Problem 57 - Applied Mathematics: For the Managerial, Life, and Social Sciences

Question

Find \(f(x)\) by solving the initial value problem. $$f^{\prime}(x)=\frac{x+1}{x} ; f(1)=1$$

Step-by-Step Solution

Answer

The specific solution to the initial value problem is \(f(x) = x + (\ln |x|)\).

Step 1: Integrate the differential equation

Integrate both sides of the equation with respect to \(x\): $$\int f^{\prime}(x) dx = \int \frac{x+1}{x} dx$$ On the left side, we have: $$ \int f^{\prime}(x) dx = f(x) + C_1$$ On the right side, notice that the integral can be separated into two simpler integrals: $$\int \frac{x+1}{x} dx = \int \frac{x}{x} dx + \int \frac{1}{x} dx = \int 1 dx + \int x^{-1} dx $$ Now, integrate these two integrals individually: $$ \int 1 dx + \int x^{-1} dx = x + (\ln |x|) + C_2$$ So, the general solution of the differential equation is: $$f(x) = x + (\ln |x|) + C$$ where \(C = C_1 - C_2\). #Step 2: Apply the initial condition to find the specific solution# We have the initial condition \(f(1) = 1\). We will use this condition to find the value of the constant \(C\).

Step 2: Substituting the initial condition

Substitute the initial condition into the general solution: $$1 = 1 + (\ln |1|) + C$$ $$1 = 1 + 0 + C$$ Solve for \(C\): $$C=0$$ Thus, the specific solution to the initial value problem is:

Step 3: Final Solution

$$f(x) = x + (\ln |x|)$$

Key Concepts

Differential Equations

Differential equations are mathematical equations that describe the relationship between a function and its derivatives. They are a cornerstone of modeling real-world phenomena in physics, engineering, biology, and economics. In our example, the problem presents a first-order differential equation, \( f'(x) = \frac{x + 1}{x} \), which involves the first derivative of the unknown function \( f(x) \). The goal is to find \( f(x) \) that satisfies this equation.

To solve such an equation, we look for an antiderivative: a function whose derivative gives us the function in the equation. This leads us into the realm of integration calculus, an indispensable tool in finding solutions to differential equations.

When we integrate both sides of a first-order differential equation, we obtain what is known as the general solution, which contains an arbitrary constant. To pin down a specific solution, an initial condition is typically provided. In our case, \( f(1) = 1 \) serves this purpose. Through integration and applying the initial condition, we can find the particular solution that fits the scenario in question.

Integration Calculus

Integration calculus is about finding the integral of functions, which is the opposite operation of taking derivatives. It allows us to find the accumulated quantity, like area under a curve, when given a rate of change. The initial value problem presents an opportunity to apply integration to find the function \( f(x) \).

In our example, we start by integrating the right-hand side of the differential equation, \( \frac{x + 1}{x} \), which can be broken down into simpler parts as \( \int 1 dx \) and \( \int x^{-1} dx \). Here, we use basic integration rules to find that \( \int 1 dx = x \) and \( \int x^{-1} dx = \ln |x| \)—the latter being an integral leading to the natural logarithm function, which is frequently encountered in integration calculus.

By combining these integrals, we construct a general solution to the differential equation. However, without the initial condition, we'd only have a family of potential functions, differentiated by various constants. The initial condition provides us with a specific value that we use to determine the exact solution.

Natural Logarithm

The natural logarithm, denoted as \( \ln \), is the inverse function of the exponential function \( e^x \). In calculus, \( \ln \) comes up often, particularly when integrating the reciprocal of \( x \) as seen in our example, \( \int \frac{1}{x} dx = \ln |x| \). It's important to remember the absolute value, as \( \ln \) is only defined for positive arguments.

The natural logarithm has unique properties that make it a vital tool in solving differential equations. For instance, the derivative of \( \ln x \) is \( \frac{1}{x} \) and, conversely, the integral that leads to \( \ln x \) arises frequently in problems involving growth and decay processes, such as interest rates and population dynamics.

In the step-by-step solution, we see the natural logarithm emerge naturally from the integration process. When the initial condition is applied to the general solution, the constant \( C \) is found to be zero, indicating that the specific solution to our initial value problem doesn't require any adjustments from the natural logarithm.