91Ó°ÊÓ

An Ordinary Differential Equation (ODE) involves functions and their derivatives. In simpler terms, an ODE is an equation that relates a function of one variable to its derivatives. These equations are crucial in describing various phenomena in physics, engineering, biology, and other fields. For instance, the change in population over time or the velocity of a moving object can often be modeled using ODEs.
When solving an ODE, the primary goal is to find a function that satisfies the equation. This function is often dependent on an independent variable, typically denoted as \( x \). For example, in the given exercise, the ODE is \( y^{\prime}(x)=3 x^{2}-3 x^{-4} \). It tells us how the function \( y \) changes as \( x \) changes.
One key aspect of solving ODEs is that they can usually be classified by their order, which is determined by the highest derivative involved. In our case, \( y^{\prime} \) indicates a first-order ODE, as it involves the first derivative of \( y \). First-order ODEs are often more straightforward to solve than higher-order ODEs.

Integration is a fundamental tool in mathematics that helps in determining the antiderivatives of functions, which are essential in solving differential equations. In simple terms, integration is the process of "adding up" tiny changes to determine the total change over an interval.
In the context of an ODE, integration is used to find the general solution by undoing the differentiation. When we integrate \( y'(x) = 3x^2 - 3x^{-4} \), we aim to find a function \( y(x) \) such that its derivative matches the right-hand side of the equation. Hence, the integral of the right-hand side gives us:

• \( \int 3x^2 \ dx \) yields \( x^3 \)
• \( \int -3x^{-4} \ dx \) results in \( \frac{-1}{x^3} \)

Thus, the integrated form, which is the general solution, becomes \( y(x) = x^3 - \frac{1}{x^3} + C \). Here, \( C \) stands as an integration constant and represents an infinite family of solutions related to different initial conditions.

After finding the general solution of a differential equation, we often aim to find a specific, or particular, solution that satisfies a given initial condition. This process involves pinpointing the exact variation of the function that matches the initial state of the system described by the ODE.
In an initial value problem, the initial condition provides the value of the function at a specific point to determine the constant, \( C \). In our exercise, the condition given is \( y(1) = 0 \). Applying this to the general solution \( y(x) = x^3 - \frac{1}{x^3} + C \), we substitute \( x = 1 \) and solve for \( C \):

• \( 0 = 1^3 - \frac{1}{1^3} + C \)
• The equation simplifies to \( 0 = 1 - 1 + C \), which leads to \( C = 0 \)

Finally, substituting \( C = 0 \) back into the general solution converts it into the particular solution: \( y(x) = x^3 - \frac{1}{x^3} \). This solution satisfies both the differential equation and the initial condition provided.