Question

Find the general solution of the following differential equations. $$v^{\prime}(t)=\frac{4}{t^{2}-4}$$

Short answer

Question: Find the general solution of the differential equation \(v^{\prime}(t)=\frac{4}{t^{2}-4}\). Answer: The general solution of the given differential equation is \(v(t) = \ln\left|\frac{t+2}{t-2}\right| + C\), where \(C\) is an arbitrary constant.

Step-by-step solution

Identify the given differential equation

We are given the first-order ordinary differential equation (ODE): $$v^{\prime}(t)=\frac{4}{t^{2}-4}$$ where \(v^{\prime}(t)\) represents the derivative of \(v(t)\) with respect to \(t\). Our task is to find the function \(v(t)\).

Integrate both sides of the equation

Integrating both sides of the equation with respect to \(t\), we get: $$\int v^{\prime}(t) dt = \int \frac{4}{t^{2}-4} dt$$

Perform the integration using the partial fraction decomposition method

To solve the right-hand side integral, we can use the partial fraction decomposition method. We can rewrite the integrand as follows: $$\frac{4}{t^{2}-4} = \frac{A}{t-2}+\frac{B}{t+2}$$ Multiplying both sides by \(t^2-4\), we get: $$4=A(t+2)+B(t-2)$$ Solving for A and B by equating coefficients, we get \(A=1\) and \(B=1\). Now, we can write the right-hand side integral as: $$\int v^{\prime}(t) dt = \int \left(\frac{1}{t-2}+\frac{1}{t+2}\right) dt$$

Complete the integration

Integrating each term separately, we get: $$v(t) + C = \int \left(\frac{1}{t-2}+\frac{1}{t+2}\right) dt = \ln|t-2| + \ln|t+2| + C$$ Rearranging the terms and combining the logarithms, we get: $$v(t) = \ln\left|\frac{t+2}{t-2}\right| + C$$

Write the general solution

The general solution of the given differential equation is: $$v(t) = \ln\left|\frac{t+2}{t-2}\right| + C$$ where \(C\) is an arbitrary constant.

Key concepts

Ordinary Differential Equation

An ordinary differential equation (ODE) is an equation that involves a function and its derivatives. The primary goal when dealing with an ODE is to find a function that satisfies this relationship. ODEs can be found in many forms and can represent a variety of real-world processes such as the cooling of an object or the decay of a radioactive material.

In our example, we are given a first-order ODE: \[ v^{\prime}(t) = \frac{4}{t^{2}-4} \] Here, the term \(v^{\prime}(t)\) denotes the first derivative of \(v(t)\) with respect to \(t\). First-order indicates that the highest derivative is the first one, which often makes the problem simpler compared to higher-order differential equations. Our task is to express \(v(t)\), the function we are looking for.

Partial Fraction Decomposition

Partial fraction decomposition is a method used to break down complex rational expressions into simpler fractions that are easier to integrate. This technique is particularly helpful when working with rational functions in the context of integration.

In the solution to our exercise, we apply this method to the fraction: \[ \frac{4}{t^{2}-4} \] Our goal is to express it in terms of simpler fractions which we can then integrate easily: \[ \frac{4}{t^{2}-4} = \frac{A}{t-2} + \frac{B}{t+2} \] Upon solving, we find that \(A = 1\) and \(B = 1\). These simpler fractions are advantageous as they align well with standard integral forms, helping us proceed with the integration process smoothly.

Integration

Integration is a fundamental concept in calculus that revolves around finding the antiderivative of a function. It serves to solve differential equations by reversing the differentiation process, thus recovering the original function from its derivative.

In step 3 of our solution, after employing partial fraction decomposition, the expression is ready for integration. We individually integrate each fraction, leading to:\[ \int \left( \frac{1}{t-2} + \frac{1}{t+2} \right) dt \] These integrals evaluate to natural logarithms: \[ \ln|t-2| + \ln|t+2| + C \] Here, \(C\) is known as the constant of integration, representing any constant value that may have been lost during the differentiation process.

The integration step is crucial to find the general solution, effectively wrapping the process of solving this ODE.

First-Order Differential Equations

First-order differential equations involve functions and their first derivatives. They are often easier to solve compared to higher-order equations because they involve fewer complications. Understanding how to handle first-order differential equations is an essential skill when studying calculus and differential equations.

The given equation in our exercise is a classic example: \[ v^{\prime}(t) = \frac{4}{t^{2}-4} \] Here, we start by integrating both sides to move from the derivative to the antiderivative (original function). Once decomposed and integrated, this method leads us directly to the general solution. The particular simplicity of first-order equations lies in this straightforward approach that often only requires elementary calculus techniques like integration. Hence, mastering first-order differential equations creates a foundation for tackling more advanced topics in mathematics.