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Chapter 3: Problem 28

The figure shows the graph of the solution of an initial value problem \(y^{\prime \prime}+a y^{\prime}+b y=0, y(0)=y_{0}, y^{\prime}(0)=y_{0}^{\prime}\). Use the information given to express the solution in the form \(y(t)=R \cos (\beta t-\delta)\), where \(0 \leq \delta<2 \pi\). Determine the constants \(a, b, y_{0}\). and \(y_{0^{-}}^{r}\) The graph has a maximum value at \((\pi / 12,1)\) and a \(t\)-intercept at \((5 \pi / 12,0)\).

Short Answer

Expert verified
Question: Express the solution in the form \(y(t) = R \cos(\beta t - \delta)\) and determine the constants \(a\), \(b\), \(y_0\), and \(y'_0\). Solution: The solution can be expressed as \(y(t) = 1 \cos(24t)\), and the constants are \(a=0\), \(b=576\), \(y_0=1\), and \(y'_0=0\).

Step by step solution

01

Identify the maximum value

The graph has a maximum value at \((\frac{\pi}{12}, 1)\). This means that when \(t = \frac{\pi}{12}\), \(y(t) = R\cos(\beta t - \delta)\) reaches its maximum value. Since the cosine function takes its maximum value at 1, we have: $$R\cos(\beta \cdot \frac{\pi}{12} - \delta) = 1$$
02

Identify the \(t\)-intercept

We are also given that the graph has a \(t\)-intercept at \((\frac{5 \pi}{12}, 0)\). This means that when \(t = \frac{5 \pi}{12}\), \(y(t)=0\). So we have: $$R\cos(\beta \cdot \frac{5 \pi}{12} - \delta) = 0$$
03

Determine \(R\), \(\beta\), and \(\delta\)

From Step 1, since \(R\cos(\beta\cdot\frac{\pi}{12} - \delta) = 1\), we can deduce that: $$\cos(\beta\cdot\frac{\pi}{12} - \delta) = 1/R$$ Thus, \(\beta\cdot\frac{\pi}{12} - \delta = 2 \pi k\) for some integer \(k\) because the cosine function has a period of \(2\pi\). From Step 2, since \(R\cos(\beta\cdot\frac{5 \pi}{12} - \delta) = 0\), we can deduce that: $$\cos(\beta\cdot\frac{5 \pi}{12} - \delta) = 0$$ Thus, \(\beta\cdot\frac{5 \pi}{12} - \delta = (2k + 1)\frac{\pi}{2}\) for some integer \(k\) because the cosine function has zeros at odd multiples of \(\frac{\pi}{2}\). Now we have a system of equations involving \(\beta\), \(\delta\), and \(k\): 1. \(\beta\cdot\frac{\pi}{12} - \delta = 2 \pi k\) 2. \(\beta\cdot\frac{5 \pi}{12} - \delta = (2k + 1)\frac{\pi}{2}\) We can solve for \(\beta\) and \(\delta\): $$\beta = \frac{2\pi}{\pi / 12} = 24$$ $$\delta = 2 \pi k - \beta\cdot\frac{\pi}{12} = - 46 \pi k$$ To keep \(\delta\) in the range \([0, 2\pi)\), we can choose \(k = 0\). So \(\delta = 0\). Finally, from Step 1, we get \(R = 1\).
04

Determine \(a\), \(b\), \(y_0\) and \(y'_0\)

Since we have found \(R = 1\), \(\beta = 24\), and \(\delta = 0\), the solution can be written as: $$y(t) = \cos(24t)$$ Now we can find the constants \(a\), \(b\), \(y_0\), and \(y'_0\): The first derivative of \(y(t)\) is: $$y'(t) = -24\sin(24t)$$ The second derivative of \(y(t)\) is: $$y''(t) = -576\cos(24t)$$ Plugging these into the given differential equation, \(y''+ay'+by=0\), we get: $$-576\cos(24t) + a(-24\sin(24t)) + b\cos(24t) = 0$$ Comparing the coefficients of \(\cos(24t)\) and \(\sin(24t)\) terms on both sides, we find: $$b - 576 = 0 \Rightarrow b = 576$$ $$-24a = 0 \Rightarrow a = 0$$ Lastly, we can find \(y_0 = y(0)\) and \(y'_0 = y'(0)\): $$y_0 = \cos(24\cdot 0) = 1$$ $$y'_0 = -24\sin(24\cdot 0) = 0$$ Hence, the constants are \(a=0\), \(b=576\), \(y_0=1\), and \(y'_0=0\).

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Key Concepts

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

Differential Equations
Differential equations are mathematical equations that describe the relationship between a function and its derivatives. These equations are fundamental in expressing physical laws and phenomena where the rate of change of a quantity is related to the quantity itself.

In the context of the problem at hand, we are dealing with a second-order linear homogeneous differential equation with constant coefficients, denoted as \(y''+ay'+by=0\). Here, \(y''\) represents the second derivative of the function \(y(t)\) with respect to time \(t\), \(a\) and \(b\) are constants, and \(y'\) is the first derivative or the rate at which \(y\) changes with time.

The solution to such an equation often involves trigonometric functions like sine and cosine or exponential functions, depending on the discriminant related to the coefficients of \(a\) and \(b\). These solutions describe wave-like or oscillatory motion, as is common in physics and engineering.
Cosine Function
The cosine function is a periodic trigonometric function that describes the horizontal coordinate of a point on the unit circle as it moves around the circle. It has a fundamental role in modeling oscillatory phenomena because of its inherent periodicity and wave-like shape.

The general form of a cosine function, as used in the exercise, is \(y(t) = R \cos(\beta t - \delta)\), where \(R\) is the amplitude, \(\beta\) is the frequency (which determines how many complete cycles occur in a unit of time), and \(\delta\) is the phase shift that determines the horizontal translation of the wave.

Understanding the cosine function's properties, such as its periodicity, amplitude, and phase shift, is essential in solving the initial value problem where the values of \(R\), \(\beta\), and \(\delta\) are to be determined based on given conditions, maximum values, and \(t\)-intercepts.
Boundary Value Problems
Boundary value problems involve finding solutions to differential equations that meet certain specified conditions, called boundary conditions. Unlike initial value problems that provide information at a single point, boundary value problems involve conditions at more than one point, often at the boundaries of the domain of interest.

In our scenario, although the problem is termed as an initial value problem, in practice we are given both an initial condition and additional information akin to a boundary condition: the graph's maximum value and a specific zero point, which the solution must pass through.

Boundary value problems are crucial in various fields such as physics, engineering, and finance for modeling steady-state conditions or the behavior of systems over an interval. Solving such a problem involves ensuring the solution satisfies the given equation and adheres to the given boundaries or conditions outlined.

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Most popular questions from this chapter

Concavity of the Solution Curve In the discussion of direction fields in Section 1.3, you saw how the differential equation defines the slope of the solution curve at a point in the ty-plane. In particular, given the initial value problem \(y^{\prime}=f(t, y), y\left(t_{0}\right)=\) \(y_{0}\), the slope of the solution curve at initial condition point \(\left(t_{0}, y_{0}\right)\) is \(y^{\prime}\left(t_{0}\right)=f\left(t_{0}, y_{0}\right)\). In like manner, a second order equation provides direct information about the concavity of the solution curve. Given the initial value problem \(y^{\prime \prime}=f\left(t, y, y^{\prime}\right), y\left(t_{0}\right)=\) \(y_{0}, y^{\prime}\left(t_{0}\right)=y_{0}^{\prime}\), it follows that the concavity of the solution curve at the initial condition point \(\left(t_{0}, y_{0}\right)\) is \(y^{\prime \prime}\left(t_{0}\right)=f\left(t_{0}, y_{0}, y_{0}^{\prime}\right)\). (What is the slope of the solution curve at that point?) Consider the four graphs shown. Each graph displays a portion of the solution of one of the four initial value problems given. Match each graph with the appropriate initial value problem. (a) \(y^{\prime \prime}+y=2-\sin t, \quad y(0)=1, \quad y^{\prime}(0)=-1\) (b) \(y^{\prime \prime}+y=-2 t, \quad y(0)=1, \quad y^{\prime}(0)=-1\) (c) \(y^{\prime \prime}-y=t^{2}, \quad y(0)=1, \quad y^{\prime}(0)=1\) (d) \(y^{\prime \prime}-y=-2 \cos t, \quad y(0)=1, \quad y^{\prime}(0)=1\)

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

Let \(y(t)\) denote the solution of the given initial value problem. Is it possible for the corresponding limit to hold? Explain your answer. $$y^{\prime \prime}+\frac{1}{t^{2}-16} y=0, \quad y(0)=1, \quad y^{\prime}(0)=1, \quad \lim _{t \rightarrow 3^{-}} y(t)=+\infty$$

(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? $$ y^{\prime \prime}+2 y^{\prime}+y=0, \quad y(1)=1, \quad y^{\prime}(1)=0 $$

(a) Find 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 \(y(t)\) as \(t \rightarrow-\infty\) and as \(t \rightarrow \infty\). Does \(y(t)\) approach \(-\infty,+\infty\), or a finite limit? $$y^{\prime \prime}-y=0, \quad y(0)=1, \quad y^{\prime}(0)=-1$$
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