91Ó°ÊÓ

A differential equation is an equation that involves an unknown function and its derivatives. In our context, the differential equation is given by \(\frac{dy}{dx} + \frac{y}{3} = 0\). This particular form is called a first-order linear differential equation because it involves the first derivative \(\frac{dy}{dx}\) and the function \(y\) raised to the first power. The goal is to solve for the function \(y\) that satisfies this relationship for all values of \(x\).
To solve this differential equation, we use a technique called **separation of variables**. This involves rearranging the equation so that all terms involving \(y\) are on one side, and all terms involving \(x\) are on the other. This simplifies the problem by allowing each variable to be integrated separately. In our case, we rearranged the equation to \(\frac{dy}{y} = -\frac{1}{3} dx\). This form directly leads to the next step of integration.

In solving differential equations, especially when seeking particular solutions, initial conditions provide specific information about the unknown function at a particular point. This helps determine the constant of integration that appears after integrating. In the given exercise, the initial condition is specified as \(y(0) = 10\).
What this means is that when \(x = 0\), the value of \(y\) must be 10. After finding the general solution, which involves an arbitrary constant, applying this initial condition allows us to solve for that constant. As such, without the initial condition, we could have a family of solutions, but knowing \(y(0) = 10\) allows us to pin down the exact solution. Thus, when substituting \(x = 0\) into \(y = A e^{-\frac{x}{3}}\), we find that \(A = 10\).
This is a crucial step because it turns the general solution into a specific solution tailored to the problem's requirements.

Integration is a mathematical process that essentially reverses differentiation. When we separate variables in a differential equation, we often end up with an expression that requires integration on both sides. In our case, the separated equation \(\frac{dy}{y} = -\frac{1}{3} dx\) involves integrating both sides to find \(y\) as a function of \(x\).
For the left side, integrating \(\int \frac{1}{y} \, dy\) gives \(\ln|y|\). On the right side, \(\int -\frac{1}{3} \, dx\) results in \(-\frac{x}{3} + C\). Here, \(C\) is the constant of integration that accounts for any constant that could be added during the integration process. These integrated results are then set equal to each other: \(\ln|y| = -\frac{x}{3} + C\).
Integration is crucial because it transforms a derivative-based equation into one involving the original function, allowing us to further manipulate it, as seen in the next step with exponentiation.

Exponentiation is the mathematical operation that reverses taking a logarithm. After integrating the differential equation, we found \(\ln|y| = -\frac{x}{3} + C\). To solve for \(y\), we need to eliminate the natural logarithm. This is done by exponentiating both sides of the equation.
When we exponentiate, we have \(|y| = e^{-rac{x}{3} + C}\). This simplifies to \(y = A e^{-rac{x}{3}}\), where \(A = e^C\) because \(e\) raised to a constant power is itself another constant. At this point, \(A\) represents an undetermined constant, which can be positive or negative based on the initial conditions. By applying the initial condition \(y(0) = 10\), we determine that \(A = 10\), leading to the specific solution \(y = 10 e^{-rac{x}{3}}\).
Exponentiation is vital in this context because it allows us to transition from a logarithmic form back to a standard algebraic expression for the function \(y\). This step completes the process of solving the differential equation.