91Ó°ÊÓ

A first order ordinary differential equation is a relationship involving an unknown function, its derivative, and an independent variable. To clarify with the textbook problem presented, the equation given is \(f^{\textprime}(s)=6 s-8 s^{3}\). Here, the unknown function is \(f\), and \(s\) is the independent variable.

The degree of the derivative in the equation indicates it's a first order differential equation, as it involves the first derivative \(f^{\textprime}(s)\) but no higher derivatives. The goal when solving such an equation is to find the original function \(f(s)\) that satisfies the equation across its domain. This is typically done through integration, a fundamental operation in calculus, which is the reverse process of differentiation.

The process of integration of functions is essential when solving differential equations. In the context of our exercise, the integration of the function \(6s - 8s^3\) with respect to \(s\) allows us to find the antiderivative, which in this case, is the original function \(f(s)\).

The integral \(\int (6s - 8s^3) ds\) is calculated using basic rules of integration. For polynomial terms like \(s\) and \(s^3\), we increase the exponent by one and divide by the new exponent. Through integration, we've found the general solution \(f(s) = 3s^2 - 2s^4 + C\), where \(C\) is the constant of integration. The constant \(C\) represents an infinite number of solutions that make the antiderivative true, and to pinpoint a specific solution, we need an initial condition.

The initial condition in differential equations is a point through which the solution curve must pass. It allows us to find the unique solution that satisfies both the differential equation and the condition.

In our exercise, we're given the initial condition \(f(2) = 3\), meaning that when \(s = 2\), the value of \(f(s)\) is 3. Applying the initial condition to the general solution \(f(s) = 3s^2 - 2s^4 + C\), we solve for the constant \(C\), using the values \(s = 2\) and \(f(s) = 3\). This provides a specific value for \(C\), which in this case turns out to be \(C = -5\). With \(C\) determined, we can state the complete unique solution \(f(s) = 3s^2 - 2s^4 - 5\), satisfying both the differential equation and the initial condition.