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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 \(y(t) = 2t\) satisfies the given differential equation.

Step by step solution

01

Find the first derivative of y(t)

To find the first derivative of the function \(y(t) = 2t\), we simply apply the power rule: \(y^{\prime}(t) = \frac{d(2t)}{dt} = 2\)
02

Find the second derivative of y(t)

Now that we have the first derivative, we can find the second derivative by taking the derivative of the first derivative: \(y^{\prime \prime}(t) = \frac{d(2)}{dt} = 0\)
03

Find the third derivative of y(t)

Next, we will find the third derivative by taking the derivative of the second derivative: \(y^{\prime \prime \prime}(t) = \frac{d(0)}{dt} = 0\)
04

Substitute the derivatives into the differential equation

Now, we will substitute the derivatives we found in Steps 1-3 into the given differential equation: \(y^{\prime \prime \prime}(t) + y^{\prime}(t) = 2\) Plug in the values: \(0 + 2 = 2\)
05

Analyze the result

Now that we have substituted the derivatives, we can determine whether the given function satisfies the differential equation: \(0 + 2 = 2\) Since this equation is true, the function \(y(t) = 2t\) does indeed satisfy the 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.

Derivatives
In calculus, a derivative represents the rate at which a function is changing at any given point. It's like looking at the function with a magnifying glass and seeing its slope or steepness at that particular point. In mathematical terms, if you have a function such as \( y(t) \), its derivative is denoted as \( y'(t) \) or \( \frac{dy}{dt} \). In simple terms, it's how much \( y \) changes as \( t \) changes slightly. This change is called the derivative.
Working with derivatives is essential in solving differential equations, calculating velocities, and understanding how things vary. When we talk about higher-order derivatives like the second and third, we are basically looking at the derivative of a derivative, exploring how the rate of change itself changes as well. Each derivative gives us more insight into the behavior of the original function.
When exploring derivatives, always remember tools like the power rule or product rule, which help streamline the process of finding derivatives of more complex functions.
Power Rule
The power rule is one of the most straightforward techniques in calculus for finding the derivative of a function that is expressed in a power form, such as \( x^n \). It states that the derivative of \( x^n \) is \( nx^{n-1} \). This rule simplifies the process of differentiation immensely.
For instance, consider the function \( y(t) = 2t \). To find its derivative using the power rule, you simply take the exponent of \( t \) (which is 1 here), multiply by the constant (2), and decrease the exponent by one. This leads to \( 2 \times t^{1-1} = 2 \times t^0 = 2 \).
Always keep in mind that the power rule applies straightforwardly to polynomials. For more complex functions, it may need to be combined with other rules like the product or chain rule. Nonetheless, it's a fundamental tool that simplifies many differentiation tasks.
First Derivative
The first derivative of a function gives a lot of insight into the function's behavior. It tells you how the function is growing or shrinking at any point. For example, for the function \( y(t) = 2t \), we used the power rule to find that \( y'(t) = 2 \). This means that the function is increasing at a constant rate of 2, regardless of the value of \( t \).
Finding the first derivative is often the first step in solving many problems in physics, engineering, and anywhere the rate of change is important. It's the foundational building block for analyzing dynamic systems.
Whenever you calculate a first derivative, check the overall trend it suggests in the original function. Is it increasing or decreasing? Such insights can then be used for predicting future values and understanding past behavior.
Second Derivative
The second derivative digs even deeper into a function, showing how the rate of change of the rate of change happens. In other words, it gives information about the function's concavity or curvature at a given point. A positive second derivative suggests a concave upwards curve, like a smile, indicating the rate of change is increasing. Conversely, a negative second derivative indicates a concave downwards curve, like a frown.
In our example with the function \( y(t) = 2t \), the second derivative \( y''(t) = 0 \) signifies no change in the rate of change. This means the original function is linear, without any bends or changes in its steepness, just a straight line.
The second derivative is crucial in understanding acceleration in physics or optimizing functions to find minima or maxima (like finding the top of a hill or the bottom of a valley). Mastery of this concept can provide comprehensive insights into the intricacies of how changes evolve over time.

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Most popular questions from this chapter

Widely used models for population growth involve the logistic equation \(P^{\prime}(t)=r P\left(1-\frac{P}{K}\right),\) where \(P(t)\) is the population, for \(t \geq 0,\) and \(r>0\) and \(K>0\) are given constants. a. Verify by substitution that the general solution of the equation is \(P(t)=\frac{K}{1+C e^{-n}},\) where \(C\) is an arbitrary constant. b. Find that value of \(C\) that corresponds to the initial condition \(P(0)=50\). c. Graph the solution for \(P(0)=50, r=0.1,\) and \(K=300\). d. Find \(\lim _{t \rightarrow \infty} P(t)\) and check that the result is consistent with the graph in part (c).

a. Show that the logistic growth rate function \(f(P)=r P\left(1-\frac{P}{K}\right)\) has a maximum value of \(\frac{r K}{4}\) at the point \(P=\frac{K}{2}\) b. Show that the Gompertz growth rate function \(f(M)=-r M \ln \left(\frac{M}{K}\right)\) has a maximum value of \(\frac{r K}{e}\) at the point \(M=\frac{K}{e}\)

a. Show that for general positive values of \(R, V, C_{i},\) and \(m_{0},\) the solution of the initial value problem $$m^{\prime}(t)=-\frac{R}{V} m(t)+C_{i} R, \quad m(0)=m_{0}$$ is \(m(t)=\left(m_{0}-C_{i} V\right) e^{-R t / V}+C_{i} V\) b. Verify that \(m(0)=m_{0}\) c. Evaluate \(\lim m(t)\) and give a physical interpretation of the result. d. Suppose \(^{t} \vec{m}_{0}^{\infty}\) and \(V\) are fixed. Describe the effect of increasing \(R\) on the graph of the solution.

Solve the differential equation for Newton's Law of Cooling to find the temperature in the following cases. Then answer any additional questions. An iron rod is removed from a blacksmith's forge at a temperature of \(900^{\circ} \mathrm{C}\). Assume that \(k=0.02\) and the rod cools in a room with a temperature of \(30^{\circ} \mathrm{C}\). When does the temperature of the rod reach \(100^{\circ} \mathrm{C} ?\)

Solution of the logistic equation Use separation of variables to show that the solution of the initial value problem $$P^{\prime}(t)=r P\left(1-\frac{P}{K}\right), \quad P(0)=P_{0}$$ $$\text { is } P(t)=\frac{K}{\left(\frac{K}{P_{0}}-1\right) e^{-n}+1}$$

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