91Ó°ÊÓ

Separation of variables is a crucial technique used to solve differential equations. It involves rearranging the equation so that each variable appears on a different side of the equation. This process simplifies solving, as each variable can be integrated independently.
For example, consider the given differential equation \( \frac{dy}{dt} = y^4 \). By separating variables, we rewrite it as \( y^{-4} \, dy = dt \).
This separation allows us to deal with the equation using integration. Here, the equation is arranged such that all terms involving \( y \) are on one side and those involving \( t \) are on the other. This is a necessary step in making these equations more manageable to solve.

Integration is a mathematical operation that is used to find a function, known as the antiderivative, from its derivative. After separating variables in a differential equation, integration is the next logical step to determine the general solution.
Each side of our separated equation, \( \int y^{-4} \, dy = \int dt \), is integrated with respect to its own variable. For \( \int y^{-4} \, dy \), the antiderivative is \(-\frac{1}{3}y^{-3} \). For \( \int dt \), it results in \( t + C \), where \( C \) represents the constant of integration.
Through integration, we find the general form of the solution to the differential equation, which in this case is \(-\frac{1}{3}y^{-3} = t + C \). Integration is vital as it transforms a rate of change into a comprehensible function.

Initial conditions are specific values provided in a problem that allow us to solve for any constants in the general solution. They personalize the solution to match the context of a given situation.
In our exercise, the initial condition given is \( y = 1 \) when \( t = 0 \). By substituting these values into the general solution, \(-\frac{1}{3}(1)^{-3} = 0 + C \), we obtain \( 1 = -3C \). Solving this yields \( C = -\frac{1}{3} \).
These initial conditions are crucial in converting a general solution into a particular solution, which precisely aligns with real-world scenarios.

The particular solution of a differential equation is the specific solution obtained after satisfying the initial conditions. Once we determine the constant \( C \), we substitute it back into the general solution.
For the given problem, by substituting \( C = -\frac{1}{3} \) into the general solution, we attain the particular solution \( y = \left(-3t + 1\right)^{-1/3} \).
This provides a definitive answer that uniquely meets the initial condition \( y = 1 \) when \( t = 0 \). The role of the particular solution is to ensure the solution's applicability to the specific scenario described.