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Chapter 9: Problem 3

Does the function \(y(t)=2 t\) satisfy the differential equation \(y^{\prime \prime \prime}(t)+y^{\prime}(t)=2 ?\)

Short Answer

Expert verified
Answer: Yes, the function \(y(t) = 2t\) does satisfy the given differential equation \(y^{\prime\prime\prime}(t) + y^{\prime}(t) = 2\).

Step by step solution

01

Identify the given function and the differential equation

We have a function \(y(t) = 2t\) and the differential equation: \(y^{\prime\prime\prime}(t) + y^{\prime}(t) = 2\).
02

First derivative of y(t)

We need to find the first derivative of the given function \(y(t)\) with respect to \(t\). The first derivative is \(y^{\prime}(t) = \frac{dy}{dt} = 2\).
03

Second derivative of y(t)

Now, we need to find the second derivative of \(y(t)\) with respect to \(t\). Since the first derivative is a constant, the second derivative will be 0: \(y^{\prime\prime}(t) = \frac{d^2y}{dt^2} = 0\).
04

Third derivative of y(t)

Similarly, as the second derivative is 0 (a constant), the third derivative will also be 0: \(y^{\prime\prime\prime}(t) = \frac{d^3y}{dt^3} = 0\).
05

Substitute the derivatives into the differential equation

Now we will substitute these derivatives into the given differential equation to check if it holds. The differential equation is \(y^{\prime\prime\prime}(t) + y^{\prime}(t) = 2\). Substituting the derivatives, we get: \(0 + 2 = 2\).
06

Check the equation for equality

As we can see, the differential equation holds since the equation \(0 + 2 = 2\) is true. Hence, the function \(y(t) = 2t\) does satisfy the given differential equation \(y^{\prime\prime\prime}(t) + y^{\prime}(t) = 2\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Differential Equation
When we talk about a differential equation, we're looking at a fundamental concept in calculus that deals with the relationship between a function and its derivatives. In simple terms, a differential equation is an equation that contains one or more functions and their derivatives. These equations describe a wide range of phenomena in fields like physics, engineering, and economics.

They come in many varieties, such as ordinary differential equations (ODEs) which have one independent variable and partial differential equations (PDEs) that involve multiple independent variables. The goal when working with a differential equation is typically to find a function or set of functions that satisfy the equation.

In our example, the function in question is of the simpler form, an ODE, where the function is defined as: \( y(t) = 2t \), and the differential equation is \( y^{\text{'''}}(t) + y'(t) = 2 \). We're seeking to determine if our given function solves this particular equation.
Derivatives
A derivative represents the rate at which a function is changing at any point. It's one of the core concepts in calculus and underpins many of the principles governing physical laws like motion and growth rates. For students first encountering calculus, it's essential to grasp that derivatives provide a means of measuring instantaneous change, similar to calculating speed or acceleration for a moving object.

The derivative process can be thought of as determining the slope of the tangent line at any point along a curve on a graph of a function. The first derivative tells you about the function's velocity, that is, how fast the y-value (output) of the function is changing with respect to the x-value (input). When solving calculus problems, finding the derivative is often a crucial step in unraveling more complex relationships between variables.
Third Derivative
Regarding third derivatives, we venture deeper into the layers of calculus. While the first derivative measures the rate of change, and the second derivative gives us information on the curvature or acceleration of the function, the third derivative, which is less commonly discussed, can provide insight into the rate at which the acceleration itself is changing. This is sometimes referred to as the 'jerk' in physics.

In the problem we are examining, the third derivative is specifically asked for in the differential equation provided. Computing this higher-order derivative, especially when it leads to zero, simplifies the equation significantly as it eliminates the term that includes it. Whenever you encounter a problem involving third derivatives or higher, it's a sign that the problem may involve complex motion or changing forces.
Calculus Problems
Solving calculus problems can be a step-by-step dance with functions and derivatives. These problems often involve finding the maximum or minimum values of functions, understanding the motion of objects, or optimizing various outcomes. Each problem requires a strategic approach to dissecting the information given and applying appropriate calculus techniques to find a solution.

In the context of differential equations, the challenge is to identify the correct function that satisfies all parts of the equation. This can involve finding derivatives, like in our exercise example with the function \( y(t) = 2t \), and then substituting them into the equation to verify if the function is indeed a solution. It's essential to be meticulous with each differentiation and algebra step, as calculus problems are often a test of both conceptual understanding and procedural fluency.

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Most popular questions from this chapter

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