91Ó°ÊÓ

Differential equations are equations that involve a function and its derivatives. Solving them often means finding a function that satisfies the equation given its derivatives. In particular, first-order differential equations involve only the first derivative of a function.

The problem provided involves a first-order linear differential equation with the form \(\frac{dQ}{dt} = 0.3Q - 120\). The goal is to find the function \(Q(t)\) that meets this equation.

Separation of variables is a technique used to solve such equations. By manipulating the equation so that all terms involving the function \(Q\) are on one side and all terms involving \(t\) are on the other, we can integrate each side separately. This allows us to solve for the function that satisfies the differential relationship between \(Q\) and \(t\).

Integration is a fundamental tool in solving differential equations, especially when separation of variables is used. Once the differential equation is separated, each side of the equation can be integrated in respect to its own variable.

For this problem, after rearranging, the left side becomes \(\int \frac{dQ}{0.3Q - 120}\) and the right side becomes \(\int dt\). The integration of the left side involves a substitution. By letting \(u = 0.3Q - 120\), we simplify the integration to get \(\frac{1}{0.3}\ln|u|\).

The right side integrates straightforwardly to \(t + C\), where \(C\) is a constant of integration. After integrating both sides, the two expressions are equated, which is a crucial step to find the general solution before applying any initial conditions.

Initial conditions are information given at a specific point which allows us to find a particular solution to a differential equation. Without this, we'd only have a family of solutions. The problem states that when \(t=0\), \(Q=50\).

After solving the general solution \(|0.3Q - 120| = C_1e^{0.3t}\), we need the initial condition to find the specific constant \(C_1\).

Substituting the initial values into the solved equation provides a concrete number for \(C_1\). In our case, substituting \(Q=50\) and \(t=0\) lets us solve for \(-105\) as \(C_1\). This irregular rightward shift personalizes the solution for this particular scenario, and by substituting \(C_1\) back, we obtain the specific solution \(Q(t) = -350e^{0.3t} + 400\).

Initial conditions thus help bridge general solutions to specific real-world scenarios.