91Ó°ÊÓ

Differential equations are a type of equation that relate a function to its derivatives. They are often used to model real-world phenomena where change is involved, such as physics, biology, and engineering.

In this exercise, we deal with a first-order differential equation which involves the derivative \( \frac{d y}{d t} \) of a dependent variable \( y \) with respect to an independent variable \( t \). The equation given is \( \frac{d y}{d t} = \frac{y}{3+t} \). By separating variables, we can rearrange the equation to isolate all \( y \)-related terms on one side and \( t \)-related terms on the other. This simplifies the problem, enabling us to solve it by integration.

Separation of variables is a fundamental technique for solving simple types of differential equations and works well when the equation can be reorganized to contain functions solely of \( y \) and \( t \). It forms the basis for many methods in solving differential equations.

Integration is a key mathematical tool used to find antiderivatives of functions, effectively reversing the process of differentiation. When we integrate both sides of a differential equation, we're finding an expression for one variable in terms of another. In this solution:

• The equation is separated as \( \int \frac{dy}{y} = \int \frac{dt}{3+t} \).
• We perform integration on both sides. The left side integrates to \( \ln |y| \), and the right side to \( \ln |3+t| + C \), where \( C \) is the integration constant.

The constant of integration, \( C \), represents the family of all possible solutions to the equation. Applying the initial conditions later helps us pin down the specific solution that fits our scenario. Integrals of natural logarithms are particularly helpful in simplifying the expressions when dealing with multiplicative separations in variables.

Initial conditions are extra pieces of information given alongside differential equations to find a specific solution that fits the problem context. They help to determine the value of the integration constant \( C \) by providing a known value of \( y \) at a certain \( t \).

In our exercise, the initial condition is \( y(0) = 1 \). After integrating the equation, we use this condition to replace \( t \) with 0 and \( y \) with 1 in the equation \( |y| = k |3+t| \).

This yields \( 1 = k \times 3 \), allowing us to solve for \( k = \frac{1}{3} \). Thus, equipping us with a specific function \( y(t) = \frac{1}{3} (3 + t) \) that satisfies both the differential equation and the initial condition.

The natural logarithm, denoted by \( \ln \), is fundamental in calculus, specifically for solving differential equations involving exponential functions. It plays a crucial role in our solution:

• The integration of \( \frac{1}{y} \) results in \( \ln |y| \).
• The integration of \( \frac{1}{3+t} \) gives \( \ln |3+t| \).

After integration, the solution involves the equation \( \ln |y| = \ln |3 + t| + C \). When both sides are exponentiated, the natural logarithms disappear because the exponential function is the inverse of the natural logarithm. This helps isolate \( y \) in terms of \( t \), leading us to the expression \( |y| = |3+t| e^C \) or \( y = \frac{1}{3}(3+t) \) after determining \( C \).

Understanding how logarithms transform back and forth with exponentials is key to solving such equations effectively.