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Rewrite the function \(y(t)\) in the form \(y(t)=R e^{a t} \cos (\beta t-\delta)\), where \(0 \leq \delta<2 \pi\). Use this representation to sketch a graph of the given function, on a domain sufficiently large to display its main features. $$ y(t)=e^{-2 t} \cos 2 t-e^{-2 t} \sin 2 t $$

Short Answer

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Question: Rewrite the given function \(y(t) = e^{-2t} \cos 2t - e^{-2t} \sin 2t\) in the form \(y(t) = R e^{at}\cos(\beta t - \delta)\) and sketch its graph, highlighting the main features such as the amplitude, frequency, and phase shift. Answer: After rewriting, the function is \(y(t) = e^{-2t}\cos(2t)\) with amplitude given by \(e^{-2t}\), frequency of \(2\), and no phase shift (\(\delta = 0\)). To sketch the graph, start at an amplitude of 1, have two oscillations within each time period of \(2\pi\), and let the amplitude decay exponentially as \(t\) increases. The graph will oscillate between positive and negative values, with the peaks getting closer to the horizontal axis as \(t\) increases.

Step by step solution

01

Rewrite the function using complex exponentials

We will use the property \(e^{i\theta} = \cos \theta + i\sin \theta\), which allows us to express the given function as a complex exponential. $$ y(t) = e^{-2t} \cos 2t - e^{-2t} \sin 2t = e^{-2t}( \cos 2t - i\sin 2t ) $$ Now, using the property \(e^{i\theta} = \cos \theta + i\sin \theta\) again, we can rewrite the expression inside the parentheses in terms of another complex exponential: $$ \cos 2t - i\sin 2t = e^{-i 2t} $$ Now the function becomes: $$ y(t) = e^{-2t}e^{-i2t} $$
02

Express the function in the desired form

Combine the exponentials and multiply them by a constant \(R\) that takes care of the initial amplitude: $$ y(t) = R e^{(-2 - i2)t} $$ We see that our function now has the form \(y(t) = R e^{at}\cos(\beta t - \delta)\) with \(a=-2\), \(\beta=2\), and \(\delta = 0\). So, $$ y(t) = R e^{-2t}\cos(2t) $$ Comparing this expression to our original function, we see that \(R=1\). Thus, our final rewritten function is: $$ y(t) = e^{-2t}\cos(2t) $$
03

Sketch the graph

To sketch the graph, first, identify the main features of the function: - Amplitude: The function amplitude varies as \(e^{-2t}\). This means the amplitude starts at 1 when \(t = 0\) and decays exponentially as \(t\) increases. - Frequency: The oscillatory part of the function is given by \(\cos(2t)\). The frequency is \(2\), which means the function completes two cycles in every time period of \(2\pi\). - Phase Shift: There is no phase shift as \(\delta = 0\). Based on these features, sketch the graph of the function \(y(t)\) starting at an amplitude of 1, with two oscillations within each time period of \(2\pi\), while the amplitude decays exponentially. The graph will oscillate between positive and negative values, with the peaks getting closer to the horizontal axis as \(t\) increases.

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Key Concepts

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

Complex Exponentials
Complex exponentials are a beautiful bridge between trigonometric functions and exponential functions. They are based on Euler's formula: \(e^{i\theta} = \cos \theta + i \sin \theta\). This formula tells us that we can express a cosine function and a sine function by using a complex exponential.
In the original problem, we used this idea to simplify the given function \(y(t) = e^{-2t} \cos 2t - e^{-2t} \sin 2t\). By combining the cosine and sine terms using their expressions in terms of complex exponentials, this results in a format of \(e^{at}\).
  • Complex exponentials can simplify mathematical expressions significantly.
  • They help in converting trigonometric identities into more manageable forms.
  • This conversion makes it easier to analyze the behavior of a function especially in the context of differential equations.
Through the process above, our original function was transformed into the simpler exponential form \(y(t) = e^{-2t} e^{-i2t}\).
Amplitude and Frequency
When dealing with functions of the form \(R e^{at} \cos(\beta t - \delta)\), it's important to understand the role of amplitude and frequency. Amplitude describes the peak value of the oscillations, while frequency refers to how often these oscillations occur in a given time period.
In the rewritten function \(y(t) = e^{-2t} \cos(2t)\), let's explore these concepts further:
  • Amplitude: Given by \(e^{-2t}\), the amplitude of this function decreases over time, due to the negative exponent in the exponential term.
  • Frequency: Determined by the \(\cos(2t)\), the frequency is 2. This means there are two complete oscillations within each \(2\pi\) time period.
The amplitude decay is exponential, contributing to the damping of the oscillations. This means the waves get smaller as time progresses. However, the frequency remains constant at 2 cycles per \(2\pi\) interval, causing the oscillations' pace to remain steady despite reducing in size.
Graph Sketching
Graph sketching helps visualize the behavior of a function over time. For the function \(y(t) = e^{-2t} \cos(2t)\), sketching is especially useful to see the changes in amplitude and the frequency of oscillations.
To sketch the graph, follow these steps:
  • Start with Amplitude: As \(t\) starts at 0, the amplitude is 1, because \(e^{-2 \cdot 0} = 1\). This is the highest point on the graph.
  • Observe Decay: As \(t\) increases, note how the amplitude decreases exponentially, causing peaks in the graph to get lower.
  • Track the Frequency: The graph will complete two oscillations within each \(2\pi\) interval because of the frequency \(\cos(2t)\).
Putting these observations together, the sketch should start at \(t = 0\) with an initial peak at 1, followed by oscillations which decrease over time. The graph will pass through the \(x\)-axis as it oscillates, gradually tapering off towards it as amplitude diminishes. This exponential decay reflects the damping effect from \(e^{-2t}\), illustrating the shrinking oscillations in sync with the persistent frequence of the cosine function.

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Most popular questions from this chapter

Rectilinear Motion with a Drag Force In Chapter 2, we considered rectilinear motion in the presence of a drag force proportional to velocity. We solved the first order linear equation for velocity and antidifferentiated the solution to obtain distance as a function of time. We now consider directly the second order linear differential equation for the distance function. A particle of mass \(m\) moves along the \(x\)-axis and is acted upon by a drag force proportional to its velocity. The drag constant is denoted by \(k\). If \(x(t)\) represents the particle position at time \(t\), Newton's law of motion leads to the differential equation \(m x^{\prime \prime}(t)=-k x^{\prime}(t)\). (a) Obtain the general solution of this second order linear differential equation. (b) Solve the initial value problem if \(x(0)=x_{0}\) and \(x^{\prime}(0)=v_{0}\). (c) What is \(\lim _{t \rightarrow \infty} x(t)\) ?

Consider the initial value problem \(t^{2} y^{\prime \prime}-t y^{\prime}+y=0, y(1)=1, y^{\prime}(1)=1\). (a) What is the largest interval on which Theorem \(3.1\) guarantees the existence of a unique solution? (b) Show by direct substitution that the function \(y(t)=t\) is the unique solution of this initial value problem. What is the interval on which this solution actually exists? (c) Does this example contradict the assertion of Theorem \(3.1\) ? Explain.

The general solution of the nonhomogeneous differential equation \(y^{\prime \prime}+\alpha y^{\prime}+\beta y=g(t)\) is given, where \(c_{1}\) and \(c_{2}\) are arbitrary constants. Determine the constants \(\alpha\) and \(\beta\) and the function \(g(t)\). $$y(t)=c_{1} e^{t}+c_{2} e^{2 t}+2 e^{-2 t}$$

For each differential equation, (a) Find the complementary solution. (b) Find a particular solution. (c) Formulate the general solution. $$ y^{\prime \prime \prime}+y=t^{3} $$

(a) Obtain the general solution of the differential equation. (b) Impose the initial conditions to obtain the unique solution of the initial value problem. (c) Describe the behavior of the solution as \(t \rightarrow-\infty\) and \(t \rightarrow \infty\). In each case, does \(y(t)\) approach \(-\infty,+\infty\), or a finite limit? $$ y^{\prime \prime}-5 y^{\prime}+6.25 y=0, \quad y(-2)=0, \quad y^{\prime}(-2)=1 $$

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