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Derivatives are a fundamental concept in calculus, representing the rate of change or slope of a function. They are essential for understanding how functions behave, especially when they change with respect to another variable. The derivative of a function at a point tells us how the function's value is changing at that very point. For example, in the function \( y = 5 + 3e^{kx} \), we differentiate it with respect to \( x \) to find how \( y \) changes as \( x \) changes.

To calculate the derivative of \( y \), we look at each part of the function:

• The constant part, 5, has a derivative of 0 since constants do not change.
• The exponential part, \( 3e^{kx} \), follows the rule that \( \frac{d}{dx}[e^{u}] = u'e^{u} \), with \( u = kx \) and \( u' = k \).
• Therefore, \( \frac{d}{dx}[3e^{kx}] = 3ke^{kx} \).

Combining these results gives us the derivative \( \frac{dy}{dx} = 3ke^{kx} \). This expression indicates how the value of \( y \) shifts as \( x \) changes.

Exponential functions are special mathematical functions where a constant base is raised to a variable exponent. They are often seen in scenarios where something grows or decays exponentially, such as populations or radioactive decay.

In the equation given, \( y = 5 + 3e^{kx} \), the term \( e^{kx} \) is an exponential function. Its properties include:

• Rapid growth or decay, depending on the sign of \( k \). If \( k > 0 \), the function grows as \( x \) increases; if \( k < 0 \), it decays.
• The constant \( e \) is approximately equal to 2.71828 and is a base of the natural logarithms, making it crucial for many natural processes.

Understanding exponential functions like \( e^{kx} \) is essential because they frequently appear in real-world applications including finance, physics, and statistics. They also allow us to model complex systems with simple mathematical expressions.

Solving equations is a process of finding the values of variables that satisfy given mathematical statements. In the context of differential equations, our goal is to identify parameters or functions that satisfy both the derivative and the original equation.

For the differential equation \( \frac{dy}{dx} = 10 - 2y \), we substituted the derivative \( \frac{dy}{dx} = 3ke^{kx} \).

After simplification, the resulting equation is \( 3ke^{kx} = -6e^{kx} \). Solving this involves:

• Recognizing that since \( e^{kx} \) is non-zero, we can divide both sides by \( e^{kx} \) without losing any solutions.
• This leaves us with the linear equation \( 3k = -6 \).
• Solving for \( k \) gives us \( k = -2 \).

The value \( k = -2 \) indicates that the original function \( y = 5 + 3e^{kx} \) will satisfy the differential equation for this specific value of \( k \). This value tells us how the exponential part needs to behave to ensure the entire function meets the rate of change specified by the differential equation.