Question

Verifying general solutions Verify that the given function is a solution of the differential equation that follows it. Assume \(C, C_{1}, C_{2}\) and \(C_{3}\) are arbitrary constants. $$y(t)=C e^{-5 t} ; y^{\prime}(t)+5 y(t)=0$$

Short answer

Question: Verify whether the given function \(y(t) = Ce^{-5t}\) is a solution to the differential equation \(y'(t) + 5y(t) = 0\). Answer: The given function \(y(t) = Ce^{-5t}\) is a solution to the differential equation \(y'(t) + 5y(t) = 0\).

Step-by-step solution

Calculate the derivative of the given function

To calculate the derivative of the given function \(y(t) = Ce^{-5t}\) with respect to \(t\), we'll differentiate the function using the chain rule. Let's compute the derivative: $$y'(t) = \frac{d}{dt}(Ce^{-5t})$$

Apply the chain rule

Applying the chain rule, we get: $$y'(t) = C(-5e^{-5t})$$ $$y'(t) = -5Ce^{-5t}$$

Substitute the function and its derivative into the differential equation

Now, we will substitute the given function \(y(t)\) and its derivative \(y'(t)\) into the given differential equation to check if it holds true: \(y'(t) + 5y(t) = 0\) Substitute \(y(t) = Ce^{-5t}\), and \(y'(t) = -5Ce^{-5t}\): $$(-5Ce^{-5t}) + 5(Ce^{-5t}) = 0$$

Simplify and check if the equation holds true

Simplify the equation and check if it holds true: $$(-5Ce^{-5t}) + 5(Ce^{-5t}) = 0$$ $$Ce^{-5t}(-5+5) = 0$$ $$0=0$$ Since both sides of the equation are equal, the given function is indeed a solution to the given differential equation.

Key concepts

Chain Rule

The chain rule is a fundamental tool in calculus, particularly when dealing with differential equations involving composite functions. It allows us to calculate the derivative of a function that is the combination of two or more functions.

In the context of the exercise, we use the chain rule to differentiate the function \(y(t) = Ce^{-5t}\). Here, the outer function is the exponential function \(e^u\), and the inner function is the linear function \(u=-5t\). The chain rule tells us to first take the derivative of the outer function with respect to the inner function, \(u\), and then multiply it by the derivative of the inner function with respect to \(t\). The formula for the chain rule is expressed as follows:

• \( \frac{dy}{dt} = \frac{dy}{du} \cdot \frac{du}{dt} \)

By applying the chain rule, we find the derivative of \(y(t)\), ensuring each step is carefully understood, enhancing our ability to handle similar problems in the future.

Derivative Calculation

Derivative calculation lies at the heart of differential equations, where we determine the rate at which a function changes. When we compute the derivative of \(y(t) = Ce^{-5t}\), we're looking for \(y'(t)\), the rate of change of \(y\) with respect to \(t\).

The step-by-step solution outlines the use of the chain rule to find \(y'(t)\), resulting in \(y'(t) = -5Ce^{-5t}\). This calculation is critical, as it provides us with a building block for verifying that our function \(y(t)\) is indeed a solution to the differential equation. Proper understanding and execution of derivative calculation are essential, as it applies to a wide range of mathematical and real-world problems.

Verifying Solutions

Verifying solutions is a verification process where we check if a given function satisfies a differential equation. For the exercise in question, we are required to confirm whether \(y(t) = Ce^{-5t}\) is a solution to \(y'(t) + 5y(t) = 0\).

Verification is completed by substituting the function and its derivative into the original differential equation and simplifying. The verification step is crucial as it either confirms the correctness of the found solution or highlights an error in the preceding steps. This practice is not just a mathematical formality but a vital part of the learning process, encouraging students to scrutinize and confirm their results at each stage of problem-solving.

Exponential Functions

Exponential functions, such as \(y(t) = Ce^{-5t}\) used in the exercise, are pervasive in various scientific fields, including biology, physics, and finance. These functions are characterized by a constant base raised to a variable exponent.

The function \(y(t) = Ce^{-5t}\) represents exponential decay, which is a process that decreases over time at a rate proportional to its current value. Understanding exponential functions is imperative as they are integral to modeling real-world phenomena. The exercise provided requires an acquaintance with the properties of exponential functions, particularly their derivatives, to solve differential equations effectively.