91Ó°ÊÓ

The rate of change in mathematics refers to how a quantity changes over time, and it is a fundamental concept in differential equations. In our exercise, the rate of change of substance A decomposing into substance B is given by the differential equation \(\frac{dy}{dt} = -ky^2\). This equation states that the speed at which substance A decomposes is proportional to the square of the amount of substance A present at any given time t.

Here, \( \frac{dy}{dt} \) represents the rate of change of substance A with respect to time, and \( k \) is a proportionality constant. By understanding the rate of change, we can predict how long it will take for substance A to decompose completely, given its initial quantity.

Separation of variables is a powerful technique used to solve differential equations where variables can be separated on different sides of the equation. In our example, the differential equation \( \frac{dy}{dt} = -ky^2\), we can rearrange this equation to group all y terms on one side and all t terms on the other side.

This process leads to:
\( \frac{1}{y^2} dy = -k \, dt \).
Once we have separated the variables, we can proceed to integrate each side of the equation. This method is incredibly useful for solving first-order differential equations and offers a straightforward path to finding the solution.

Integration is the process of finding the antiderivative or the integral of a function. In our differential equation, after separating the variables, we integrate both sides to find the solution. On integration we get:
\( \begin{aligned} \text{Left side:} \ \frac{1}{y^2} dy &\to -\frac{1}{y} \ \text{Right side:} \ \to -kt + C \ \text{Combining both:} \ -\frac{1}{y} = -kt + C \ \text{Rearrange to solve for y:} \ y = \frac{1}{kt + C} \end{aligned} \).
Here, we solved the integration to express the amount of substance A in terms of time t. The integration constant C can be determined using initial conditions.