/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 29 (a) Find two familiar functions ... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

(a) Find two familiar functions that solve the equation \(d^{2} y / d t^{2}=-y\). Which one starts with \(y(0)=1\) and \(y^{\prime}(0)=0\) ? (b) This second-order equation \(y^{\prime \prime}=-y\) produces a vector equation \(u^{\prime}=A u\) : $$ u=\left[\begin{array}{l} y \\ y^{\prime} \end{array}\right] \quad \frac{d u}{d t}=\left[\begin{array}{l} y^{\prime} \\ y^{\prime \prime} \end{array}\right]=\left[\begin{array}{rr} 0 & 1 \\ -1 & 0 \end{array}\right]\left[\begin{array}{l} y \\ y^{\prime} \end{array}\right]=A u $$ Put \(y(t)\) from part (a) into \(u(t)=\left(y, y^{\prime}\right)\). This solves Problem 6 again.

Short Answer

Expert verified
(a) Solutions are \( \cos(t) \) and \( \sin(t) \); \( y(t) = \cos(t) \) meets \( y(0)=1 \), \( y'(0)=0 \). (b) \( u(t) = \begin{bmatrix} \cos(t) \\ -\sin(t) \end{bmatrix} \).

Step by step solution

01

Identify Solutions to the Differential Equation

The equation given is \( \frac{d^2 y}{dt^2} = -y \). This is a second-order differential equation, and its familiar solutions are the sine and cosine functions. These functions satisfy the equation: if \( y(t) = \cos(t) \) or \( y(t) = \sin(t) \), then the second derivative \( \frac{d^2 y}{dt^2} = -y(t) \).
02

Apply Initial Conditions to Find Specific Function

We are given the initial conditions \( y(0) = 1 \) and \( y'(0) = 0 \). If we use \( y(t) = \cos(t) \), this satisfies \( y(0) = \cos(0) = 1 \). Also, the derivative \( y'(t) = -\sin(t) \) gives \( y'(0) = -\sin(0) = 0 \). Hence, \( y(t) = \cos(t) \) satisfies these conditions.
03

Construct the Vector Equation using Matrix Multiplication

Given the system \( u = \begin{bmatrix} y \ y' \end{bmatrix} \) and \( \frac{du}{dt} = \begin{bmatrix} y' \ y'' \end{bmatrix} \), we see the matrix \( A = \begin{bmatrix} 0 & 1 \ -1 & 0 \end{bmatrix} \) describes the system of equations. Applying \( y(t) = \cos(t) \), we find that \( u(t) = \begin{bmatrix} \cos(t) \ -\sin(t) \end{bmatrix} \).
04

Verify the Solution Satisfies the Differential Equation

Substitute \( u(t) = \begin{bmatrix} \cos(t) \ -\sin(t) \end{bmatrix} \) into \( \frac{du}{dt} = A u \). This gives us \( \begin{bmatrix} -\sin(t) \ -\cos(t) \end{bmatrix} = \begin{bmatrix} 0 & 1 \ -1 & 0 \end{bmatrix} \begin{bmatrix} \cos(t) \ -\sin(t) \end{bmatrix} \), which verifies because setting \( u'(t) = A u(t) \) results in the correct expressions. Each calculated derivative matches the matrix multiplication, confirming the solution.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Second-order differential equations
Second-order differential equations involve derivatives up to the second degree and are essential in modeling various physical phenomena. The given equation, \( \frac{d^2 y}{dt^2} = -y \), is a simple harmonic oscillator, a type of second-order linear differential equation. In these equations, the second derivative of a function is directly related to the function itself. This characteristic leads to periodic solutions, such as sine and cosine functions, that frequently arise in physics and engineering.
When solving second-order differential equations like this, we often encounter familiar forms like \( y(t) = A \cos(t) + B \sin(t) \), where \( A \) and \( B \) are constants determined by initial conditions. Finding particular solutions involves checking if the function and its derivatives satisfy the original equation. Here, both \( y(t) = \cos(t) \) and \( y(t) = \sin(t) \) satisfy \( \frac{d^2 y}{dt^2} = -y \), proving them as valid solutions without needing further adjustment beyond initial conditions.
Matrix representation
Matrix representation simplifies differential equations, especially useful for systems of equations. For the equation \( y'' = -y \), a convenient vector form is used: \( u = \begin{bmatrix} y \ y' \end{bmatrix} \). This transforms the differential equation into a system of first-order differential equations \( \frac{du}{dt} = Au \), where
\( A = \begin{bmatrix} 0 & 1 \ -1 & 0 \end{bmatrix} \), a matrix encapsulating the relationships between \( y \) and its derivative.
Matrix \( A \) relates the time derivative of \( u \), \( \frac{du}{dt} \), to the current state \( u \). This representation is especially powerful in numerical computations and theoretical analysis, as it allows leveraging linear algebra techniques to solve differential equations. The notation \( u'(t) = Au(t) \) provides a clear path to verifying solutions by plugging in familiar functions and their derivatives into the equation form to ensure left and right sides match.
Initial conditions
Initial conditions are critical in differential equations as they help pin down a specific solution from a family of possibilities. For the problem at hand, the initial conditions are \( y(0) = 1 \) and \( y'(0) = 0 \). These give us a starting point that helps determine the constants \( A \) and \( B \) in the general solution \( y(t) = A \cos(t) + B \sin(t) \).
Applying the initial conditions, we can see that \( y(t) = \cos(t) \) fits \( y(0) = 1 \) because \( \cos(0) = 1 \) and \( y'(t) = -\sin(t) \) gives \( y'(0) = 0 \) as \( -\sin(0) = 0 \). Initial conditions not only specify parameters of the solution but also ensure the uniqueness of the solution for the specific scenario described.
Sine and cosine functions
Sine and cosine functions are fundamental in periodic motion descriptions, such as oscillating systems modeled by second-order differential equations. In this context, these functions serve as solutions because the second derivative of sine or cosine returns the negative of the original function, satisfying the harmonic oscillator equation \( \frac{d^2 y}{dt^2} = -y \).
Specifically, \( \cos(t) \) and \( \sin(t) \) naturally arise in equations modeling waveforms and oscillations due to this unique property. Their periodic nature makes them excellent candidates for representing repeated cycles of motion or periodic signals. Understanding their role in differential equations gives insight into solving problems related to mechanical vibrations, electrical circuits, and even quantum mechanics.
  • Both functions are solutions to the differential equation and play a role in everything from simple pendulum movements to complex electronic circuits.
  • They provide the basis for Fourier analysis, which decomposes complex signals into sinusoidal components.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

From their trace and determinant, at what time \(t\) do the following matrices change between stable with real eigenvalues, stable with complex eigenvalues, and unstable? $$ A_{1}=\left[\begin{array}{ll} 1 & -1 \\ t & -1 \end{array}\right], \quad A_{2}=\left[\begin{array}{cc} 0 & 4-t \\ 1 & -2 \end{array}\right] . \quad A_{3}=\left[\begin{array}{rr} t & -1 \\ 1 & t \end{array}\right] $$

The eigenvalues of \(A\) are 1 and 9 , the eigenvalues of \(B\) are \(-1\) and 9 : $$ A=\left[\begin{array}{ll} 5 & 4 \\ 4 & 5 \end{array}\right] \quad \text { and } \quad B=\left[\begin{array}{ll} 4 & 5 \\ 5 & 4 \end{array}\right] $$ Find a matrix square root of \(A\) from \(R=S \sqrt{\Lambda} S^{-1}\). Why is there no real matrix square root of \(B\) ?

The identity transformation takes every vector to itself: \(T x=x\). Find the corresponding matrix, if the first basis is \(v_{1}=(1,2), v_{2}=(3,4)\) and the second basis is \(w_{1}=(1,0), w_{2}=(0,1)\). (It is not the identity matrix!)

Multiplying term by term, check that \((I-A)\left(I+A+A^{2}+\cdots\right)=I\). This series represents \((I-A)^{-1}\). It is nonnegative when \(A\) is nonnegative, provided it has a finite sum; the condition for that is \(\lambda_{\max }<1 .\) Add up the infinite series, and confirm that it equals \((I-A)^{-1}\), for the consumption matrix $$ A=\left[\begin{array}{lll} 0 & 1 & 1 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{array}\right] \quad \text { which has } \lambda_{\max }=0 $$

(a) How many degrees of freedom are there in a real symmetric matrix, a real diagonal matrix, and a real orthogonal matrix? (The first answer is the sum of the other two, because \(A=Q \Lambda Q^{\mathrm{T}}\).) (b) Show that 3 by 3 Hermitian matrices \(A\) and also unitary \(U\) have 9 real degrees of freedom (columns of \(U\) can be multiplied by any \(e^{i \theta}\) ).

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.