91Ó°ÊÓ

A differential equation is an equation that involves the derivatives of a function. These equations describe how a particular quantity changes with respect to another quantity, often time. They play a fundamental role in mathematics, as they model real-world phenomena such as motion, growth, decay, and much more.
In the given problem, we have a first-order differential equation denoted as \( \frac{dy}{dt} = \frac{2t+1}{2y-2} \). This equation means that the rate of change of \( y \) with respect to \( t \) (time) is dependent on both \( t \) and \( y \) itself. The goal is to find the function \( y(t) \) that satisfies this relationship.
Working through these problems involves specific methods, and one of the common techniques used is separation of variables. This approach entails manipulating the equation to bring similar quantities to one side, facilitating easier integration to find \( y \). In our example, we achieve this by cross-multiplying components of the equation.

Integration is a fundamental concept in calculus that is used to calculate areas, volumes, and to solve differential equations. In differential equations like our example, integration allows us to find the original functions when their derivatives are known.
After separating variables, the equation becomes \( (2y-2) \, dy = (2t+1) \, dt \). The next step is to integrate each side independently.

• For the left side: \( \int (2y-2) \, dy \) simplifies to \( y^2 - 2y + C_1 \), where \( C_1 \) is a constant of integration.
• For the right side: \( \int (2t+1) \, dt \) simplifies to \( t^2 + t + C_2 \).

Once integrated, we combine the constants, leading to the equation \( y^2 - 2y = t^2 + t + C \), where \( C = C_2 - C_1 \). Solving these equations via integration is critical in predicting the behavior of a system over time.

An initial value problem involves finding a function that satisfies a differential equation and fulfills initial conditions. Initial conditions specify the value of the function at a particular point, giving clarity to the solution's trajectory and making the solution unique.
In our scenario, we are given the initial condition \( y(0) = -1 \). This means when \( t = 0 \), \( y \) must equal \(-1\). We apply this condition after integrating the separated variables, leading to \((y^2 - 2y) = (t^2 + t + C)\).
Substituting \( t = 0 \) and \( y = -1 \) into the integrated equation, we find \( C \). As calculated, \( (-1)^2 - 2(-1) \) simplifies to 3, which gives us \( C = 3 \). Incorporating this value back helps us find a solution that is consistent with the initial conditions.
Solving for \( y \) using the quadratic formula leads to two potential solutions. However, we choose the branch that satisfies both the quadratic and initial conditions, resulting in the explicit function \( y = 1 - \sqrt{t^2 + t + 4} \). Initial value problems are crucial in ensuring that mathematically derived solutions are applicable to specific situations.