Question

Finding general solutions Find the general solution of each differential equation. Use \(C, C_{1}, C_{2}, \ldots\) to denote arbitrary constants. $$v^{\prime}(t)=\frac{4}{t^{2}-4}$$

Short answer

The general solution to the given differential equation is: $$v(t) = \ln\left|\frac{t-2}{t+2}\right| + C$$

Step-by-step solution

Identify the given differential equation

The given differential equation is: $$v'(t) = \frac{4}{t^2 - 4}$$

Solve the differential equation

Integrate both sides of the equation with respect to \(t\): $$\int v'(t) dt = \int \frac{4}{t^2 - 4} dt$$

Perform the integration

We can solve the right-hand side of the equation using partial fraction decomposition. The factorization for the denominator is: $$t^2 - 4 = (t-2)(t+2)$$ Using partial fraction decomposition, we can rewrite the integrand as: $$\frac{4}{(t-2)(t+2)} = \frac{A}{t-2} + \frac{B}{t+2}$$ To find the values of A and B, multiply both sides of the equation by the denominator \((t-2)(t+2)\): $$4 = A(t+2) + B(t-2)$$ Setting \(t = 2\): $$4 = 4A \implies A = 1$$ Setting \(t = -2\): $$4 = -4B \implies B = -1$$ Now substitute the values of A and B back into the integrand: $$\frac{4}{(t-2)(t+2)} = \frac{1}{t-2} - \frac{1}{t+2}$$ Now, integrate the right-hand side of the equation with respect to \(t\): $$\int v'(t) dt = \int \left(\frac{1}{t-2} - \frac{1}{t+2}\right) dt$$ $$v(t) = \ln|t-2| - \ln|t+2| + C$$

Write the general solution

Now we can write the general solution for the given differential equation as: $$v(t) = \ln\left|\frac{t-2}{t+2}\right| + C$$ The general solution to the given differential equation is: $$v(t) = \ln\left|\frac{t-2}{t+2}\right| + C$$

Key concepts

Partial Fraction Decomposition

Partial fraction decomposition is a useful algebraic method for breaking down complex rational expressions. When you have a fraction where the denominator can be factored into simpler polynomials, partial fraction decomposition allows you to express this fraction as a sum of simpler fractions. This is particularly helpful when integrating expressions of this type.

Let's consider the expression \( \frac{4}{t^2 - 4} \). Notice that the denominator can be factored as \( (t-2)(t+2) \). Using partial fraction decomposition, we aim to express this as \( \frac{A}{t-2} + \frac{B}{t+2} \).

To find the values of \( A \) and \( B \), multiply the entire equation by the original denominator \((t-2)(t+2)\) to eliminate the fractions, resulting in \(4 = A(t+2) + B(t-2)\). By substituting convenient values for \( t \) (like \( t=2 \) and \( t=-2 \)), we can solve for \( A \) and \( B \) easily. This technique simplifies the process of integration and is a powerful tool when solving differential equations.

Integration Techniques

When faced with an integral, such as \( \int \frac{4}{t^2 - 4} \, dt \), knowing different integration techniques can be crucial. After using partial fraction decomposition, the integrand becomes two simpler functions: \( \int \left( \frac{1}{t-2} - \frac{1}{t+2} \right) \, dt \). Each term now involves a basic natural logarithm integration.

The integral \( \int \frac{1}{t-2} \, dt \) results in \( \ln|t-2| \), and similarly, \( \int \frac{1}{t+2} \, dt \) leads to \( \ln|t+2| \). This approach brings us to a point where we can comfortably find the solution by applying basic integration rules.

Mastering these techniques is crucial for anyone dealing with complex integrals, as they greatly simplify the calculation and allow us to find the solution to differential equations efficiently.

General Solution

The general solution of a differential equation not only solves the equation but also includes all possible solutions. In the context of the given exercise, the general solution is the expression \( v(t) = \ln\left|\frac{t-2}{t+2}\right| + C \). This form includes the constant \( C \), representing arbitrary constants resulting from the integration process.

In solving initial value problems, this constant becomes crucial as it adjusts the solution to meet initial conditions or specific scenarios. For general solutions, however, it signifies that the solution encompasses a family of functions.

Understanding the nature of general solutions is essential when studying differential equations, as it explains how different conditions can affect the particular solution of a problem. Always remember to check for domain restrictions based on the initial equation to ensure that the solution is valid across the intended range of values.