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Calculus is a fundamental branch of mathematics that focuses on change. In calculus, we analyze how things change over time or in different conditions. There are two main types of calculus: differential calculus and integral calculus. Differential calculus, which is the most relevant to differential equations, revolves around understanding how a function changes as its inputs change. Calculus is essential in a variety of fields from physics to economics, as it provides the tools needed for modeling and solving problems involving dynamic systems.

In the context of differential equations, which are equations that relate a function to its derivatives, calculus helps us understand the relationship between changing quantities. Calculus allows us to solve these equations to find unknown functions based on provided conditions.

For the exercise we are discussing, knowing how to calculate derivatives using calculus is key to verifying whether a function is a solution to a given differential equation. The process involves finding the derivatives of the function, which reveals the rules by which changes in the function are related to changes in its inputs.

Derivatives are a core concept in calculus, representing the rate at which a function is changing at any given point. If you think of a function as describing some kind of motion, the derivative tells us the speed or rate of that motion at each point in time.

To explore derivatives practically, consider the function given in the exercise: \(u(t) = C_{1}t^2 + C_{2}t^3\).

• **First Derivative**: This helps us find the instantaneous rate of change of the function, calculated as \(u'(t) = 2C_{1}t + 3C_{2}t^2\). It's like finding the speed of a car at a particular moment.


• **Second Derivative**: This is the derivative of the derivative, or the rate of change of the rate of change. It's found to be \(u''(t) = 2C_{1} + 6C_{2}t\). In our car analogy, this would represent the acceleration – how the speed is changing over time.

Being comfortable with finding derivatives is crucial for verifying solutions to differential equations, as it allows us to substitute these derivatives back into the equation to see if it holds true.

Verification of solutions is the process of proving whether the function you have is a solution to a differential equation. This involves substituting the function, its first derivative, and its second derivative back into the equation and simplifying the terms.

Here, the given function \(u(t) = C_{1}t^2 + C_{2}t^3\) and its derivatives \(u'(t) = 2C_{1}t + 3C_{2}t^2\) and \(u''(t) = 2C_{1} + 6C_{2}t\) are placed into the differential equation: \(t^2 u''(t) - 4t u'(t) + 6 u(t) = 0\).

• Substitute each derivative and the function into the equation.
• Simplify each component by distributing terms and combining like terms.
• Finally, verify if the left side of the equation simplifies to zero, confirming the function satisfies the differential equation.

In this problem, simplification showed that all terms canceled out, leaving zero on both sides of the equation. Because the equation holds true, we verify that the function is indeed a solution to the differential equation. Understanding how to perform these steps accurately allows us to validate solutions effectively.