f(t) \), we need to isolate it. We do this by exponentiating both sides to remove the natural logarithm: \( f = e^{t+C} \). This can be rewritten as \( f = e^t \cdot e^C \). We can redefine \( e^C \) as a new arbitrary constant \( C_1 \), meaning \( f(t) = C_1 e^t \).

Determine Arbitrary Constants

The solution \( f(t) = C_1 e^t \) contains one arbitrary constant, \( C_1 \). This corresponds with the order of the given differential equation, which is first-order. Hence, the general solution of a first-order differential equation typically contains one arbitrary constant.

Key concepts

First-order differential equations

First-order differential equations are the simplest type of differential equation you will encounter. They involve an unknown function and its first derivative. In our exercise, the equation is \( \frac{df}{dt} = f \). This means that the rate of change of our function \( f(t) \) is directly proportional to \( f \) itself.

This type of equation can be encountered in various fields such as physics, biology, and finance. They often model exponential growth or decay. Recognizing a first-order differential equation is crucial because it sets the stage for how we go about solving it. Solving means finding an expression for \( f(t) \), our unknown function, in terms of \( t \).

When you approach such an equation, you'll often use a technique known as "separation of variables," which is a method that simplifies the equation into a form that can be easily integrated, as seen in the given example.

General solution

The general solution of a differential equation is a family of functions that includes all possible solutions. For our first-order equation \( \frac{df}{dt} = f \), we aim to find a solution for \( f(t) \) that satisfies this equation for any initial condition.

When we derive the general solution, we incorporate an arbitrary constant, which accounts for all potential solutions. In our example, after integrating, we end up with \( \ln |f| = t + C \). Solving for \( f \), we get \( f(t) = C_1 e^t \), where \( C_1 \) is an arbitrary constant representing all possible vertical shifts of the solution curve.

This expression means there's an entire set of functions, depending on \( C_1 \), that could potentially satisfy the original differential equation. Each choice of \( C_1 \) gives a specific solution that passes through a different point on the graph. The general solution is an important concept because it gives us a complete picture of all potential behavior of the system modeled by the equation.

Integration constants

Integration constants are introduced when you integrate a differential equation to find its solution. They come from the process of indefinite integration, where the antiderivative of a function is found. The integration constant \( C \) allows for a whole family of solutions, as it accounts for all possible shifts or translations of the solution curve.

In the given exercise, when we integrated \( \frac{1}{f} \, df = dt \), we obtained \( \ln |f| = t + C \). The constant \( C \) reflects the uncertainty of the initial value of the function or where precisely the function curve might intersect the axes.

When we solved for \( f(t) \), we redefined \( e^C \) as a new constant \( C_1 \), so our function became \( f(t) = C_1 e^t \). This reflects how free we are to choose \( C_1 \), given any initial set of conditions. In any first-order differential equation, this integration constant plays a vital role, ensuring solutions fit various initial conditions we might encounter.

Variables separation

Variable separation is a powerful method used in solving first-order differential equations. The technique involves rearranging the equation so that all terms involving the unknown function \( f(t) \) and its differential \( df \) are on one side, and all the terms involving \( t \) and \( dt \) are on the other.

In the solution to our exercise, the equation \( \frac{df}{dt} = f \) was transformed to \( \frac{df}{f} = dt \). This step isolated each variable, allowing us to integrate separately straightforwardly. The left-hand side, involving \( f \), and the right-hand side, involving \( t \), are both integrated to yield a solvable expression.

This approach often simplifies the process and is particularly useful for exponential growth or decay equations. It's an essential tool that, once mastered, can make handling these mathematical problems efficient and less daunting. Mastery of variable separation will prove beneficial not only in basic problems but also in more complex applications you may encounter in advanced studies.