91Ó°ÊÓ

Differential equations are equations that involve derivatives of a function. They express a relationship between the function itself and its rates of change. In our exercise, the differential equation is given by \( \frac{dy}{dt} = 2 \sqrt{1-y^{2}} \), where \( y \) is a function of \( t \). Here, \( \frac{dy}{dt} \) represents the rate at which \( y \) changes with respect to \( t \).
When solving a differential equation, our goal is to find the unknown function, which, in this case, would be \( y = \sin(2t) \). The solution function must satisfy the original equation for all values within a given domain.

• The differential equation defines a dynamic process.
• The function \( y \) is evaluated at various points \( t \), called the solution.

The significance of understanding differential equations is immense as they are used to model real-world phenomena like motion, heat, growth, and more.

Solution Verification is a critical step in ensuring that a proposed solution to a differential equation is correct. It involves checking whether the function that you propose as a solution actually satisfies the differential equation when substituted back into it.

In the exercise, we needed to confirm that \( y = \sin(2t) \) is a valid solution for the given differential equation. This was done by first finding the derivative of the function, which was \( \frac{dy}{dt} = 2\cos(2t) \).

Next, substitute \( \cos(2t) \) back into the differential equation \( \frac{dy}{dt} = 2 \sqrt{1 - y^2} \) and check if both sides match:

• With \( y = \sin(2t) \), calculate \( \sqrt{1-y^2} = \cos(2t) \).
• The substitution yields the same expression on both sides: \( 2\cos(2t) = 2\cos(2t) \).

This step assures us that the chosen function satisfies the differential relationship posed by the equation.

Initial conditions are crucial in uniquely determining the solution to a differential equation. They specify the value of the solution function at a particular point, meaning they provide additional information that "anchors" the solution to a specific pathway among potentially many.

For the exercise, the initial condition was given as \( y(0) = 0 \). It asks whether the function \( y = \sin(2t) \) passes through the specified point at \( t = 0 \). To verify this:

• Substitute \( t = 0 \) into \( y = \sin(2t) \).
• Calculate: \( y(0) = \sin(2 \times 0) = \sin(0) = 0 \).

The calculation shows that the proposed solution satisfies the initial condition. This means that \( y = \sin(2t) \) is not only a solution to the differential equation but also adheres to the specific initial value given. Initial conditions narrow down the general solution to a unique solution related to the initial scenario of the problem.