Question

Show that the solution to \(f^{\prime}=a f\) satisfying \(f\left(x_{0}\right)=k\) is \(f(x)=k e^{a\left(x-x_{0}\right)}\).

Short answer

The solution is \(f(x) = k e^{a(x-x_0)}\).

Step-by-step solution

Understand the Problem Statement

We are given a differential equation \(f^{\prime} = af\) with an initial condition \(f(x_0) = k\). We need to show that the solution to this differential equation is \(f(x) = k e^{a(x-x_0)}\).

Review Exponential Growth

The differential equation \(f^{\prime} = af\) is known as an exponential growth differential equation. The general solution to this type of equation is of the form \(f(x) = C e^{ax}\), where \(C\) is a constant. Our task involves identifying this constant using the provided initial condition.

Apply the Initial Condition

To find the constant \(C\), we use the initial condition \(f(x_0) = k\). Substitute \(x = x_0\) into the general solution: \(f(x_0) = C e^{ax_0} = k\). Solve for \(C\), we find \(C = k e^{-ax_0}\).

Substitute Back into the Solution

Substitute \(C = k e^{-ax_0}\) back into the general solution \(f(x) = C e^{ax}\) to get \(f(x) = k e^{-ax_0} e^{ax}\). This simplifies to \(f(x) = k e^{a(x-x_0)}\), because \(e^{-ax_0} \cdot e^{ax} = e^{ax - ax_0}\).

Verification

Check if \(f(x) = k e^{a(x-x_0)}\) satisfies both the differential equation and the initial condition. Differentiating \(f(x)\), we get \(f^{\prime}(x) = a k e^{a(x-x_0)}\), which is \(a f(x)\), satisfying \(f^{\prime} = a f\). Also, when \(x = x_0\), \(f(x_0) = k e^{a(x_0 - x_0)} = k\), satisfying the initial condition.

Key concepts

Exponential Growth

Exponential growth is a fundamental concept in differential equations often described by the equation \( f'(x) = a f(x) \). This indicates that the rate of change of the function \( f(x) \) is proportional to its current value. In mathematical models, exponential growth represents populations or quantities that increase rapidly.

• Here, \( a \) is a constant that determines the rate of growth or decay. If \( a > 0 \), the function exhibits exponential growth. Conversely, if \( a < 0 \), it exhibits exponential decay.
• The general form of the solution to this equation normally looks like \( f(x) = C e^{ax} \), where \( e \) is the base of natural logarithms and \( C \) is an arbitrary constant determined by initial conditions.

Understanding this balance between the rate of change and the current value is key.

Initial Conditions

Initial conditions are crucial in solving differential equations, as they allow us to find specific solutions from a general form. In our problem, we start with the general solution \( f(x) = C e^{ax} \). However, this general solution contains an unknown constant \( C \).
To determine \( C \), we apply the initial condition \( f(x_0) = k \). This condition notes the function's value at a certain point, providing the necessary information to solve for \( C \).

• Substitute \( x = x_0 \) into the general form to establish \( f(x_0) = C e^{ax_0} = k \).
• Solving for \( C \), we get \( C = k e^{-ax_0} \). This changes the general solution into a specific one tailored to the initial condition given.

As a result, initial conditions help us connect theory to reality by pinpointing specific solutions.

General Solution

The general solution in differential equations like \( f'(x) = a f(x) \) is a formula that represents the entire family of potential solutions. It embodies the sum of all possible outcomes based on varying constants, like \( C \), which we need to determine through the initial conditions.
In the example problem, our general solution starts as \( f(x) = C e^{ax} \). Once we incorporate initial conditions, we modify this to fit particular scenarios.

• By applying the calculated constant \( C = k e^{-ax_0} \), we insert it back into the general formula, resulting in \( f(x) = k e^{a(x-x_0)} \).
• This specific form of the function satisfies both the initial condition and the differential equation.

The beauty of this process lies in the conversion from an abstract formula to a concrete solution, tying broad mathematical theory to specific factual outcomes.